(\(\dfrac{2}{2a-b}\)+\(\dfrac{6b}{b^{2^{ }}-4a^2}\)-\(\dfrac{4}{2a+b}\)):(1+\(\dfrac{4a^{2^{ }}+b^{2^{ }}}{4a^{2^{ }}-b^2}\))
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12 tháng 12 2021
\(M=a+\dfrac{4a+2ab+2b+b^2+4a-2ab-2b+b^2-4a}{\left(2-b\right)\left(2+b\right)}\\ M=a+\dfrac{4a+2b^2}{\left(2-b\right)\left(2+b\right)}=\dfrac{4a-ab^2+4a+2b^2}{\left(2-b\right)\left(2+b\right)}\\ M=\dfrac{8a-ab^2+2b^2}{4-b^2}\)
Ta có \(8a-b^2\left(a-2\right)=8a-\dfrac{a^2\left(a-2\right)}{\left(a+1\right)^2}=\dfrac{8a^3+16a^2+8a-a^3+2a^2}{\left(a+1\right)^2}=\dfrac{7a^3+18a^2+8a}{\left(a+1\right)^2}\)
\(4-b^2=4-\dfrac{a^2}{\left(a+1\right)^2}=\dfrac{4a^2+8a+4-a^2}{\left(a+1\right)^2}=\dfrac{3a^2+8a+4}{\left(a+1\right)^2}\)
\(\Leftrightarrow M=\dfrac{7a^3+18a^2+8a}{3a^2+8a+4}=\dfrac{a\left(7a+4\right)\left(a+2\right)}{\left(3a+2\right)\left(a+2\right)}=\dfrac{a\left(7a+4\right)}{3a+2}\)
21 tháng 5 2022
\(=\left(\dfrac{2}{2a-b}-\dfrac{6b}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{4}{2a+b}\right):\dfrac{4a^2-b^2+4a^2+b^2}{4a^2-b^2}\)
\(=\dfrac{4a+2b-6b-8a+4b}{\left(2a-b\right)\left(2a+b\right)}\cdot\dfrac{\left(2a-b\right)\left(2a+b\right)}{8a^2}\)
\(=\dfrac{-4a}{8a^2}=\dfrac{-1}{2a}\)
DX
23 tháng 12 2022
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
21 tháng 5 2022
\(A=\dfrac{4a+2b-6b-8a+4b}{\left(2a-b\right)\left(2a+b\right)}:\dfrac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)
\(=\dfrac{-4a}{\left(2a-b\right)\left(2a+b\right)}\cdot\dfrac{\left(2a-b\right)\left(2a+b\right)}{8a^2}=\dfrac{-1}{2a}\)
26 tháng 5 2022
\(B=\left(\dfrac{a}{\left(a-4\right)\left(a+4\right)}-\dfrac{a-4}{a\left(a+4\right)}\right):\dfrac{2a-4}{a\left(a+4\right)}-\dfrac{a}{a-4}\)
\(=\dfrac{a^2-\left(a-4\right)^2}{a\left(a-4\right)\left(a+4\right)}\cdot\dfrac{a\left(a+4\right)}{2\left(a-2\right)}-\dfrac{a}{a-4}\)
\(=\dfrac{a^2-a^2+8a-16}{a-4}\cdot\dfrac{1}{2\left(a-2\right)}-\dfrac{a}{a-4}\)
\(=\dfrac{8\left(a-2\right)}{2\left(a-2\right)}\cdot\dfrac{1}{a-4}-\dfrac{a}{a-4}\)
\(=\dfrac{4}{a-4}-\dfrac{a}{a-4}=-1\)
2 tháng 7 2021
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
\(=\left(\dfrac{2\left(2a+b\right)-6b-4\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\right):\dfrac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)
\(=\dfrac{4a+2b-6b-8a+4b}{8a^2}\)
\(=\dfrac{-4a}{8a^2}=\dfrac{-1}{2a}\)