mình chịu.

<=> \(x^2-25=10x+35-2x^2-7x\)

<=> \(3x^2-3x-60=0\)

<=> \(x^2-x-20=0\)

<=> \(\left(x-5\right)\left(x+4\right)=0\)

<=> \(\orbr{\begin{cases}x=5\\x=-4\end{cases}}\)

Vay \(x\in\left\{-4;5\right\}\)

Chuc ban hoc tot 

\(x^2-25=\left(5-x\right)\left(2x+7\right)\)

\(\Leftrightarrow x^2-25=10x-2x^2+35-7x\)

\(\Leftrightarrow3x^2-3x-60=0\)

Ta có \(\Delta=3^2+4.3.60=729,\sqrt{\Delta}=27\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3+27}{6}=5\\x=\frac{3-27}{6}=-4\end{cases}}\)

\(ĐKXĐ:x\ne-1;x\ne\frac{2}{3}\)

\(pt\Leftrightarrow\frac{7x-2\left(x+1\right)+\left(3x-2\right)}{\left(3x-2\right)\left(x+1\right)}=1\)

\(\Leftrightarrow7x-2\left(x+1\right)+\left(3x-2\right)=\left(3x-2\right)\left(x+1\right)\)

\(\Leftrightarrow8x-4=3x^2-2x+3x-2\)

\(\Leftrightarrow3x^2-7x+2=0\)

\(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{7+5}{6}=2\\x=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)

Tự cho đkxđ nha!!!

<=> \(\frac{x+1-x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2}{3x-2}\)

<=> \(\frac{3x-2}{\left(3x-2\right)\left(x+1\right)}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(3x-2\right)\left(x+1\right)}\)

<=> \(\frac{7x-2x-2-3x+2}{\left(3x-2\right)\left(x+1\right)}=0\)

<=> \(\frac{2x}{\left(3x-2\right)\left(x+1\right)}=0\)

=> 2x = 0

<=> x = 0 (TM)

Vậy ...

\(1-\frac{x}{x+1}=\frac{7x}{\left(3x-2\right)\left(x+1\right)}+\frac{2}{2-3x}\)

\(\left(x+1\right)\left(3x-2\right)\left(2-3x\right)-x\left(3x-2\right)\left(2-3x\right)=7x\left(2-3x\right)+2\left(x+1\right)\left(3x-2\right)\)

\(-9x^2+12x-4=16x-15x^2-4\)

\(-9x^2+12x=16x-15x^2\)

\(9x^2-12x+16x-15x^2=0\)

\(-6x^2+4x=0\)

\(-2x\left(3x-2\right)=0\)

\(Th1:-2x=0\Leftrightarrow x=0\)

\(Th2:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)

\(\frac{x}{2}-\left(\frac{3}{5}x-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)

\(\Rightarrow\frac{5x-6x+26+14+7x}{10}=0\Rightarrow6x+40=0\Rightarrow x=-\frac{20}{3}\)

Thấy câu 2b hay hay mình làm nha bạn còn câu kia bạn tính bth, câu gttđ xét 2th

Làm bài ạ:

(x-5²):2=32:(x-5)

=>(x-5)².(x-5)=32.2

=>(x-5)³=64

=>(x-5)³=4³

=>x-5=4

=>x=9

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

a>Tìm đkxđ

b>Rút gọn

\(a,\frac{x^2+2.x.5+5^2+x^2-2.x.5+5^2}{x^2+25}\)

\(=\frac{2\left(x^2+25\right)}{x^2+25}=2\)

Vậy giá trị biểu thức không phụ thuộc vào x

\(b,\frac{4x^2+20x+25+25x^2-20x+4}{x^2+1}\)

\(=\frac{29\left(x^2+1\right)}{x^2+1}=29\)

Vậy gt biểu thức không phụ thuộc vào x

a)  \(\frac{\left(x+5\right)^2+\left(x-5\right)^2}{x^2+25}\)

   \(=\frac{x^2+10x+25+x^2-10x+25}{x^2+25}\)

   \(=\frac{2\left(x^2+25\right)}{x^2+25}=2\)

\(\Rightarrow\)đpcm

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

\(=\frac{4x^2+20x+25+25x^2-20x+4}{x^2+1}\)

\(=\frac{29\left(x^2+1\right)}{x^2+1}=29\)

\(\Rightarrow\)đpcm

Mình cảm ơn nhiều.

a)

\(\frac{7x-3}{x-1}=\frac{2}{3}\\ \Leftrightarrow\frac{21x-9}{3\cdot\left(x-1\right)}-\frac{2x-2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow21x-9-2x+2=0\\ \Leftrightarrow19x-7=0\\ \Rightarrow x=\frac{7}{19}\)

Vậy \(x=\left\{\frac{7}{19}\right\}\) là nghiệm của phương trình.

b)

\(\frac{1}{x-2}+3=\frac{3-x}{x-2}\\ \Leftrightarrow\frac{1}{x-2}+\frac{3x-6}{x-2}-\frac{3-x}{x-2}=0\\ \Leftrightarrow1+3x-6-3+x=0\\ \Leftrightarrow4x-8=0\\ \Rightarrow x=\frac{8}{4}=2\)

Mà \(ĐKXĐ:x\ne2\\ \Rightarrow x\in\varnothing\)

Hay phương trình vô nghiệm.

c)

\(\frac{8-x}{x-7}-8=\frac{1}{x-7}\\ \Leftrightarrow\frac{8-x}{x-7}-\frac{8x-56}{x-7}-\frac{1}{x-7}=0\\ \Leftrightarrow8-x-8x+56-1=0\\ \Leftrightarrow63-9x=0\\ \Rightarrow x=\frac{63}{9}=7\)

Mà \(ĐKXĐ:x\ne7\\ \Rightarrow x\in\varnothing\)

Hay phương trình vô nghiệm.

d)

\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\\ \Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\\ \Leftrightarrow\left(x+5\right)^2-\left(x-5\right)^2-20=0\\ \Leftrightarrow x^2+10x+25-x^2+10x-25-20=0\\ \Leftrightarrow20x-20=0\\ \Rightarrow x=1\)

Vậy \(x=1\) là nghiệm của phương trình.

e)

\(\frac{x}{2\cdot\left(x-3\right)}+\frac{x}{2\cdot\left(x+1\right)}=\frac{2x}{\left(x+1\right)\cdot\left(x-3\right)}\\ \Leftrightarrow\frac{x^2+x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}+\frac{x^2-3x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}-\frac{4x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}=0\\ \Leftrightarrow x^2+x+x^2-3x-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\cdot\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Mà \(ĐKXĐ:x\ne\left\{-1;3\right\}\\ \Rightarrow x=0\)

Vật \(x=0\) là nghiệm của phương trình.

f)

\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\\ \Leftrightarrow\frac{12x+120}{12x\cdot\left(x+10\right)}+\frac{12x}{12x\cdot\left(x+10\right)}-\frac{x^2+10x}{12x\cdot\left(x+10\right)}=0\\ \Leftrightarrow12x+120+12x-x^2-10x=0\\ \Leftrightarrow14x+120-x^2=0\\ \Leftrightarrow x^2-14x-120=0\\ \Leftrightarrow x^2+6x-20x-120=0\\ \Leftrightarrow x\cdot\left(x+6\right)-20\cdot\left(x+6\right)=0\\ \Leftrightarrow\left(x-20\right)\cdot\left(x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-20=0\\x+6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)

Vậy \(x=\left\{20;-6\right\}\) là ngiệm của phương trình.