alimov-11-2003-gdz- (Алгебра - 10-11 класс - Алимов), страница 14

Ответ: x = (− 1)l+ , l∈Z .312 424) 4 sin x cos x cos 2 x = cos 4 x , 2 sin 2 x cos 2 x = cos 4 x , sin 4 x = cos 4 x ,ππ nππ nπ4 x = + nπ, n ∈ Z , x =+, n ∈ Z . Ответ: x =+, n∈Z .416 416 4sin 4 x =№ 11911) sin4x–cos4x + 2cos2x = cos2x, (sin2x–cos2x)(sin2x+cos2x)+cos2x+sin2x= 0,-cos2x + 1=0, cos2x = 1, 2х = 2πn, x=πn, n∈Z. Ответ: х = 2nπ, n ∈ Z.2) 2sin2x–cos4x=1–sin4x, cos4x–sin4x=2sin2x–1, cos2x–sin2x = sin2x – cos2x,2cos2x – 2sin2x = 0, cos2x = 0, 2 x =Ответ: x =ππ πn+ πn, n ∈ Z , x = +, n∈Z .24 2π nπ+, n∈Z .4 2№ 11921) sin3x cos x + cos3x sin x = cos2x, sin x cos x(sin2x+cos2x)=cos2x – sin2x,sin2x – cos2x + sin x ⋅ cos x = 0,sin 2 x2cos xa1 =2+sin x− 1 = 0 , tg2x + tg x – 1 = 0, tg x = a, a2 + a – 1 = 0,cos x−1± 1+ 4 −1± 5=,221)tgx =−1+ 5−1+ 5+ nπ, n ∈ Z ;, x = arctg222)tgx =−1− 51− 5+ lπ, l ∈ Z ., x = − arctg22−1 + 51− 5+ nπ, n ∈ Z , x = − arctg+ lπ, l ∈ Z ;222) 2 + cos2x + 3sinx ⋅ cosx = sin2x, cos2x – sin2x + 3sinx ⋅ cosx = -2,2cos2x+2sin2x+cos2x–sin2x+3sinx⋅cosx = 0, 3cos2x+sin2x+3sinx cosx = 0,3 + tg2x+3tgx=0, tgx=a, a2+3a+3 = 0, D < 0, следовательно, решений нет.Ответ: решений нет.Ответ: x = arctg№ 11931) 4sin2x – 8sinx ⋅ cosx + 10cos2x = 3,4sin2x – 3sin2x – 8sinx ⋅ cosx + 10cos2x – 3cos2x = 0,sin2x – 8sinx ⋅ cosx + 7cos2x = 0, tg2x – 8tgx + 7 = 0, a2 – 8a + 7 = 0,183a1 = 1, a2 = 7, tgx = 1, x =π+ nπ, n ∈ Z , tgx = 7, x = arctg7 + lπ, l ∈ Z.4π+ nπ, n ∈ Z , x = arctg7 + lπ, l ∈ Z;42) 3sin2x – 2sinx ⋅ cosx = 1, 3sin2x – 2sinx ⋅ cosx – sin2x – cos2x = 0,2sin2x – 2sinx ⋅ cosx – cos2x = 0, 2tg2x – 2tgx – 1 = 0, tgx = a, 2a2–2a–1=0,Ответ: x =a1 =21± 31± 1+ 2 1± 31± 3+ nπ, n ∈ Z .=, tgx =, x = arctg2222Ответ: x = arctg1± 3+ nπ, n ∈ Z .2№ 11941) sin5x = sin3x, sin5x – sin3x = 0, 2sinx ⋅ cos4x = 0,⎡ x = nπ, n ∈ Zπ lπ⎡sin x = 0 ; ⎢.

Ответ: x=nπ, n∈Z, x = + , l ∈ Z ;π lπ⎢⎣cos 4 x = 0 ⎢ x = + , l ∈ Z8 48 4⎣2) cos6x + cos2x = 0, 2cos4x ⋅ cos2x = 0π nπ⎡⎢x = 8 + 4 , n ∈ Z⎡cos 4 x = 0⎢⎣cos 2 x = 0 ; ⎢⎢ x = π + lπ , l ∈ Z4 2⎣⎢π nππ lπОтвет: x = +, n∈Z , x = + , l∈Z ;8 44 2⎛π⎞3) sin3x + cos7x = 0, sin 3 x + sin ⎜ + 7 x ⎟ = 0 ,⎝2⎠⎛π⎞⎛π⎞2 sin ⎜ + 5 x ⎟ ⋅ cos⎜ + 2 x ⎟ = 0 ,44⎝⎠⎝⎠⎡ ⎛π⎞⎢sin ⎜ + 5 x ⎟ = 0⎠⎢ ⎝4;⎢ ⎛π⎞cos+2x0=⎜⎟⎢⎠⎣ ⎝4π nπ⎡⎡π⎢ x = − 20 + 5 , n ∈ Z⎢ 4 + 5 x = nπ, n ∈ Z; ⎢⎢π⎢ + 2 x = π + lπ, l ∈ Z ⎢ x = π + lπ , l ∈ Z⎢⎣ 4⎢⎣28 2π lπ+ , l∈Z ;8 2⎛π⎞4) sinx = cos5x, sinx – cos5x = 0, sin x − sin ⎜ − 5 x ⎟ = 0 ,⎝2⎠Ответ: x = −π nπ+, n∈Z ,20 5π⎞⎛π⎞⎛2 sin ⎜ 3x − ⎟ ⋅ cos⎜ − 2 x ⎟ = 04⎠⎝⎝4⎠184x=⎡ ⎛π⎞π nππ⎡⎡x=+, n∈Z⎢sin ⎜ 3 x − ⎟ = 0 ⎢3 x − = nπ, n ∈ Z⎢4⎝⎠12 34⎢; ⎢; ⎢π lππ⎢ ⎛π⎞⎢⎢π⎢cos⎜ 4 − 2 x ⎟ = 0 ⎣⎢ 4 − 2 x = 2 + lπ, l ∈ Z ⎣⎢ x = − 8 + 2 , l ∈ Z⎠⎣ ⎝π nππ lπОтвет: x =+, n∈Z , x = − + , l ∈Z .12 38 2№ 11951) sinx + sin5x = sin3x, 2sin3x ⋅ cos2x – sin5x = 0, sin3x(2cos2x – 1) = 0,nπ⎡⎢x = 3 , n ∈ Znππ lπОтвет: x =, n∈Z , x = ± + , l∈Z ;⎢πlπ36 2⎢x = ± + , l ∈ Z6 2⎣⎢2) cos7x – cos3x = 3sin5x, -2sin5x⋅sin2x–3sin5x=0, sin5x(2sin2x + 3) = 0,⎡sin 5 x = 0nπ⎢, n∈Z .3; x=⎢sin 2 x = −52⎣⎡sin 3 x = 0⎢1⎢cos 2 x =2⎣№ 11961(sin 8 x + sin 10 x ) = 1 (sin 4 x + sin 10 x ) ,22sin8x – sin4x = 0, 2sin2x ⋅ cos6x = 0,nπ⎡x=, n∈Znππ lπ⎡sin 2 x = 0 ⎢2Ответ: x =;, n∈Z , x =+ , l∈Z ;⎢⎢⎣cos 6 x = 0πlπ2126⎢x =+ , l∈Z12 6⎣⎢12) sinxcos5x = sin9x ⋅ cos3x,(− sin 4 x + sin 6 x ) = 1 (sin 6 x + sin 12 x ) ,22sin12x + sin4x = 0, 2sin8x ⋅ cos4x = 0,1) cosx ⋅ sin9x = cos3x ⋅ sin7x,⎡x=⎡sin 8 x = 0 ⎢⎢⎣cos 4 x = 0 ; ⎢⎢x =⎢⎣nπ, n∈Znππ lπ8Ответ: x =, n∈Z , x = + , l ∈Z .π lπ88 4+ , l∈Z8 4№ 11971) 5 + sin2x = 5(sinx + cosx), 4 + (sinx + cosx)2 = 5(sinx + cosx),⎛ ⎛π ⎞⎞cosx + sinx = t 2 ⎜⎜ cos⎜ x − ⎟ ⎟⎟ = t , t2 – 5t + 4 = 0, D = 25 – 16 = 9,4 ⎠⎠⎝⎝π⎞t45+3⎛= 4 , cos⎜ x − ⎟ === 2 2 > 1 - нет решений,24⎠22⎝π⎞t15−3⎛=,t2 == 1 , cos⎜ x − ⎟ =4⎠222⎝t1 =185π π⎡⎢ x − 4 = 4 + 2πn ⎡ x = π + 2πn, n ∈ Z⎢;⎢2⎢ x − π = − π + 2πn ⎢⎣ x = 2πn, n ∈ Z44⎣⎢2) 2 + 2cosx = 3sinx ⋅ cosx + 2sinx,1 3+ cos 2 x − 2 sin x cos x + sin 2 x + 2(cos x − sin x ) = 0 ,2 23(cosx – sinx)2 + 4(cosx – sinx) + 1 = 0, cosx – sinx = 0,⎛ ⎛π ⎞⎞2 ⎜⎜ cos⎜ x + ⎟ ⎟⎟ = t , 3t2 + 4t + 1 = 0, D = 4 – 3 = 1,4 ⎠⎠⎝⎝()⎡π3πt1 ⎢ x + 4 = 4 + 2πnπ⎞−2 − 1⎛, ⎢=−t1 == −1 , cos⎜ x + ⎟ =34⎠⎝22 ⎢ x + π = − 3π + 2πn⎢⎣44π⎡⎢ x = 2 + 2πn, n ∈ Z t2 = −2 + 1 = − 1 , cos⎛⎜ x + π ⎞⎟ = −1 ,⎢ x = -π + 2πn, n ∈ Z334⎠ 3 2⎝⎣x+⎛ 1 ⎞⎛π1 ⎞π⎟⎟ + 2πn, n ∈ Z , x = − + arccos⎜= ± arccos⎜⎜ −⎟⎜ 3 3 ⎟ + 2πn, n ∈ Z .44⎝ 3 2⎠⎠⎝Ответ: x =x=−π+ 2πn; x = π + 2πn; n ∈ Z ,2⎛ 1 ⎞π⎟ + 2πn, n ∈ Z .+ ar cos⎜⎜⎟4⎝3 3 ⎠№ 11985x53x ⋅ cos + 2 sin x ⋅ cos x = 0 ,22225 ⎛x3 ⎞5 ⎛x⎞sin x⎜ cos + cos x ⎟ = 0 , sin x⎜ 2 cos x ⋅ cos ⎟ = 0 ,2 ⎝22 ⎠2 ⎝2⎠1) sinx + sin2x + sin3x + sin4x = 0, 2 sin2⎡⎢ x = 5 nπ, n ∈ Z⎢⎢ x = π + lπ, l ∈ Z⎢2⎢ x = π + 2mπ, m ∈ Z⎢⎣2πОтвет: nπ, n ∈ Z , x = + lπ, l ∈ Z , x = π + 2mπ, m ∈ Z;525x532) cosx+cos2x+cos3x+cos4x=0, 2 cos x ⋅ cos + 2 cos x ⋅ cos x = 0 ,22225 ⎛x3 ⎞5xcos x⎜ cos + cos x ⎟ = 0 , 2 cos x ⋅ cos x ⋅ cos = 0 ,222 ⎝22 ⎠⎡ 5⎢sin 2 x = 0⎢cos x = 0 ;⎢⎢cos x = 0⎢⎣2186π 2⎡⎢ x = 5 + 5 nπ, n ∈ Z⎢⎢ x = π + lπ, l ∈ Z⎢2⎢ x = π + 2mπ, m ∈ Z⎢⎣π 2πОтвет: x = + nπ, n ∈ Z , x = + lπ, l ∈ Z .5 525⎡⎢cos 2 x = 0⎢cos x = 0 ;⎢⎢cos x = 0⎢⎣2№ 1199sin 2 3 x− 4 sin 2 3 x = 0 , cos3x ≠ 0,cos 2 3xsin23x – 4sin23x ⋅ cos23x = 0, sin23x – 4sin23x(1 – sin23x) = 0,4sin43x – 3sin23x = 0, sin23x(4sin23x – 3) = 0,nπ⎡⎡sin 3 x = 0⎢x = 3 , n ∈ Z⎢; ⎢⎢sin 3 x = ± 3 ⎢l⎛ π ⎞2 ⎢3 x = (− 1) ⎜ ± 3 ⎟ + lπ, l ∈ Z⎣⎢⎝⎠⎣nππ lππ lπОтвет: x =, n ∈ Z , x = (− 1)l + , l ∈ Z , x = (− 1)l +1 + , l ∈ Z39 39 31) tg23x – 4sin23x = 0,2) sinxtgx = cosx + tgx,sin x sin 2 x − cos 2 x − sin xsin 2 x= cos x +,=0,cos xcos xcos x⎡sin 2 x − 1 + sin 2 x − sin x = 0 ⎡ 2 sin 2 x − sin x − 1 = 0⎢cos x ≠ 0⎢cos x ≠ 0⎣⎣⎧⎡π⎧⎡sin x = 1⎪⎢ x = + 2πl , l ∈ Z2⎪⎢1⎪⎢π⎪⎢sin x = −⎪⎢(x=− 1)m +1 + mπ, m ∈ Z⎨⎨⎣2⎢6⎣⎪⎪ππ⎪ x ≠ + nπ, n ∈ Z ⎪2⎩⎪⎩ x ≠ 2 + nπ, n ∈ ZπОтвет: x = (− 1)m +1 + mπ, m ∈ Z .61 ⎞cos x ⎛ cos x + 1 ⎞cos 2 x + cos x⎛3) ctgx⎜ ctgx +=1,⎟ =1,⎜⎟ = 1,sin x ⎠sin x ⎝ sin x ⎠sin 2 x⎝⎧⎡cos 2 x + cos x − sin 2 x = 0⎪⎢ 2⎨⎣⎢sin x ≠ 0⎪2⎩2 cos x + cos x − 1 = 0⎡cos x = −1⎢1⎢cos x = +2⎣⎡ x = π + 2nπ , n ∈ Zπ⎢⎢⎣ x = ± 3 + 2lπ , l ∈ Z187⎧⎡ x = π + 2nπ, n ∈ Zπ⎪⎪⎢⎨⎢ x = ± + 2lπ, l ∈ Z3⎪⎣⎪⎩ x ≠ mπ, m ∈ Z4) 4ctg 2 x = 5 −Ответ: x = ±π+ 2lπ, l ∈ Z .394 cos 2 x 5 sin x − 9 ⎧4 cos2 x − 5 sin 2 x + 9 sin x = 0,, ⎨=sin x sin 2 xsin x⎩sin x ≠ 0⎧4 − 9 sin 2 x + 9 sin x = 0 ⎧9 sin 2 x − 9 sin x − 4 = 0, 9sin2x - 9sinx – 4 = 0,⎨⎨⎩sin x ≠ 0⎩sin x ≠ 049 ± 81 + 144 9 ± 151, sin x = , x ≠ φ, sin x = − ,=3318181⎧1⎪ x = (− 1)n +1 arcsin + nπ, n ∈ Zx = (− 1)n +1 arcsin + nπ, n ∈ Z , ⎨33⎪⎩ x ≠ mπ, m ∈ Z1Ответ:x = (− 1)n +1 arcsin + nπ, n ∈ Z .39a2–9a–4 = 0, a 1 =2№ 12001) tg2x = 3tgx()sin 2 x 3 sin x 2 sin x ⋅ cos 2 x − 3 sin x cos 2 x − sin 2 x,=0,=cos 2 xcos xcos 2 x − sin 2 x cos x()⎧⎪2 sin x ⋅ cos x − 3 sin x ⋅ cos x + 3 sin x = 0,⎨⎪⎩cos x cos 2 x − sin 2 x ≠ 02(2)233– sinxcos x + 3sin x = 0, – sinx (1 – sin2 x) + 3sin3x = 0, 4sin3 x – sinx = 0,sinx(4sin2x – 1) = 0⎧⎡⎪⎢ x = mπ, m ∈ Z⎪⎢⎡⎪⎢ x = (− 1)l π + lπ, l ∈ Z⎢ x = mπ, m ∈ Z⎪⎢6⎡sin x = 0⎢⎪⎪⎢πl+1πl⎢+ lπ, l ∈ Z1 ⎢ x = (− 1) + lπ, l ∈ Z ⎨⎢ x = (− 1)6⎣⎢sin x = ±6⎢⎪2⎣⎢πl +1 π⎪⎧⎢ x = (- 1) 6 + lπ, l ∈ Z ⎪⎪⎪ x ≠ + nπ, n ∈ Z⎣2⎪⎨π kπ⎪⎪ x ≠ + , k ∈ Z4 2⎩⎪⎪⎩Ответ: x = mπ, m ∈ Z , x = (− 1)l2) ctg2x = 2ctgx,188ππ+ lπ, l ∈ Z , x = (− 1)l +1 + lπ, l ∈ Z ;66cos 2 x 2 cos xcos 2 x4 cos 2 x=,−=0sin 2 xsin x2 sin x ⋅ cos x 2 sin x ⋅ cos x⎧cos 2 x − sin 2 x − 4 cos 2 x = 0⎪, cos2x – sin2x – 4cos2x=0, 3cos2x+sin2x = 0,⎨⎧sin 2 x ≠ 0⎪⎨sin x ≠ 0⎩⎩1.

Ответ: решений нет.2πtgx − tg4 =2,+π1 + tgx ⋅ tg43cos2x + 1 – cos2x = 0, 2cos2x + 1 = 0, cos 2 x = −πtgx + tgπ⎞π⎞⎛⎛43) tg ⎜ x + ⎟ + tg ⎜ x − ⎟ = 2 ,π4⎠4⎠⎝⎝1 − tgxtg4tg + 1 tgx − 11 + 2tgx + tg 2 x + tgx − tg 2 x − 1 + tgx − 2 + 2tg 2 x+−2 = 0,=01 − tgx 1 + tgx1 − tg 2 x2tg 2 x + 4tgx1 − tg 2 x⎧⎪2tg 2 x + 4tgx = 0= 0, ⎨, 2tg2x + 4tgx = 0,⎪⎩1 − tg 2 x ≠ 0⎧⎡ x = nπ, n ∈ Z⎡tgx = 0 ⎪⎢ x = − arctg 2 + πl , l ∈ Z⎢⎣tgx = −2 ⎨⎣⎪1 − tg 2 x ≠ 0⎩Ответ:x = nπ, n ∈ Z,x = -arctg2 + πl, l ∈ Z.4) tg(2x + 1)ctg(x + 1) = 1, tg(2x + 1) = tg(x + 1), tg(2x + 1) – tg(x + 1) = 0,sin(2 x + 1 − x − 1)sin x⎧ x = πn, n ∈ Z= 0,= 0, ⎨cos(2 x + 1)cos(x + 1)cos(2x + 1)cos(x + 1)⎩cos(2 x + 1)cos(x + 1) ≠ 0Ответ:х = πn, n ∈ Z.№ 12011) cosx = 3x – 1Построим графики функцийу = cosx и y = 3x – 1:x≈122) sinx = 0,5x3x ≈ ±1; x = 01893) cos x = xy = cosx, y = xx≈124) cosx = x2y = cosx, y = x2x ≈ ±0,8№ 12021) x + 8 > 4 – 3x, 4x > -4, x > -1;2) 3x + 1 – 2(3 + x) < 4x + 1, 3x + 1 – 6 – 2x – 4x – 1 < 0, -3x < 6, x > -2.№ 12034 − 3x 5 − 2 x−< 2 , 3(4–3х)–2(5–2х)–2⋅24<0, 12–9x–10+4x–48 < 0,812-5x – 46 < 0, x > -46/5;5x − 7 x + 22)−≥ 2 б 7(5x – 7) – 6(x + 2) – 42 ⋅ 2 ≥ 0,6735x – 49 – 6x – 12 – 84 ≥ 0, 29x ≥ 145, x ≥ 5.1)№ 12041)5x − 4>07x + 5⎧x>⎧5 x − 4 > 0 ⎪⎪а) ⎨⎨⎩7 x + 5 > 0 ⎪ x >⎩⎪⎧5 x − 4 < 0б) ⎨⎩7 x + 5 < 0Ответ:19045 x > 4/5−5744⎡⎧⎪⎪ x < 5⎢x > 5x < -5/7 ⎢⎨⎢x < − 5⎪x < − 5⎪⎩⎢⎣775⎞ ⎛4⎛⎞x ∈ ⎜ − ∞;− ⎟ U ⎜ ;+∞ ⎟ ;7⎠ ⎝5⎝⎠2)3 x + 10>040 − x10⎧⎛ 10⎞⎧3 x + 10 > 0 ⎪ x > −а) ⎨⎨3 x ∈ ⎜ − ;40 ⎟⎩40 − x > 0 ⎪ x < 40⎝ 3⎠⎩10⎧⎧3 x + 10 < 0 ⎪ x < −б) ⎨⎨3 х∈φ⎩40 − x < 0 ⎪ x > 40⎩⎛ 10⎞Ответ: x ∈ ⎜ − ;40 ⎟ .⎝ 3⎠x+23)>05 − 4x⎧ x > −2⎧x + 2 > 0 ⎪а) ⎨5⎨⎩5 − 4 x > 0 ⎪ x <4⎩⎡⎛ 10⎞⎢ x ∈ ⎜ − ;40 ⎟ .3⎝⎠⎢⎣⎢ x ∈ φ−2< x <⎧ x < −2⎧x + 2 < 0 ⎪б) ⎨5 х∈φ⎨⎩5 − 4 x < 0 ⎪ x >4⎩5Ответ: − 2 < x < .48− x4)>06 + 3x545⎡⎢− 2 < x < 4⎢x ∈ φ⎣⎧8 − x > 0 ⎧ x < 8а) ⎨-2 < x < 8⎨⎩6 + 3 x > 0 ⎩ x > −2⎧8 − x < 0 ⎧ x > 8б) ⎨х∈φ⎨⎩6 + 3x < 0 ⎩ x < −2Ответ: -2 < x < 8.№ 12051)3 − 2x<03x − 23⎧⎪⎪ x > 23⎧3 − 2 x > 0; x > ; б) ⎨⎨22⎩3 x − 2 > 0⎪x >⎪⎩32⎞ ⎛3⎛⎞Ответ: x ∈ ⎜ − ∞; ⎟ U ⎜ ;+∞ ⎟ .3⎠ ⎝2⎝⎠⎧3 − 2 x < 0а) ⎨⎩3 x − 2 < 0⎧⎪⎪ x <⎨⎪x <⎪⎩32; x< 22331915⎧x>10 − 4 x⎧10 − 4 x < 0 ⎪⎪2 ; x> 52)< 0 ; а) ⎨⎨9x + 22⎩9 x + 2 < 0 ⎪ x > − 2⎪⎩955⎧⎡x<x>⎢2⎧10 − 4 x > 0 ⎪⎪2 ; x<−2б) ⎨⎢⎨9 ⎢x < − 2⎩9 x + 2 > 0 ⎪ x < − 2⎪⎩99⎣⎢2⎞ ⎛5⎛⎞Ответ: x ∈ ⎜ − ∞;− ⎟ U ⎜ ;+∞ ⎟ .9⎠ ⎝2⎝⎠18 − 7 x18 − 7 x<0;>03)− 4x2 − 14x2 + 1181818 – 7x > 0; x <(4x2 + 1 > 0 при любых значениях х).

Ответ: x <.77№ 12061)5x + 45 x + 4 − 4( x − 3)x + 16< 4;<0;<0;x−3x−3x −3⎧ x + 16 > 0⎧ x + 16 < 0 ⎧ x > −16⎧ x < −16или ⎨; -16 < x < 3;⎨ x − 3 < 0 или ⎨ x − 3 > 0 ; ⎨ x < 3⎩⎩⎩⎩x > 322 − 1(x − 4)6− x2)<1;<0;<0;x−4x−4x−4⎧6 − x < 0⎧6 − x > 0 ⎧ x > 6⎧x < 6⎨ x − 4 > 0 или ⎨ x − 4 < 0 ; ⎨ x > 4 или ⎨ x < 4 x > 6 или x < 4;⎩⎩⎩⎩−4 x − 1022 − 4( x + 3)≤0,3)≤ 4,≤0,x+3x+3x+3⎧− 4 x − 10 ≤ 0 или ⎧− 4 x − 10 ≥ 0 ; ⎧ x ≥ −2,5 или ⎧ x ≤ −2,5⎨ x > −3⎨ x < −3⎨x + 3 > 0⎨x + 3 < 0⎩⎩⎩⎩x ≥ -2,5 или x < -3.№ 12071 ⎞⎛1⎞11⎛1) 8х2 – 2х – 1 < 0, ⎜ x + ⎟⎜ x − ⎟ < 0 , − < x <4 ⎠⎝2⎠42⎝++−142) 5x2 + 7x ≤ 0,–12+7− ≤x≤0.5192+−75–0№ 12081)⎧ (x − 3)(x + 3)< 0 (x − 3)(x + 3)⎪<0равносильно<0,⎨ (x − 2 )(x + 2 )(x − 2)(x + 2)x2 − 4⎪⎩ x ≠ ±2x2 − 9++–-3+-2–23x ∈ (−3;−2) U (2;3) ;2) (2х2 + 3)(х + 4)3 > 0, 2x2 + 3 > 0 при любых х;(x + 4)3 > 0 равносильно x + 4 > 0; x > -4.№ 1209()⎧⎪(3 x − 15) x 2 + 5 x − 14 ≥ 0≥0, ⎨ 2,⎪⎩ x + 5 x − 14 ≠ 0x 2 + 5 x − 14(3х – 15)(х – 2)(х + 7) ≥ 0, (х – 5)(х – 2)(х + 7) ≥ 0,3 x − 151)+–+-72x ∈ (−7;2 ) U [5;+∞ ) ;2)x −1<0,x + 4x + 22(+−2− 2()(x − 1)(x − (2 +−2+ 2) ())()2 x+2− 2 < 0,+)x ∈ − ∞;−2 − 2 U − 2 + 2 ;1–1()()⎧⎪ x 2 + 2 x − 8 x 2 − 2 x − 3 > 0x + 2x − 8> 0 равносильно ⎨ 22⎪⎩ x − 2 x − 3 ≠ 0x − 2x − 323)5⎧⎪(x − 1) x 2 + 4 x + 2 < 0,⎨ 2⎪⎩ x + 4 x + 2 ≠ 0(х – 1)(х2 + 4х + 2) < 0,––(x – 2)(x + 4)(x + 1)(x –3) > 0++–-4-1x ∈ (−∞;−4 ) U (−1;2 ) U (3;+∞ ) .+2–3№ 1210lg(x2 + 8x + 15).