Đề có vấn đề à

x²=2x-10=y² (1)

xét x²=2x-10

(=) x²-2x+10=0

(=) (x-1)²+9=0 (vô lí)

=) x vô nghiệm

-=) y vô nghiệm

vậy ko có nghiệm x,y thỏa mãn pt(1)

k cho mk va kb vs mk nhé

chúc bnn học tốt

mik 2k5 nha

sorry mik 2k7 

cho e 3 k đc ko chị xu 

DELL hiểu

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\frac{bc+ac+ba}{abc}=0\Leftrightarrow bc+ac+ba=0\Leftrightarrow c.\left(a+b\right)=-ba\Leftrightarrow a+b=\frac{-ab}{c}\)

\(b+c=-\frac{bc}{a},a+c=\frac{-ac}{b}\)

thay vô là đc :") lazzy~~

ĐKXĐ : \(\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}}\)

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

\(\Leftrightarrow A=\left[\frac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{5\left(x-1\right)}{2x}\)

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

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

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

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

b) Xét x = -1 không thỏa mãn ĐKXĐ nên ta xét x = 3

Thay x = 3 vào A ta có :

\(A=\frac{5\left(5\cdot3-1\right)}{2\cdot3\cdot\left(3+1\right)}=\frac{35}{12}\)

c) Để A = 2 thì :

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

\(\Leftrightarrow4x\left(x+1\right)=5\left(5x-1\right)\)

\(\Leftrightarrow4x^2+4x=25x-5\)

\(\Leftrightarrow4x^2+4x-25x+5=0\)

\(\Leftrightarrow4x^2-21x+5=0\)

\(\Leftrightarrow4x^2-20x-x+5=0\)

\(\Leftrightarrow4x\left(x-5\right)-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{4}\end{cases}}\)( thỏa mãn ĐKXĐ )

Vậy....