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VV
2
3 tháng 7 2018
\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}=\frac{\left(x-1\right).\left(x-2\right)}{x^2-4}-\frac{\left(x+2\right)^2}{x^2-4}+\frac{x^2+12}{x^2-4}\)
\(=\frac{x^2-3x+2}{x^2-4}-\frac{x^2+4x+4}{x^2-4}+\frac{x^2+12}{x^2-4}=\frac{x^2-7x+10}{x^2-4}=\frac{\left(x-2\right).\left(x-5\right)}{\left(x-2\right).\left(x+2\right)}=\frac{x-5}{x+2}\)
3 tháng 7 2018
\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\)\(\frac{x^2+12}{4-x^2}\)\(ĐKXĐ\): \(x\ne\pm2\)
\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)\(-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}\)\(+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2-2x-x+2-x^2-4x-4+x^2+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x\left(x-2\right)-5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-5\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-5}{x+2}\)
5 tháng 8 2016
\(A=\frac{1}{x-2}+\frac{1}{x+2}+\frac{x^2}{x^2-4}\)
\(ĐKXĐ:x\ne\pm2\)
\(a,A=\frac{x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x+2+x-2+x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x+x^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x\left(2+x\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x}{x-2}\)
\(b,A=\frac{x}{x-2}\)
\(=\frac{x-2+2}{x-2}\)
\(=\frac{x-2}{x-2}+\frac{2}{x-2}\)
\(=1+\frac{2}{x-2}\)
\(\text{Để A có giá trị nguyên thì:2⋮ x-2}\)
\(\text{hay }x-2\inƯ\left(2\right)=\left\{-1;1;-2;2\right\}\)
\(\Rightarrow x\in\left\{1;3;0;4\right\}\left(tm\right)\)
\(\text{Vậy }x\in\left\{1;3;0;4\right\}\) \(\text{thì A có giá trị nguyên.}\)
19 tháng 8 2019
\(1,ĐK:x\ne0;x\ne\pm6\)
\(A=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right].\frac{\left(x+6\right)\left(x-6\right)}{12\left(x^2+1\right)}\)
\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x}.\frac{1}{12\left(x^2+1\right)}\)
\(=\frac{12\left(x^2+1\right)}{x}.\frac{1}{12\left(x^2+1\right)}=\frac{1}{x}\)
\(2,A=\frac{1}{x}=\frac{1}{\frac{1}{\sqrt{9+4\sqrt{5}}}}=\sqrt{9+4\sqrt{5}}\)
12 tháng 2 2020
Cho tam giác ABC vuông tại B có góc B1=B2 ; Â=60o, kẻ BH vuông góc với AC (H thuộc AC). Qua B kẻ đường thẳng d song song với AC.
a) Tính góc ABH.
b) Chứng minh đường thẳng d vuông góc với BH.
26 tháng 11 2017
M = 1/(x+1).(x+2) + 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/x+5
= 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 = 1/x+1
k mk nha
30 tháng 1 2022
\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
BT
9 tháng 11 2017
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right).\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
\(=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right).\left(x^4+\frac{\left(1+x^2\right)\left(1-x^2\right)}{1+x^2}\right)\)
\(=\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\left(x^4+1-x^2\right)\)
\(=\frac{x^2-2}{x^2+1}\).
30 tháng 3 2021
a) ĐKXĐ : x ≠ ±2
\(=\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(=\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]\div\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\div\frac{6}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\times\frac{x+2}{6}=\frac{-1}{x-2}\)
b) Để A < 0 thì -1/x-2 < 0
=> x - 2 > 0 <=> x > 2
Vậy với x > 2 thì A < 0
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
\(A=\frac{x-1}{x+2}-\frac{x+2}{x-2}-\frac{x^2+12}{4-x^2}\) ĐKXĐ: \(x\ne\pm2\)
\(=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2-2x-x+2-x^2-4x-4+x^2+12}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-7x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x^2-2x-5x+10}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x\left(x-2\right)-5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{\left(x-5\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\frac{x-5}{x+2}\)