Giúp Mk mí bài nha...........Mơn nhìu
GIẢI CÁC PHƯƠNG TRÌNH
1/ \(\dfrac{x+2}{x+1}\) + \(\dfrac{3}{x-2}\) = \(\dfrac{3}{x^2-x-2}\) +1
2/\(\dfrac{x+6}{x-5}\) + \(\dfrac{x-5}{x+6}\) = \(\dfrac{2x^2+23x+61}{x^2+x-30}\)
3/ \(\dfrac{6}{x-5}\) + \(\dfrac{x+2}{x-8}=\dfrac{18}{\left(x-5\right)\left(8-x\right)}-1\)
4/ \(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\)
5/\(\dfrac{3}{x+1}-\dfrac{1}{x-2}=\dfrac{9}{\left(x+1\right)\left(2-x\right)}\)
6/ \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)
7/\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\)
8/ \(\dfrac{x+2}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)
9/ \(\dfrac{1+a}{1-x}=1-a\) (a là hằng số)
10/ \(\dfrac{x}{2a+x}+\dfrac{2a+x}{2a-x}=\dfrac{8a^2}{x^2-4a^2}\) (a là hằng số)
11/ \(\dfrac{2a-3b}{x-2a}+\dfrac{3b-2a}{x-3b}=0\) (a là hằng số)
1: \(\Leftrightarrow\left(x+2\right)\left(x-2\right)+3\left(x+1\right)=3+x^2-x-2\)
\(\Leftrightarrow x^2-x+1=x^2-4+3x+3=x^2+3x-1\)
=>-4x=-2
hay x=1/2
2: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)
\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x^2+23x+61=2x^2+2x+11\)
=>21x=-50
hay x=-50/21
3: \(\Leftrightarrow6\left(x-8\right)+\left(x+2\right)\left(x-5\right)=-18-\left(x-5\right)\left(x-8\right)\)
\(\Leftrightarrow6x-48+x^2-3x-10+18+x^2-13x+40=0\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0(nhận) hoặc x=5(loại)