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Ta có :\(\frac{4}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x}\)
\(=\frac{4\left(x-2\right)\left(4-x\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}+\frac{2\left(x+2\right)\left(4-x\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)\(+\frac{\left(5x-6\right)\left(x+2\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{24x-8x^2-32}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}+\frac{4x-2x^2+16}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)\(+\frac{5x^2+4x-12}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{24x-8x^2-32+4x-2x^2+16+5x^2+4x-12}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
\(=\frac{32x-5x^2-28}{\left(x+2\right)\left(x-2\right)\left(4-x\right)}\)
A=(1/x-2 - (2x/(2-x)(2+x) - 1/2+x) ) *(2-x)/x
=(1/x-2 - x^2+5x-2/(2-x)(2+x))*2-x/x
=(-x^3-4x^2+12x/(x-2)(2-x)(2+x))*2-x/x
= - x(x-2)(x+6)(2-x)/x(x-2)(2-x)(2+x)
= - x+6/x+2
Tiếp
\(=\left(\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}\right).\left(\frac{x^2+x+1}{2x+1}\right)=\left(\frac{x^2+x+1}{x^2-1}\right)=1+\frac{x+2}{x^2-1}\)
a) ĐKXĐ: \(x\ne-1;x\ne2\)
Ta có: \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
⇔\(\frac{1}{x+1}-\frac{5}{x-2}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(x-2-5x-5+15=0\)
⇔\(-4x+8=0\)
⇔\(-4x=-8\)
⇔\(x=\frac{-8}{-4}=2\)(loại)
Vậy: x không có giá trị
b) ĐKXĐ: \(x\ne0;x\ne\frac{3}{2}\)
Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
⇔\(\frac{x}{\left(2x-3\right)\cdot x}-\frac{3}{x\left(2x-3\right)}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}=0\)
⇔\(x-3-10x+15=0\)
⇔\(-9x+12=0\)
⇔\(-9x=-12\)
⇔\(x=\frac{-12}{-9}=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
c) ĐKXĐ:\(x\ne3;x\ne1\)
Ta có: \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2\left(x-3\right)}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{4}{x-3}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}-\frac{4}{x-3}=0\)
⇔\(\frac{6}{x-1}-\frac{8}{x-3}=0\)
⇔\(\frac{6\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=0\)
⇔\(6\left(x-3\right)-8\left(x-1\right)=0\)
⇔6x-18-8x+8=0
⇔-2x-10=0
⇔-2(x+5)=0
Vì 2≠0 nên x+5=0
hay x=-5
Vậy: x=-5
A = (\(\frac{x}{x+1}\) + \(\frac{1}{x-1}\) ) : (\(\frac{2x+2}{x-1}\) - \(\frac{4x}{x^2-1}\) )
A = (\(\frac{x}{x+1}\) + \(\frac{1}{x-1}\) ) : ( \(\frac{2x+2}{x-1}\) - \(\frac{4x}{\left(x+1\right)\left(x-1\right)}\) )
\(\Rightarrow\) MTC: (x+1)(x-1)
A = ( \(\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\) + \(\frac{x+1}{\left(x+1\right)\left(x-1\right)}\) ) : (\(\frac{2\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\) - \(\frac{4x}{\left(x+1\right)\left(x-1\right)}\) )
A = \(\frac{x^2+1}{\left(x+1\right)\left(x-1\right)}\) : \(\frac{2x^2+2}{\left(x+1\right)\left(x-1\right)}\)
A = \(\frac{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}{2\left(x^2+1\right)\left(x+1\right)\left(x-1\right)}\)
A = \(\frac{1}{2}\)
mệt rồi :v ngủ =)))
