\(\dfrac{5x-5}{\left(x+1\right)^2}:\dfrac{20x^2-20}{3x+3}\\ =\dfrac{5\left(x-1\right)}{\left(x+1\right)^2}:\dfrac{20\left(x^2-1\right)}{3\left(x+1\right)}\\ =\dfrac{5\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{3\left(x+1\right)}{20\left(x^2-1\right)}\\ =\dfrac{3}{4\left(x+1\right)^2}\)

A= \(\frac{5x-5}{\left(x+1\right)^2}.\left(\frac{3+3x}{20-20x}\right)\left(\text{Đ}K\text{XĐ}:x\ne\pm1\right)\)

\(=\frac{5\left(x-1\right)}{\left(x+1\right)^2}.\frac{3\left(x+1\right)}{20\left(1-x\right)}=\frac{-3}{4\left(x+1\right)}\)

Vậy với \(x\ne\pm1\) thì A= \(\frac{-3}{4\left(x+1\right)}\)

a. 5x.(12x+7)-3x.(20x-5)=-150

x=-3

b. ( 2x-1).(3-x)+(x+4).(2x-5)=20

x=43/10

c. 9x²-1+(3x-1)²=0

x=1/3

d. 3x.(x-2)-(3x+2).(x-1)=7

x=-5/2

e. (2x-1)²-(2x+5).(2x-5)=20

x=3/2

f. 4x²-5=4

x=3/2

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

d) 5x(12x + 7) – 3x(20x – 5) = - 100

⇒60x²+35x-60x²+15=-100

⇒35x+15=-100

⇒35x=-100-15

⇒35x=-115

⇒x=\(\dfrac{-115}{35}\)

⇒x=\(\dfrac{-23}{7}\)

d: ta có: \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

   x²-4x+5=0

=>(x-2)²+1=0

=>(x-2)² =-1

=> pt vô nghiệm

   (x²+5x)(x³+3x²-18x)=0

=>\(\int^{x^2+5x=0}_{x^3+3x^2-18x=0}=>\int^{\int^{x=0}_{x=-5}}_{x=3;x=0;x=-6}\)