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а)4б) 4 12 − 9 x − 2 x 2 ≥ 2 ;12 – 9x – 2x2 ≥ 16; –2x2 – 9x – 4 ≥ 0;2x2 + 9x + 4 ≤ 0; D = 81 – 4⋅2⋅4 = 72; x =++––4−⎡−9 ± 71; x1 = –4; x2 = − .42x121⎤Ответ: ⎢ −4; − ⎥ .2⎦⎣3.3.В08. а) 4 x 4 − 9 x − 9 ≥ 2 x 2 ; 4x4 – 9x – 9 ≥ 4x4; –9x ≥ 9; x ≤ –1.Ответ: x ≤ –1.б)9 x 4 − 8 x − 6 ≥ 3 x 2 ; 9x4 – 8x – 6 ≥ 9x4; –8x ≥ 6; x ≤−63; x≤− .8434Ответ: x ≤ − .3x 4 + 52 x 2 − 135 ≥ 2 x 2 + 3 ; 3x4 + 52x2 – 135 ≥ 4x4 + 6 ⋅ 2x2 + 9;D–x4 + 40x2 – 144 ≥ 0; x4 – 40x2 + 144 ≤ 0;= 400 – 144 = 162, x2 = 36 и x2 = 4;43.3.В09.

а)4 ≤ x2 ≤ 36, x ∈ [–6; –2] ∪ [2; 6]. Ответ: [–6; –2] ∪ [2; 6].б) 8 x 4 + 53x 2 − 84 ≥ 3x 2 + 4 ; 8x4 + 53x2 – 84 ≥ 9x4 + 24x2 + 16;–x4 + 53x2 – 24x2 – 84 – 16 ≥ 0; –x4 + 29x2 – 100 ≥ 0; x4 – 29x2 + 100 ≤ 0;D = 841 – 400 = 441; x2 = 4 и x2 =25; 4 ≤ x2 ≤ 25, x ∈ [–5; –2] ∪[2; 5].Ответ: x ∈ [–5; –2] ∪ [2; 5].3.3.В10. а) 3 −64 + 3x < x − 4 ; –64 + 3x < x3 – 3⋅x2⋅4 3⋅x⋅42 – 43;264 + 3x < x3 – 12x2 + 48x – 64; –x3 + 12x2 – 48x + 3x < 0; x3 – 12x2 + 45x > 0;x(x2 – 12x + 45) = 0; x = 0 или x2 – 12x + 45 = 0; D = 144 – 4⋅45 < 0.+–0168xОтвет: (0; +∞).б) 3 27 − 2 x < x + 3 ; 27 – 2x < x3 + 3x2⋅3 + 3⋅x⋅32 + 33;27 – 2x < x3 + 9x2 + 27x + 27; –x3 – 9x2 – 29x < 0; x3 + 9x2 + 29x > 0;x(x2 + 9x + 29) > 0; x = 0 или x2 + 9x + 29 = 0; D = 81 – 4⋅29 < 0.+–x03.3.В11. а) 4 x − 3 4 x ≥ 10 . ПустьОтвет: (0; +∞).x = t . –3t + 4t2 – 10 ≥ 0; 4t2 – 3t – 10 ≥ 0;44t2 – 3t – 10 ≥ 0; 4t2 – 3t – 10 = 0; D = 9 + 4⋅10⋅4 = 132; t =t2 = −б)10— нет решений; x = 24 = 16.

Ответ: [16; +∞).8x + 4 6 x ≥ 21 ; t2 + 4t – 21 ≥ 0; t = 16 + 4⋅21 = 102; t =33 ± 13; t1 = 2;8t1 = –7 — нет решений; t2 = 3;Ответ: [729; +∞).6x =t⇒6−4 ± 10;2x = 3 ⇒ x = 36 = 729.3x 2 − 17 x + 14 ≤ 2 ;3.3.В12. а)⎡14⎞; +∞ ⎟ ; 3x2 – 17x + 14 ≤ 4;3⎣⎠17 ± 1322223x – 17x + 10 ≤ 0; D = 17 – 4⋅3⋅10 = 13 ; x =; x1 = 5; x2 = .63ОДЗ: 3x2 – 17x + 14 ≥ 0; x ∈ (–∞; 1] ∪ ⎢++–x523⎡2⎤⎡14⎤Ответ: ⎢ ; 1⎥ ∪ ⎢ ; 5⎥ .⎣3 ⎦ ⎣ 3⎦⎡ 7⎞4 x 2 + 23 x + 28 ≤ 3 ; ОДЗ: 4x2 + 23x + 28 ≥ 0; x ∈ (–∞; –4] ∪ ⎢ − ; +∞ ⎟ ;⎣ 4⎠б)4x2 + 23x + 28 ≤ 9; 4x2 + 23x + 19 ≤ 0; D = 232 – 4⋅4⋅19 = 152;x=−23 ± 1538193; x1 = − = − = −4 ; x2 = –1.8844++––4,75x–1⎡ 7⎣ 4⎤⎦Ответ: [–4,75; –4] ∪ ⎢ − ; −1⎥ .Уровень С.3.3.С01 а)x 2 − 5 < x2 – 7;⎪⎧ x − 5 ≥ 02D: ⎨2⎪⎩ x − 7 ≥ 02⎪⎧ x ≥ 5; ⎨2⎪⎩ x ≥ 7;x57x ∈ (–∞; − 7 ] ∪ [ 7 ; +∞); x2 – 5 < x4 – 14x2 + 49; –x4 + 15x2 – 54 < 0;− 7− 5169x4 – 15x2 + 54 > 0; D = 225 – 4⋅54 = 32; x2 =15 ± 3; x12 = 9 ⇒ x = ±3;2x22 = 6 ⇒ x = ± 6 — не принадлежат области значений.–++x3–3x ∈ (–∞; –3) ∪ (3; +∞).

Ответ: x ∈ (–∞; –3) ∪ (3; +∞).б) x 2 + 15 < x 2 + 3 ; x2 + 15 > 0; x2 + 3 > 0; x2 + 15 < x4 + 2⋅3⋅x2 + 9;x2 + 15 < x4 + 6x2 + 9; –x4 – 5x2 + 6 < 0; x4 + 5x2 – 6 > 0; D = 25 + 5⋅6 = 49;x2 =−5 ± 7; x12 = –6 — не имеет решений; x22 = 1 ⇒ x = ±1.2–++x–11Ответ: (–∞; –1) ∪ (1; +∞).223.3.С02 а) 3x − 18 x − 3 ≥ 3x − 19 x + 20 ;D = 3x2 – 19x + 20 ≥ 0;x=19 ± 1141; x1 = 5; x2 = = 1 .633++–x5131⎤⎛Значит, x ∈ ⎜ −∞; 1 ⎥ ∪ [5; +∞). 3x2 – 18x – 3 ≥ 3x2 – 19x + 20; x ≥ 23.3⎦⎝Ответ: x ≥ 23.14 x 2 − 21x − 4 ≥ 4 x 2 − 25 x + 25 ; D: 4x2 – 25x + 25 ≥ 0;25 ± 1510 51D = 252 – 4⋅4⋅25 = 152; x =; x1 = 5; x2 = = = 1 .88 44++–б)1,25x5Значит, x ∈ (–∞; 1,25] ∪ [5; +∞).

4x2 – 21x – 4 ≥ 4x2 – 25x + 25;4x ≥ 29; x ≥2911; x ≥ 7 . Ответ: x ≥ 7 .4443.3.С03. а) 6 x 2 − 14 x − 24 ≥| 3x + 1 | ; 6x2 – 14x – 24 ≥ (3x + 1)2;6x2 – 14x – 24 ≥ 9x2 + 6x + 1; –3x2 – 20x – 25 ≥ 0; 3x2 + 20x + 25 ≤ 0;−20 ± 101052; x1 = –5; x2 = − = − = −1 .6633+D = 202 – 4⋅3⋅25 = 102; x =+––5б)170−123x⎡⎣2⎤Ответ: ⎢ −5; −1 ⎥ .32⎦2−7 x − 29 x + 25 ≥| x − 4 | ; –7x – 29x + 25 ≥ x – 8x + 16;2–8x2 – 21 + 9 ≥ 0; 8x2 + 21x – 9 ≤ 0; D = 212 + 4⋅8⋅9 = 272; x =x1 = –3; x2 = 0,375.+––3−21 ± 27;16+0,375xОтвет: [–3; 0,375].3.3.С04.−3 x + 35−3 x + 35−3 x + 35− 16 ≥ 0 ;≥ 16 ;≥4;5x − 35x − 35x − 3−3 x + 35 − 16(5 x − 3)−3 x + 35 − 80 x + 48−83 x + 83≥0;≥0;≥0;5x − 35x − 35x − 335x – 3 ≠ 0; 5x ≠ 3; x ≠ ; –83x + 83 = 0; x = 1.5––+x135⎛3 ⎤Ответ: ⎜ ; 1⎥ .⎝5 ⎦а)−4 x + 33−4 x + 33−4 x + 33−1 ≥ 0 ;≥1;≥1 ;2x − 32x − 32x − 3−6 x + 36≥ 0 ; 2x – 3 ≠ 0; 2x ≠ 3; x ≠ 1,5; –x + 6 = 0; x = 6.2x − 3––+x1,56б)−4 x + 33 − 2 x + 3≥0;2x − 3Ответ: (1,5; 6].3.3.С05.

а)2 x + 7777 ⎤ ⎛ 52 x + 77⎛⎞≤ 3 ; ОДЗ:≥ 0 ; x ∈ ⎜ −∞; − ⎥ ∪ ⎜ − ; +∞ ⎟ ;2⎦ ⎝ 22x + 52x + 5⎝⎠2 x + 772 x + 77 − 9(2 x + 5)2 x + 77 − 18 x − 45≤0;≤0;−9 ≤ 0 ;2x + 52x + 52x + 5−16 x + 325≤ 0 ; 2x + 5 ≠ 0; 2x ≠ –5; x ≠ − ; –16x + 32 = 0; x = 2;2x + 521⎞77 ⎤⎛⎛x ∈ ⎜ −∞; −2 ⎟ ∪ [2; +∞ ) . Ответ: x ∈ ⎜ −∞; − ⎥ ∪ (2; +∞).2⎠2⎦⎝⎝б)4 x + 914 x + 9191 ⎤ ⎛ 3⎛⎞≤ 3 ; ОДЗ:≥ 0 ; x ∈ ⎜ −∞; − ⎥ ∪ ⎜ − ; +∞ ⎟ ;4x + 34⎦ ⎝ 44x + 3⎝⎠4 x + 914 x + 91 − 9(4 x + 3)4 x + 91 − 36 x − 27≤0;≤ 0;−9 ≤ 0 ;4x + 34x + 34x + 3−32 x + 643≤ 0 ; 4x+3 ≠ 0; 4x ≠ –3; x ≠ − ; –32x + 64 = 0; x = 24x + 343⎞91 ⎤⎛⎛x ∈ ⎜ −∞; − ⎟ ∪ [2; +∞ ) . Ответ: x ∈ ⎜ −∞; − ⎥ ∪ [2; +∞) .4⎠4⎦⎝⎝1713.3.С06.2 x 2 − 15 x + 28 ≤ x − 2 ;а)2⎡x ≥ 4⎪⎧ 2 x − 15 x + 28 ≥ 0 ⎧ x ∈ (−∞; 3,5] ∪ [4; +∞); ⎨⇒ ⎢;x2≥⎪⎩ x − 2 ≥ 0⎩⎣ x ∈ [2; 3,5]15 ± 1142; x1 = 4; x2 = = 3 = 3,5 ;D = 152 – 4⋅2⋅28 = 1; x =444++–x3,54D: ⎨2x2 – 15x + 28 ≤ x2 – 4x + 4; x2 – 11x + 24 ≤ 0; D = 121 – 4⋅24 = 52;x=11 ± 5; x1 = 8; x2 = 3.2++–3x8Ответ: [3; 3,5] ∪ [4; 8].22 x − 11x + 15 ≤ x − 1 ;б)⎧⎪ 2 x 2 − 11x + 15 ≥ 0 ⎧ x ∈ (−∞; 2,5] ∪ [3; +∞)⎡x ≥ 3D: ⎨; ⎨⇒ ⎢;≥x1⎪⎩ x − 1 ≥ 0⎩⎣ x ∈ [1; 2,5]2x2 –11x + 15 = 0; D = 121 – 4⋅2⋅15 = 1;x=11 ± 15; x1 = 3; x2 = = 2,5 .42++–2,52x3222x – 11x + 15 ≤ x – 2x + 1; x – 9x + 14 ≤ 0; D = 81 – 4⋅14 = 25;x=9±5; x1 = 7; x2 = 2.2+–2+x7Ответ: [2; 2,5] ∪ [3; 7].−3x 2 − 5 x + 12 ≥ x + 3 ;1⎧⎧⎪ −3x 2 − 5 x + 12 ≥ 0 ⎪ −3 ≤ x ≤ 11; ⎨D: ⎨3 ⇒ –3 ≤ x ≤ 1 ;3⎪⎩ x + 3 ≥ 0⎪ x ≥ −3⎩3.3.С07.

а)3x2 + 5x – 12 ≤ 0; D = 25 + 4⋅3⋅12 = 132; x =x1 = –3; x2 =+8 41= =1 .6 33–−5 ± 13;6+x13–3x2 – 5x + 12 ≥ x2 + 6x + 9; –4x2 – 11x + 3 ≥ 0; 4x2 + 11x – 3 ≥ 0;–31721D = 121 + 4⋅4⋅3 = 132; x =+–−11 ± 131; x1 = –3; x2 = .84+x14–3⎡1⎤Ответ: x ∈ ⎢ −3; ⎥ .4⎦⎣− x2 − 5x − 4 ≥ x + 4 ;б)2⎪⎧ − x − 5 x − 4 ≥ 0 ⎧ −4 ≤ x ≤ −1; ⎨⇒ –4 ≤ x ≤ 1; x2 + 5x + 4 ≤ 0;⎪⎩ x + 4 ≥ 0⎩ x ≥ −4−5 ± 3D = 25 – 4⋅4 = 12; x =; x1 = –; x2 = –1;2D: ⎨++–x–4–122–x – 5x – 4 ≥ x + 8x + 16; –2x – 13x – 20 ≥ 0; 2x2 + 13x + 20 ≤ 0;−13 ± 3105; x1 = –4; x2 = − = − = −2,5 .D = 169 – 4⋅2⋅20 = 9; x =442++–x–2,5–42Ответ: x ∈ [–4; –2,5].3.3.С08.а) (x2 – 8x + 12) −2 x 2 + 11x − 15 ≤ 0;D: –2x2 + 11x – 15 ≥ 0; 2x2 – 11x + 15 ≤ 0; D = 121 – 4⋅2⋅15 = 1;x=11 ± 110 5; x1 = 3; x2 = = = 2,5 ;44 2++–x32,52x ∈ [2,5; 3]; x – 8x + 12 ≤ 0; D = 64 – 4⋅12 = 42; x =++–622x8± 4; x1 = 6; x2 = 2.2Ответ: x ∈ [2,5; 3].б) (x – 7x + 6) −3x − 4 x + 4 ≤ 0;D: –3x2 – 4x + 4 ≥ 0; 3x2 + 4x – 4 ≤ 0; D = 16 + 4⋅3⋅4 = 82;x=2−4 ± 82; x1 = –2; x2 = ;63++––2⎡⎣x ∈ ⎢ −2;23x2⎤ 27±5; x – 7x + 6 ≤ 0; D = 49 – 4⋅6 = 25; x =; x1 = 6; x2 = 1.⎥3⎦2173++–x61x ∈ [1; 6].

Ответ: решений нет.3.3.С09.а)x+328 − 9 x − 4 x 2 ≥ 0 ;x+42⎪⎧ 28 − 9 x − 4 x ≥ 0 ⎧ x ∈ [−4; 1,75]; ⎨⇒ (–4; 1,75];⎪⎩ x + 4 ≠ 0⎩ x ≠ −4D: ⎨–4x2 – 9x + 28 ≥ 0; 4x2 + 9x – 28 ≤ 0; 4x2 + 9x – 28 = 0; D = 81 + 4⋅4⋅28 = 232;−9 ± 23; x1 = –4; x2 = 1,75;8x=++––4+x1,75+––4x+3≥0;x+4x–3(x + 3)(x + 4) ≥ 0; x ∈ (–∞; –4) ∪ [–3; +∞). Ответ: [–3; 1,75].б)x+4−35 − 19 x − 2 x 2 ≥ 0 ;x+7⎧⎪ −35 − 19 x − 2 x 2 ≥ 0 ⎧ −14 ≤ x ≤ −55⎤⎛; ⎨⇒ x ∈ ⎜ −7; − ⎥ ;2⎦⎝⎪⎩ x + 7 ≠ 0⎩ x ≠ −7D: ⎨–2x2 – 19x – 35 ≥ 0; 2x2 + 19x + 35 ≤ 0; 2x2 + 19x + 35 = 0;D = 192 – 4⋅2⋅35 = 92; x =–+–7x+4≥ 0;x+7+−x52+––7−19 ± 95; x1 = –7; x2 = − ;2⋅22+–4x⎡⎣5⎤(x + 4)(x + 7) ≥ 0; x ∈ (–∞; –7) ∪ [–4; +∞).

Ответ: ⎢ −4; − ⎥ .2⎦⎧ 3x − 4≥ 0 ⎧(3x − 4)(2 x − 1) ≥ 04 − 3x3x − 4⎪3.3.С10. а); ⎨;+ 11> 24 ; ⎨ 2 x − 12x −12x −1⎪2 x − 1 ≠ 0 ⎩2 x ≠ 1⎩⎧1⎤ ⎡ 1⎛⎞⎪⎪ x ∈ ⎜ −∞; ⎥ ∪ ⎢1 ; +∞ ⎟2⎦ ⎣ 3⎝⎠ ⇒ x ∈ ⎛ −∞; 1 ⎞ ∪ ⎡1 1 ; +∞ ⎞ ;⎨⎜⎟⎟2 ⎠ ⎣⎢ 3⎝⎠⎪x ≠ 1⎪⎩2174+12Пусть+–113x3x − 43x − 4= −t 2 ;= t , тогда −2x −12x −1–t2 + 11t – 24 > 0; t2 – 11t + 24 < 0; t2 – 11t + 24 = 0; D = 121 – 4⋅24 = 52;11 ± 53x − 4; t1 = 8; t2 = 3;=8;22x −13x − 43x − 4 − 64(2 x − 1)1)=0;− 64 = 0 ;2x −12x −1t=3x – 4 – 128x + 64 = 0; –125x = –60; x = 0,48;2)3x − 43x − 43x − 4 − 9(2 x − 1)= 3;=0;−9 = 0 ;2x −12x −12x −113⎛1⎝3⎞⎠3x – 4 – 18x + 9 = 0; –15x = –5; x = . Значит, x ∈ ⎜ ; 0, 48 ⎟ .⎛1⎞Ответ: x ∈ ⎜ ; 0, 48 ⎟ .⎝3⎠б)⎧ 2x −1≥01 − 2x2x −1⎪;+5> 6 ; D: ⎨ 4 x + 14x + 14x +1⎪4 x + 1 ≠ 0⎩⎧1⎤ ⎡1⎛⎞⎪⎪ x ∈ ⎜ −∞; − 4 ⎥ ∪ ⎢ 2 ; +∞ ⎟⎝⎦ ⎣⎠ ⇒ x ∈ ⎛ −∞; − 1 ⎞ ∪ ⎡ 1 ; +∞ ⎞ .⎨⎜⎟⎟4 ⎠ ⎣⎢ 21⎝⎠⎪x ≠ −⎪⎩4++–x112Пусть22x −12x −1= −t 2 ;= t , тогда −4x + 14x + 1–t2 + 5t – 6 > 0; t2 – 5t + 6 < 0; D = 25 – 4⋅6 = 1; t =5 ±1; t1 = 3; t2 = 2.22x −12x −12x −12 x − 1 − 9(4 x + 1)=9;−9 = 0 ;=0;= 3;4x + 14x +14x +14x + 11052x – 1 – 36x – 9 = 0; –34x = 10; x = − = − ;34172x −12x −12x −12 x − 1 − 4(4 x + 1)=2;При t = 2;= 4;−4 = 0 ;=0;4x + 14x + 14x + 14x + 1При t = 3;2x – 1 – 16x – 4 = 0; –14x = 5; x = −5⎞5⎛ 5; x∈⎜− ; − ⎟ .141714⎝⎠1755⎞⎛ 5; − ⎟.1417⎝⎠Ответ: x ∈ ⎜ −3.3.С11.

а) 2 3x − 11 < x − 1 ;D: 3x-11 ≥ 0; 3x ≥ 11; x ≥112; x ≥ 3 ; 4(3x – 11) < x2 – 2x + 1;3312x – 44 < x2 – 2x + 1; x2 + 14x – 45 < 0; x2 – 14x + 45 > 0;D = 196 – 4 ⋅ 45 = 16; x =+14 ± 4; x1 = 9; x2 = 5.2+–x95⎡ 3⎣ 2⎞⎠Ответ: ⎢3 ; 5 ⎟ ∪ (9; +∞).б) 2 6 x + 7 < x + 7 ;7616D: 6x + 7 ≥ 0; 6x ≥ –7; x ≥ − ; x ≥ −1 ; 4(6x + 7) < x2 + 14x + 49;24x + 28 – x2 – 14x – 49 < 0; 10x – x2 – 21 < 0; x2 – 10x + 21 > 0;x2 – 10x + 21 = 0; D = 100 – 4 ⋅ 21 = 42; x =+–+x7310 ± 4; x1 = 7; x2 = 3.21Ответ: x ∈ [ −1 ; 3) ∪ (7; +∞).63.3.С12.а)4 x 2 − 15 x + 14 < 8 x − 5 x 2 ;D: 4x2 – 15x + 14 ≥ 0; D = 225 – 4⋅4⋅14 = 1; x =–++15 ± 17; x1 = 2; x2 = ;84x23143⎤⎛x ∈ ⎜ −∞; 1 ⎥ ∪ [ 2; +∞ ) ; 4x2 – 15x + 14 < 8x – 5x2; 9x2 – 23x + 14 < 0;4⎦⎝23 ± 528 145; x1 = = = 1 ; x2 = 1;D = 232 – 4⋅9⋅14 = 52; x =1818 99+–1+519x5⎞⎛x ∈ ⎜1; 1 ⎟ .

Ответ:9⎠⎝5⎞⎛⎜1; 1 ⎟ .9⎠⎝б) 2 x 2 − 23x + 66 < 24 x − 5 x 2 ;D: 2x2 – 23x + 66 ⋅ 0; 2x2 – 23x + 66 = 0; D = 232 – 4⋅2⋅66 = 1;x=17623 ± 122 11; x1 = 6; x2 == = 5,5 ;442–++x65,52x ∈ (–∞; 5,5] ∪ [6; +∞); 2x – 23x + 66 < 24x – 5x2; 7x2 – 47x + 66 < 0;47 ± 1966 335; x1 === 4 ; x2 = 2;1414 77+D = 472 – 4⋅7⋅66 = 192; x =+–245⎞⎛x ∈ ⎜ 2; 4 ⎟ . Ответ:7⎠⎝x575⎞⎛⎜ 2; 4 ⎟ .7⎠⎝Уровень D.3.3.D01.а) 5 x + 205 − 2 x + 32 > 3 ;⎧5 x + 205 ≥ 0 ⎧ x ≥ −41; ⎨⇒ x ∈ [–32; +∞);ОДЗ: ⎨⎩ x + 32 ≥ 0⎩ x ≥ −325 x + 41 − 2 x + 32 > 3 ;5 x + 41 > 3 + 2 x + 32 ;5(x + 41) > 9 + 4x + 128 + 12 x + 32 ; x + (205 – 137) > 12 x + 32 ;2⎪⎧ x + 136 x + 4624 > 144 x + 4608;⎪⎩ x + 68 > 0x + 68 > 12 x + 32 ; ⎨⎧⎪ x 2 − 8 x + 16 > 0;⎨⎪⎩ x + 68 > 02⎪⎧( x − 4) > 0⇒ x ∈ (–68; 4) ∪ (4; +∞).⎨⎪⎩ x > −68Ответ: x ∈ [–32; 4) ∪ (4; +∞).б) 5 x + 115 − 2 x + 19 > 2 ;⎡ x ≥ −19x ≥ –19;ОДЗ: ⎢⎣ x ≥ −235 x + 23 − 2 x + 19 > 2 ;5(x + 23) > 4 + 4(x + 19) + 8 x + 19 ; x + 115 – 80 > 8 x + 19 ; x + 35 > 8 x + 19 ;2⎪⎧ x + 70 x + 1225 > 64( x + 19);⎨⎪⎩ x + 35 > 0⎧⎪ x 2 + 70 x + 1225 − 64 x − 1216 > 0;⎨⎪⎩ x > −352⎪⎧ x + 6 x + 9 > 0;⎨⎪⎩ x > −352⎪⎧( x + 3) > 0; x ∈ (–35; –3) ∪ (–3; +∞).

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