Rút gọn biểu thức sau:
B= \(\frac{x^2}{x-2}\)(\(\frac{x^2+4}{x}-4\)) -\(^{x^2}\)
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Y
7 tháng 8 2023
\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)
\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)
\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)
Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.
14 tháng 10 2021
\(B=\dfrac{1-4\sin^2x\cdot\cos^2x}{\sin^2x+2\sin x\cdot\cos x+\cos^2}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}+2\sin x\cdot\cos x\\ B=\dfrac{1-4\sin^2x\cdot\cos^2x+4\sin^2x\cdot\cos^2x}{2\sin x\cdot\cos x}=\dfrac{1}{2\sin x\cdot\cos x}\)
3 tháng 2 2017
ĐK: x khác +-2
\(C=\left(\frac{2}{x+2}-\frac{x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{x-2}\right).\left(\frac{x-2}{x^2-4+6-x^2}\right)\\ \)
\(C=\frac{2\left(x-2\right)-x+\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}.\left(\frac{x-2}{2}\right)=\frac{2\left(x-1\right)\left(x-2\right)}{2.\left(x-2\right)\left(x+2\right)}\)
\(C=\frac{x-1}{x+2}\)
3 tháng 2 2017
C=[2/(x+2)-x/(x^2-4)-1/(2-x)]:[x+2+(6-x^2)/(x-2)]
=[2/(x+2)-x/(x-2)(x+2)-(-1)/(x-2)]:[x+2+(6-x^2)/(x-2)]
=[2x-4-x+x+2/(x-2)(x+2)]:[(x^2-4+6-x^2)/(x-2)]
=2x-2/(x-2)(x+2) . (x-2)/2
=2(x-1)/(x-2)(x+2) . (x-2)/2
=x-1/x+2
L
1
17 tháng 7 2018
\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}\) ĐKXĐ: \(x\ne\pm2\)
\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2-2x-x+2-x^2-4x-4+x^2+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x\left(x-2\right)-5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-5\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-5}{x+2}\)
TA
rút gọn biểu thức A =\(\left(\frac{4x}{x^2-4}+\frac{2x-4}{x+2}\right).\frac{x+2}{2x}+\frac{2}{2-x}\)
0
BT
9 tháng 11 2017
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right).\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
\(=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right).\left(x^4+\frac{\left(1+x^2\right)\left(1-x^2\right)}{1+x^2}\right)\)
\(=\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\left(x^4+1-x^2\right)\)
\(=\frac{x^2-2}{x^2+1}\).
\(B=\frac{x^2}{x-2}.\left(\frac{x^2+4}{x}-4\right)-x^2\)
\(B=\frac{-2x^3+4x^2}{x^2-2x}\)
\(B=\frac{-2x^2+4x}{x-2}\)
\(B=\frac{2x\left(-x+2\right)}{x-2}\)
\(B=-2x\)
bn có thể giải chi tiết hơn được không ạ