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Ta có:
\(VT=\left[\dfrac{16a-a^2-\left(3+2a\right)\left(a+2\right)-\left(2-3a\right)\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}\right]:\dfrac{a-1}{a^3+4a^2+4a}\)
\(=\dfrac{16a-a^2-3a-6-2a^2-4a-2a+4+3a^2-6a}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}\)
\(=\dfrac{a-2}{\left(a-2\right)\left(a+2\right)}.\dfrac{a\left(a+2\right)^2}{a-1}=\dfrac{a\left(a+2\right)}{a-1}\left(a\ne\pm2;a\ne1\right)\)
\(=a-\dfrac{a\left(a+2\right)}{a-1}=\dfrac{a^2-a-a^2-2a}{-1}=\dfrac{-3a}{a-1}=\dfrac{3a}{1-a}=VP\left(đpcm\right)\)
\(A=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{\left(2a-1\right)^2}{2a+1}\cdot\dfrac{1}{\left(2a-1\right)\left(2a+1\right)}\right)\cdot\left(\dfrac{4a\left(a+1\right)+1}{4a^2}\right)-\dfrac{1}{2a}\)
\(=\left(\dfrac{-\left(2a-1\right)}{2a+1}+\dfrac{2a-1}{\left(2a+1\right)^2}\right)\cdot\dfrac{4a^2+4a+1}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(2a-1\right)\left(2a+1\right)}{\left(2a+1\right)^2}\cdot\dfrac{\left(2a+1\right)^2}{4a^2}-\dfrac{1}{2a}\)
\(=\dfrac{-\left(4a^2-1\right)}{4a^2}-\dfrac{2a}{4a^2}\)
\(=\dfrac{-4a^2-2a+1}{4a^2}\)
\(P=\left(\dfrac{1}{ax-2}+\dfrac{1}{ax+2}+\dfrac{2ax}{a^2x^2+4}+\dfrac{4a^3x^3}{a^2x^4}\right)\cdot\dfrac{a^4x^4+16}{a^4x^4}\)
\(=\left(\dfrac{ax+2+ax-2}{a^2x^2-4}+\dfrac{2ax}{a^2x^2+4}+\dfrac{4a^3x^3}{a^4x^4}\right)\cdot\dfrac{a^4x^4+16}{a^4x^4}\)
\(=\left(\dfrac{2ax\left(a^2x^2+4\right)+2ax\left(a^2x^2-4\right)}{a^4x^4-16}+\dfrac{4a^3x^3}{a^4x^4}\right)\cdot\dfrac{a^4x^4+16}{a^4x^4}\)
\(=\left(\dfrac{4a^3x^3}{a^4x^4-16}+\dfrac{4a^3x^3}{a^4x^4}\right)\cdot\dfrac{a^4x^4+16}{a^4x^4}\)
\(=\dfrac{8a^7x^7-64a^3x^3}{a^4x^4\left(a^4x^4-16\right)}\cdot\dfrac{a^4x^4+16}{a^4x^4}=\dfrac{\left(8a^7x^7-64a^3x^3\right)\left(a^4x^4+16\right)}{a^8x^8\left(a^4x^4-16\right)}\)
\(=\dfrac{8a^3x^3\left(a^4x^4-8\right)\left(a^4x^4+16\right)}{a^8x^8\left(a^4x^4-16\right)}=\dfrac{8\left(a^4x^4-8\right)\left(a^4x^4+16\right)}{a^5x^5\left(a^4x^4-16\right)}\)
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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.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
\(B=\left(\dfrac{a}{\left(a-4\right)\left(a+4\right)}-\dfrac{a-4}{a\left(a+4\right)}\right):\dfrac{2a-4}{a\left(a+4\right)}-\dfrac{a}{a-4}\)
\(=\dfrac{a^2-\left(a-4\right)^2}{a\left(a-4\right)\left(a+4\right)}\cdot\dfrac{a\left(a+4\right)}{2\left(a-2\right)}-\dfrac{a}{a-4}\)
\(=\dfrac{a^2-a^2+8a-16}{a-4}\cdot\dfrac{1}{2\left(a-2\right)}-\dfrac{a}{a-4}\)
\(=\dfrac{8\left(a-2\right)}{2\left(a-2\right)}\cdot\dfrac{1}{a-4}-\dfrac{a}{a-4}\)
\(=\dfrac{4}{a-4}-\dfrac{a}{a-4}=-1\)