1) Решите уравнение: 4х(х+3)=4-3х 2) Решите неравенство: 3х<5(x+1)-10<8 4) Запишите...

1.
4х(х + 3) = 4 - 3х
4х² + 12х = 4 - 3х
4х² + 12х - 4 + 3х = 0
4х² + 15х - 4 = 0
a=4 ; b = 15 ; с = - 4
D = b² - 4ac = 15² - 4*4*(-4) = 225 + 64 = 289 = 17²
D>0 - два корня уравнения
х₁ = ( - b - √D)/2a = (-15 - 17)/(2*4) = -32/8 = - 4
x₂ = (- b +√D)/2a = (-15+17)/(2*4) = 2/8 = 1/4 = 0.25

2.
3x < 5(x+1) - 10<8<br>3x < 5x + 5 - 10 <8<br>3x < 5x - 5 < 8
{ 5x - 5 > 3x
{ 5x - 5 < 8<br>
{ 5x - 3x > 5
{ 5x < 8 + 5

{ 2x > 5
{ 5x < 13

{ x > 5/2
{ x < 13/5

{ x > 2.5
{ x < 2.6
2.5< x < 2.6
x∈ (2.5 ; 2,6)

4.
4√2 = √32
√33
6 = √36
4√2 < √33 < 6 <br>ответ : 4√2

5.
$\frac{b+4}{b^{2}+16} * ( \frac{b+4}{b-4} + \frac{b-4}{b+4} )= \frac{b+4}{b^{2}+16} * \frac{(b+4)(b+4)+(b - 4)(b - 4)}{(b-4)(b+4)} = \\ \\ =\frac{b+4}{b^{2}+16} * \frac{(b + 4)^2 + (b - 4)^2}{(b-4)(b+4)} = \frac{b+4}{b^{2}+16} * \frac{b^2+8b +16 + b^2 -8b+16}{(b - 4)(b+4)} = \\ \\ =\frac{b+4}{b^{2}+16} * \frac{2b^2+2*16}{(b-4)(b+4)} = \frac{(b+4) * 2 *(b^2+16)}{(b^2+16)*(b-4)(b+4)} = \frac{2}{b - 4} \\ \\ \\ b= 3.75 \\ \frac{2}{3.75 - 4} = \frac{2}{-0.25} = -8$