\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)

Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)

\(b,\frac{2x-\frac{4-3x}{5}}{15}=\frac{7x-\frac{x-3}{2}}{5}-x+1\)

\(\Leftrightarrow\frac{2x}{15}-\frac{4-3x}{75}=\frac{7x}{5}-\frac{x-3}{10}-x+1\)

\(\Leftrightarrow\frac{10.2x}{150}-\frac{2\left(4-3x\right)}{150}=\frac{30.7x}{150}-\frac{15\left(x-3\right)}{150}-\frac{150\left(x-1\right)}{150}\)

\(\Leftrightarrow2x-8+6x=210x-15x+45-150x+150\)

\(\Leftrightarrow-19x=203\)

\(\Leftrightarrow x=-\frac{203}{19}\)

Vậy ............

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

a) \(\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)

\(\Rightarrow\dfrac{5\left(x+5\right)}{15}-\dfrac{3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)}{\left(x-3\right)\left(x+5\right)}-\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Rightarrow\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

* Với \(5\left(x+5\right)-3\left(x-3\right)=0\),

Ta có được đẳng thức đúng

=> 5x + 25 - 3x + 9 = 0

=> 2x + 34 = 0

=> 2x = -34

=> x = -17

* Với 5( x+5 ) - 3 (x-3 ) \(\ne\)0, ta có

\(\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Rightarrow\dfrac{1}{15}=\dfrac{1}{\left(x-3\right)\left(x+5\right)}\)

\(\Rightarrow\left(x-3\right)\left(x+5\right)=15\)

\(\Rightarrow x^2+5x-3x-15-15=0\)

\(\Rightarrow x^2+2x-30=0\)

=> \(\left(x+1-\sqrt{31}\right)\left(x+1+\sqrt{31}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{31}\\x=-1-\sqrt{31}\end{matrix}\right.\)

\(a)\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)(ĐKXĐ: \(x\ne3,x\ne-5\))

\(\Leftrightarrow\dfrac{x+5}{3}-\dfrac{x-3}{5}-\dfrac{5}{x-3}+\dfrac{3}{x+5}=0\\ \Leftrightarrow\dfrac{5\left(x-3\right)\left(x+5\right)^2-3\left(x-3\right)^2\left(x+5\right)-75\left(x+5\right)+45\left(x-3\right)}{15\left(x-3\right)\left(x+5\right)}=0\\ \Leftrightarrow\dfrac{2x^3+38x^2+8x-1020}{15\left(x-3\right)\left(x+5\right)}=0\\ \Leftrightarrow2x^3+38x^2+8x-1020=0\\ \Leftrightarrow\left(x+17\right)\left(x^2+2x-30\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+17=0\\x^2+2x-30=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\left(TM\right)\\x=-1+\sqrt{31}\left(TM\right)\\x=-1-\sqrt{31}\left(TM\right)\end{matrix}\right.\)

Vậy....

\(c)\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)

(ĐKXĐ: \(x\ne-1,x\ne-\dfrac{2}{3}\))

\(\Leftrightarrow\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}-1=0\\ \Leftrightarrow\dfrac{2x\left(3x^2+5x+2\right)-7x\left(3x^2-x+2\right)-\left(3x^2-x+2\right)\left(3x^2+5x+2\right)}{\left(3x^2-x+2\right)\left(3x^2+5x+2\right)}=0\\ \Leftrightarrow\dfrac{-27x^3+10x^2-18x-9x^4-4}{\left(3x^2-x+2\right)\left(3x^2+5x+2\right)}=0\\ \Leftrightarrow-9x^4-27x^3+10x^2-18x-4=0\)

\(\Leftrightarrow-\left(3x^2-2x+2\right)\left(3x^2+11x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-\left(3x^2-2x+2\right)=0\\3x^2+11x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-\left(3x^2-2x+2\right)=0\left(VN\right)\\x=\dfrac{-11-\sqrt{97}}{6}\left(TM\right)\\x=\dfrac{-11+\sqrt{97}}{6}\left(TM\right)\end{matrix}\right.\)

Vậy....

\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)

\(\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x\left(x+1\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)

Ta có : `(x-1)/x -1/(x+1) =(2x-1)/(x(x+1))`

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}-\dfrac{x}{x\left(x+1\right)}=\dfrac{2x-1}{x\left(x+1\right)}\)

`=> x^2 +x -x-1 -x-2x+1=0`

`<=> x^2 -3x =0`

`<=> x(x-3)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=3\end{matrix}\right.\)

__

`(x+2)(5-3x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\3x=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)

__

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

\(\Leftrightarrow\dfrac{20\left(1-2x\right)}{12}+\dfrac{6x}{12}=\dfrac{9\left(x-5\right)}{12}-\dfrac{24}{12}\)

`<=> 2x- 40x + 6x = 9x - 45 -24`

`<=>  2x- 40x + 6x-9x + 45 +24=0`

`<=>-41x+69=0`

`<=>-41x=-69`

`<=> x=69/41`

Cậu tách 2 câu 1 lượt mn trl nhanh hơn đó ạ

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)

\(\Leftrightarrow x+3=7x+35\)

\(\Leftrightarrow-6x=32\)

\(\Leftrightarrow x=-\frac{16}{3}\)

b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)

\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)

\(\Leftrightarrow6x-3=-6x-10\)

\(\Leftrightarrow12x=-7\)

\(\Leftrightarrow x=-\frac{7}{12}\)

c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)

\(\Leftrightarrow\left(x+1\right)^2=6^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)

d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)

\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)

\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)

\(\Leftrightarrow2x=2\Leftrightarrow x=1\)

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

\(\Leftrightarrow x^2-2x+x-2=x^2+5x-3x-15\)

\(\Leftrightarrow-3x=-13\)

\(\Leftrightarrow x=\frac{13}{3}\)

f)\(\frac{x+4}{x+5}=\frac{x+6}{x+7}\Leftrightarrow\left(x+4\right)\left(x+7\right)=\left(x+5\right)\left(x+6\right)\)

\(\Leftrightarrow x^2+11x+28=x^2+11x+30\)

\(\Leftrightarrow0x=2\)

\(\Leftrightarrow x\in\varnothing\)

khó quá mk mới học lớp 6 nên k giải đc thông cảm cho mk nha

có ai giúp mk giải 2 bài này vs 

quy dong tung phan thuc mot di ban