ĐKXĐ: x<>5; x<>-6

\(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)

\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>23x+61=2x+61

=>x=0

a, đk : x khác 5;-6 

\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm) 

b, đk : x khác 1;3 

\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)

pt vô nghiệm 

a, đk : x khác 5;-6 

(tm) 

b, đk : x khác 1;3 

x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)

pt vô nghiệm 

a: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)

\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>x=0(nhận)

b: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)

\(\Leftrightarrow x^2+2x-15=x^2-1-8\)

=>2x-15=-9

=>2x=-6

hay x=-3(nhận)

1: \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)

\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>23x+61=2x+61

hay x=0

2: \(\dfrac{6}{x-5}+\dfrac{x+2}{x-8}=\dfrac{18}{\left(x-5\right)\left(8-x\right)}-1\)

\(\Leftrightarrow6x-48+x^2-3x-10=-18-x^2+13x-40\)

\(\Leftrightarrow x^2+3x-58+x^2-13x+58=0\)

\(\Leftrightarrow2x^2-10x=0\)

=>2x(x-5)=0

=>x=0

c: \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)

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

\(\Leftrightarrow x^3-3x^2-x^2+3x-x^3-3x^2+7x^2-3x=0\)

\(\Leftrightarrow x^2=0\)

hay x=0

bài của p hay trog sgk
 

a) điều kiện xác định : \(x\ne2;x\ne-1\)

ta có : \(\dfrac{x+2}{x+1}+\dfrac{3}{x-2}=\dfrac{3}{x^2-x-2}+1\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3+x^2-x-2}{\left(x+1\right)\left(x-2\right)}\)

\(\Rightarrow x^2-4+3x+3=x^2-x+1\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\left(tmđk\right)\)

vậy \(x=\dfrac{1}{2}\)

b) điều kiện xác định : \(x\ne5;x\ne-6\)

ta có : \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)

\(\Leftrightarrow\dfrac{\left(x+6\right)^2+\left(x-5\right)^2}{\left(x-5\right)\left(x+6\right)}=\dfrac{2x^2+23x+61}{\left(x-5\right)\left(x+6\right)}\)

\(\Rightarrow x^2+12x+36+x^2-25x+25=2x^2+23x+61\)

\(\Leftrightarrow-13x=23x\Leftrightarrow x=0\left(tmđk\right)\)

vậy \(x=0\)

a: ĐKXĐ: \(x\notin\left\{4;-4\right\}\)

\(\dfrac{7}{4x+16}=\dfrac{7}{4\left(x+4\right)}=\dfrac{7\left(x-4\right)}{4\left(x+4\right)\left(x-4\right)}\)

\(\dfrac{11}{x^2-16}=\dfrac{11\cdot4}{4\left(x^2-16\right)}=\dfrac{44}{4\left(x-4\right)\left(x+4\right)}\)

b: \(\dfrac{6}{x\left(x+3\right)^2};\dfrac{x-3}{2x\left(x+3\right)^2}\)

ĐKXĐ: \(x\notin\left\{0;-3\right\}\)

\(\dfrac{6}{x\left(x+3\right)^2}=\dfrac{6\cdot2}{2x\left(x+3\right)^2}=\dfrac{12}{2x\left(x+3\right)^2}\)

\(\dfrac{x-3}{2x\left(x+3\right)^2}=\dfrac{x-3}{2x\left(x+3\right)^2}\)

c: \(\dfrac{-6}{1-x};\dfrac{3x}{x^2+x+1};\dfrac{x^2-3x+5}{x^3-1}\)

ĐKXĐ: \(x\ne1\)

\(-\dfrac{6}{1-x}=\dfrac{6}{x-1}=\dfrac{6\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{6x^2+6x+6}{\left(x-1\right)\left(x^2+x+1\right)}\)

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

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

d: \(\dfrac{17}{5x};\dfrac{24}{x-2y};\dfrac{x-y}{8y^2-2x^2}\)

ĐKXĐ: \(x\ne0;x\ne\pm2y\)

\(\dfrac{17}{5x}=\dfrac{17\cdot2\left(x-2y\right)\left(x+2y\right)}{5x\cdot2\cdot\left(x-2y\right)\left(x+2y\right)}=\dfrac{34\left(x^2-4y^2\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\dfrac{24}{x-2y}=\dfrac{24\cdot10x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{240x\left(x+2y\right)}{10x\left(x-2y\right)\left(x+2y\right)}\)

\(\dfrac{x-y}{8y^2-2x^2}=\dfrac{-\left(x-y\right)}{2x^2-8y^2}=\dfrac{-\left(x-y\right)}{2\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{-5x\left(x-y\right)}{10x\left(x-2y\right)\left(x+2y\right)}=\dfrac{-5x^2+5xy}{10x\left(x-2y\right)\left(x+2y\right)}\)

sorry, em mới học lớp 6 thui

mình cũng giải câu này rồi nhưng  ko biết đúng ko

2)x+6/x-5 + x-5/x+6 = 2x²+23x+61/x²+x-30

dkxd:x khắc 5;x khác-6

mc:(x-5)(x+6)

2x²+2x+61 =2x²+23x+61

2x=23x

2x=0 suy ra x=0

23x=0 suyra x=0 

s={0}

3)6/x-5 + x+2/x-8 = 18/9(x-5)(8-x) - 1

dkxd: x khác 5 ; x khác -8

mc(x-5)(x-8) 

3x+x²-58 =36x-x²+264

3x-58=36x+264  

3x-58=0 suy ra x=58/3

36x+264=0 suy ra x=-22/8

s={58/3;-22/3}