Bài 3:
a: ĐKXĐ: x<>1
\(\frac{7x-3}{x-1}=\frac23\)
=>3(7x-3)=2(x-1)
=>21x-9=2x-2
=>21x-2x=-2+9
=>19x=7
=>\(x=\frac{7}{19}\)
b: ĐKXĐ: x<>-1
\(\frac{2\left(3-7x\right)}{1+x}=\frac12\)
=>4(3-7x)=x+1
=>12-28x=x+1
=>-29x=1-12=-11
=>\(x=\frac{11}{29}\)
c: ĐKXĐ: x<>2
\(\frac{1}{x-2}+3=\frac{3-x}{x-2}\)
=>\(\frac{1+3\left(x-2\right)}{x-2}=\frac{3-x}{x-2}\)
=>3(x-2)+1=3-x
=>3x-6+1=3-x
=>4x=3+6-1=8
=>x=2(loại)
d: ĐKXĐ: x<>7
\(\frac{8-x}{x-7}-8=\frac{1}{x-7}\)
=>\(\frac{8-x-8\left(x-7\right)}{x-7}=\frac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
=>x=7(loại)
e: ĐKXĐ: x∉{5;-5}
\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
=>\(\frac{\left(x+5\right)^2-\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\)
=>\(\left(x+5\right)^2-\left(x-5\right)^2=20\)
=>\(x^2+10x+25-x^2+10x-25=20\)
=>20x=20
=>x=1(nhận)
f: ĐKXĐ: x∉{1;-1}
\(\frac{1}{x-1}+\frac{2}{x+1}=\frac{x}{x^2-1}\)
=>\(\frac{x+1+2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x}{\left(x-1\right)\left(x+1\right)}\)
=>x+1+2(x-1)=x
=>2x-2+x+1=x
=>3x-1=x
=>2x=1
=>x=1/2(nhận)
g: ĐKXĐ: x∉{-1;3}
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\)
=>\(\frac{x\left(x+1\right)+x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{4x}{2\left(x+1\right)\left(x-3\right)}\)
=>x(x+1)+x(x-3)=4x
=>\(x^2+x+x^2-3x-4x=0\)
=>\(2x^2-6x=0\)
=>2x(x-3)=0
=>x(x-3)=0
=>x=0(nhận) hoặc x=3(loại)
h: ĐKXĐ: x∉{4;-4}
\(5+\frac{76}{x^2-16}=\frac{2x-1}{x+4}-\frac{3x-1}{4-x}\)
=>\(\frac{5\left(x^2-16\right)+76}{\left(x-4\right)\left(x+4\right)}=\frac{\left(2x-1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}+\frac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)
=>\(5\left(x^2-16\right)+76=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)\)
=>\(2x^2-9x+4+3x^2+11x-4=5x^2-80+76\)
=>2x=-4
=>x=-2(nhận)
i: ĐKXĐ: x∉{0;6}
\(\frac{90}{x}-\frac{36}{x-6}=2\)
=>\(\frac{45}{x}-\frac{18}{x-6}=1\)
=>\(\frac{45\left(x-6\right)-18x}{x\left(x-6\right)}=1\)
=>x(x-6)=45(x-6)-18x=27x-270
=>\(x^2-6x-27x+270=0\)
=>\(x^2-33x+270=0\)
=>(x-15)(x-18)=0
=>x=15(nhận) hoặc x=18(nhận)
k: ĐKXĐ: x∉{0;-10}
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
=>\(\frac{x+10+x}{x\left(x+10\right)}=\frac{1}{12}\)
=>x(x+10)=12(2x+10)
=>\(x^2+10x=24x+120\)
=>\(x^2-14x-120=0\)
=>(x-20)(x+6)=0
=>x=20(nhận ) hoặc x=-6(nhận)
l: ĐKXĐ: x∉{0;3}
\(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\)
=>\(\frac{x\left(x+3\right)-\left(x-3\right)}{x\left(x-3\right)}=\frac{3}{x\left(x-3\right)}\)
=>x(x+3)-(x-3)=3
=>\(x^2+3x-x+3=3\)
=>\(x^2+2x=0\)
=>x(x+2)=0
=>x=0(loại) hoặc x=-2(nhận)
m: ĐKXĐ: x∉{2;-2}
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
=>3(x-2)-2(x+2)+8=0
=>3x-6-2x-4+8=0
=>x-2=0
=>x=2(loại)
n: ĐKXĐ: x∉{-2;3}
\(\frac{3}{x+2}-\frac{2}{x-3}=\frac{8}{\left(x-3\right)\left(x+2\right)}\)
=>3(x-3)-2(x+2)=8
=>3x-9-2x-4=8
=>x-13=8
=>x=21(nhận)
