Question

Обобщающие задан 21. Решите неравенство.
1) (2x+3)2-7х<(2x-1)(2x+1);
2) (x-3)2-(3+x)(3-X)>2(1-x)?;
3) (x-1)2–2(x+1)(х+2)+(x+3)?<5x;
нужно​

Answer 1

1)(2x+3)²-7х<(2x-1)(2x+1)

4x²+12x+9-7x<4x²-1

4x²-4x²+12x-7x<-1-9

5x<-10

x<-2

x∈(-∞;-2)

2)(x-3)2-(3+x)(3-X)>2(1-x)

x²-6x+9-9+x²>2-x

x²+x²-6x+x>2

2x²-5x-2>0

D=b²-4ac=25+4*2*2=25+16=41

$\sqrt{D} =\sqrt{41}$

$x_{1}=\frac{-b-\sqrt{D} }{2a} = \frac{5-\sqrt{41} }{4}$

$x_{2}=\frac{-b+\sqrt{D} }{2a} = \frac{5+\sqrt{41} }{4} \\$

3)(x-1)2–2(x+1)(х+2)+(x+3)?<5x

x²-2x+1-2(x²+2x+x+2)+x+3<5x

x²-2x-1-2x²-4x-2x-4+x+3<5x

x²-2x²-2x-4x-2x+x-5x+3-4-1<0

-x²-12x-2<0

D=b²-4ac=144-4*(-2)*(-1)=144-8=136

$\sqrt{D} =2\sqrt{34}$

$x_{1} = \frac{-b-\sqrt{D}}{2a} = \frac{12-2\sqrt{34} }{-2} = \frac{-2(-6+\sqrt{34})}{-2} = -6+\sqrt{34}$

$x_{2} = \frac{-b+\sqrt{D}}{2a} = \frac{12+2\sqrt{34} }{-2} = \frac{-2(-6-\sqrt{34})}{-2} = -6-\sqrt{34}$