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Ta có : \(a=\frac{x}{x^2-x+1}\Rightarrow\frac{1}{a}=\frac{x^2-x+1}{x}\)
\(\Rightarrow\frac{1}{a^2}=\frac{x^4+x^2+1-2x^3+2x^2-2x}{x^2}\)
\(\Rightarrow\frac{1}{a^2}=\frac{x^4+x^2+1}{x^2}-\frac{2x\left(x^2-x+1\right)}{x^2}\)(1)
mà \(A=\frac{x^2}{x^4+x^2+1}\Rightarrow\frac{1}{A}=\frac{x^4+x^2+1}{x^2}\)
\(\left(1\right)\Leftrightarrow\frac{1}{a^2}=\frac{1}{A}-2.\frac{x^2-x+1}{x}\)
\(\Leftrightarrow\frac{1}{a^2}=\frac{1}{A}-2.\frac{1}{a}\)
\(\Leftrightarrow\frac{1}{A}=\frac{1}{a^2}+\frac{2}{a}=\frac{2a+1}{a^2}\)
\(\Rightarrow A=\frac{a^2}{2a+1}\)
- a/ [x/x^2-4 -2(x+2)/x^2-4 +x-2/x^2-4]:[x^2-4/x+2 +10-x^2/x+2] =(x-2x-4+x-2/x^2-4):(x^2-4+10-x^2/x+2) = - 6/x^2-4 nhân với x+2/x^2-4+10-x^2= - 6/(x+2)(x-2) nhân với x+2/6= - 1/x-2.
c/đễ A<0 <=> -1/X-2 <0 <=> x-2<0 <=>x<2
a) A = \(\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
A = \(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
A = \(\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
A = \(-\frac{6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
A = \(-\frac{6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}\)
A = \(-\frac{6}{6\left(x-2\right)}\)
A = \(-\frac{1}{x-2}\)
b) |x| = \(\hept{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
+) với x = 1/2, ta có:
A = \(-\frac{1}{\frac{1}{2}-2}=\frac{2}{3}\)
+) với x = -1/2, ta có:
A = \(-\frac{1}{\left(-\frac{1}{2}\right)-2}=\frac{2}{5}\)