giúp mik vs
b: ĐKXĐ: x<>5; x<>-6
\(\frac{x+6}{x-5}+\frac{x-5}{x+6}=\frac{2x^2+23x+61}{x^2+x-30}\)
=>\(\frac{\left(x+6\right)^2+\left(x-5\right)^2}{\left(x-5\right)\left(x+6\right)}=\frac{2x^2+23x+61}{\left(x+6\right)\left(x-5\right)}\)
=>\(\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)
=>\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
=>2x+61=23x+61
=>-21x=0
=>x=0(nhận)
c: ĐKXĐ: x<>5; x<>8
\(\frac{6}{x-5}+\frac{x+2}{x-8}=\frac{18}{\left(x-5\right)\left(8-x\right)}-1\)
=>\(\frac{6\left(x-8\right)+\left(x+2\right)\left(x-5\right)}{\left(x-5\right)\left(x-8\right)}=\frac{-18}{\left(x-5\right)\left(x-8\right)}-\frac{\left(x-5\right)\left(x-8\right)}{\left(x-5\right)\left(x-8\right)}\)
=>6(x-8)+(x+2)(x-5)=-18-(x-5)(x-8)
=>\(6x-48+x^2-3x-10=-18-\left(x^2-13x+40\right)\)
=>\(x^2+3x-58=-18-x^2+13x-40=-x^2+13x-58\)
=>\(2x^2-10x=0\)
=>2x(x-5)=0
=>x(x-5)=0
=>\(\left[\begin{array}{l}x=0\left(nhận\right)\\ x=5\left(loại\right)\end{array}\right.\)
d: ĐKXĐ: x<>1; x<>-1
\(\frac{x-4}{x-1}+\frac{x+4}{x+1}=2\)
=>\(\frac{\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=2\)
=>\(\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)=2\left(x^2-1\right)\)
=>\(x^2-3x-4+x^2+3x-4=2x^2-2\)
=>\(2x^2-8=2x^2-2\)
=>-8=-2(loại)
e: ĐKXĐ: x<>-1; x<>2
\(\frac{3}{x+1}-\frac{1}{x-2}=\frac{9}{\left(x+1\right)\left(2-x\right)}\)
=>\(\frac{3\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-9}{\left(x+1\right)\left(x-2\right)}\)
=>3(x-2)-(x+1)=-9
=>3x-6-x-1=-9
=>2x-7=-9
=>2x=-2
=>x=-1(loại)
f: ĐKXĐ: x<>3; x<>-3
\(\frac{x^2-x}{x+3}-\frac{x^2}{x-3}=\frac{7x^2-3x}{9-x^2}\)
=>\(\frac{\left(x^2-x\right)\left(x-3\right)-x^2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}\)
=>\(\left(x^2-x\right)\left(x-3\right)-x^2\left(x+3\right)=-7x^2+3x\)
=>\(x^3-3x^2-x^2+3x-x^3-3x^2=-7x^2+3x\)
=>\(-7x^2+3x=-7x^2+3x\)
=>0x=0(luôn đúng)
=>x∈R\{3;-3]
