Answer:

Bài 1:

\(a\left(b^2-2ab+1\right)=a.b^2-a.2ab+a.1=ab^2-2a^2b+a\)

\(\left(2y+1\right).\left(2y-1\right)=\left(2y\right)^2-1^2=4y^2-1\)

Bài 2:

\(a^2-6a=a\left(a-6\right)\)

\(b^2+2b+1-c^2=\left(b^2+2b+1\right)-c^2=\left(b+1-c\right)\left(b+1+c\right)\)

Bài 3:

\(\left(3ab^2+6a^2b-9ab\right):3ab=b+2a-3\)

Sắp xếp: \(x^3-x^2-7x+3:x-3=x^2+4x+19\) dư 60

4x^2 - 8x+ 4= 9(x-2)^2

<=> 4x^2 - 8x+ 4= 9x^2- 36x+36

<=> 5x^2 - 26x+ 32= 0

=> x= 16/5 và x= 2

xin

a) Rút gọn C

\(C=\left[\frac{x^2+3x+2}{\left(x+2\right)\left(x-1\right)}-\frac{x^2+x}{x-1}\right].\left(\frac{1}{x+1}+\frac{1}{x-1}\right)\)

\(\Rightarrow C=\left[\frac{x^2+2x+x+2}{\left(x+2\right)\left(x-1\right)}-\frac{x^2+x}{x-1}\right].\left(\frac{x-1+x+1}{\left(x+1\right)\left(x-1\right)}\right)\)

\(\Rightarrow C\left[\frac{x\left(x+2\right)+\left(x+2\right)}{\left(x+2\right)\left(x-1\right)}-\frac{x^2+x}{x-1}\right].\frac{2x}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow C=\left[\frac{\left(x+2\right)\left(x+1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{x^2+x}{x-1}\right].\frac{2x}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow C=\left[\frac{x+1-x^2-x}{x-1}\right].\frac{2x}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow C=\left(\frac{1-x^2}{x-1}\right).\frac{2x}{\left(x+1\right)\left(x-1\right)}\)

\(\Rightarrow C=\left(\frac{-\left(x^2-1\right)}{x-1}\right).\frac{2x}{\left(x^2-1\right)}\)

\(\Rightarrow C=\frac{-2x}{x-1}\)

Vậy BT C khi rút gọn =-2x/x-1

b) Để C=2/3 Ta có:

\(\frac{-2x}{x-1}=\frac{2}{3}\)

\(\Rightarrow-2x.3=\left(x-1\right).2\)

\(\Rightarrow-6x=2x-2\)

\(\Rightarrow-6x-2x+2=0\)

\(\Rightarrow-8x+2=0\)

\(\Rightarrow-8x=-2\)

\(\Rightarrow x=\frac{-2}{-8}=\frac{1}{4}\)

Vậy x=1/4 thì C=2/3

a) đkxđ

\(\hept{\begin{cases}x+5\ne0\\2x^2+10\ne0\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne-5\\2x^2\ne-10\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne5\\x^2\ne-5\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne5\\x\ne\sqrt{5};\sqrt{-5}\\x\ne0\end{cases}}\)

Vậy .....

b) Rút gọn

\(\Rightarrow A=\frac{x^2+2x}{x+5}+\frac{50-5x}{2x\left(x+5\right)}+\frac{x-5}{x}\)

MTC: 2x(x+5)

\(\Rightarrow A=\frac{\left(x^2+2x\right).2x+50-5x+2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}\)

\(\Rightarrow A=\frac{2x^3+4x+50-5x+2.\left(x^2-25\right)}{2x\left(x+5\right)}\)

\(\Rightarrow A=\frac{2x^3+4x+50-5x+2x^2-50}{2x\left(x+5\right)}\)

\(\Rightarrow A=\frac{2x^3-x+2x^2}{2x\left(x+5\right)}\)

\(\Rightarrow A=\frac{x\left(2x^2-1+2x\right)}{2x\left(x+5\right)}\)

\(\Rightarrow A=\frac{2x^2-1+2x}{x\left(x+5\right)}\)

Vậy với ( ghi đkxđ) thì A=_____

c) Thay x=-4 vào A =_____ ta có:

\(A=\frac{2x^2-1+2x}{x\left(x+5\right)}=\frac{2.\left(-4\right)^2-1+2.\left(-4\right)}{4.\left(4+5\right)}=\frac{-1}{36}\)

Vậy với x=-4 thì A=-1/36

d) 

Ta có để giá trị A=-3/2 thì

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

\(\Rightarrow2.\left(2x^2-1+2x\right)=-3x\left(x+5\right)\)

\(\Rightarrow4x^2-2+4x=-3x^2-15x\)

\(\Rightarrow4x^2-2+4x+3x^2-15x=0\)

\(\Rightarrow7x^2-2-11x=0\)

Ta có: vì 7x^2 > 0

=> 7x^2-2-11x > 0 ( k/tm)

=> o có gt nào của x để A=-3/2

( chờ xíu mình check b lại uwu)