cho \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
voi \(x\ne\pm2\)
a) rut gon A
b)tim x de \(A-\frac{1}{x}=-1\)
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Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
\(Q=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
\(\Leftrightarrow\) \(Q=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(2-x\right)}+\frac{5}{\left(x+3\right)\left(2-x\right)}+\frac{-1}{\left(x+3\right)\left(2-x\right)}\)
\(\Rightarrow\) \(Q=\left(x-2\right)\left(x+2\right)+5-1\)
\(\Leftrightarrow\) \(Q=x^2-4+5-1\)
\(\Leftrightarrow\) \(Q=x^2\)
Thay \(Q=\frac{-3}{4}\) ta được:
\(x^2=\frac{-3}{4}\)
Vì \(\frac{-3}{4}>0\forall x\)
\(\Rightarrow\) Pt vô nghiệm
Vậy không có giả trị nào của x thỏa mãn \(Q=\frac{-3}{4}\)
Chúc bn học tốt!!
a) \(A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
\(A=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{4x^2-8x+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)
\(a,A=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{5x-6}{4-x^2}\)
\(=\frac{4x}{x+2}+\frac{2}{x-2}+\frac{6-5x}{x^2-4}\)
\(=\frac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{6-5x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x^2-8x+2x+4+6-5x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x^2-11x+10}{\left(x-2\right)\left(x+2\right)}\)