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\(=\left(\dfrac{2}{2a-b}-\dfrac{6b}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{4}{2a+b}\right):\dfrac{4a^2-b^2+4a^2+b^2}{4a^2-b^2}\)
\(=\dfrac{4a+2b-6b-8a+4b}{\left(2a-b\right)\left(2a+b\right)}\cdot\dfrac{\left(2a-b\right)\left(2a+b\right)}{8a^2}\)
\(=\dfrac{-4a}{8a^2}=\dfrac{-1}{2a}\)
\(A=\dfrac{4a+2b-6b-8a+4b}{\left(2a-b\right)\left(2a+b\right)}:\dfrac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)
\(=\dfrac{-4a}{\left(2a-b\right)\left(2a+b\right)}\cdot\dfrac{\left(2a-b\right)\left(2a+b\right)}{8a^2}=\dfrac{-1}{2a}\)
e) = \(\dfrac{3}{2\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\)
= \(\dfrac{3x}{2x\left(x+3\right)}\) - \(\dfrac{x-6}{2x\left(x+3\right)}\) = \(\dfrac{3x-x+6}{2x\left(x+3\right)}\)
= \(\dfrac{2x-6}{2x\left(x+3\right)}\)
= \(\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\)
c) = \(\dfrac{2\left(a^3-b^3\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)\left(a^2-2ab+b^2\right)}{3\left(a+b\right)}\) . \(\dfrac{6\left(a+b\right)}{a^2-2ab+b^2}\)
= \(\dfrac{-2\left(a+b\right)}{1}\) . \(\dfrac{2}{1}\) = -4 (a+b)
a: \(=\dfrac{x\left(y-1\right)-\left(y-1\right)}{y\left(1-z\right)-\left(1-z\right)}=\dfrac{\left(y-1\right)\left(x-1\right)}{\left(1-z\right)\left(y-1\right)}=\dfrac{x-1}{1-z}\)
b: \(=\dfrac{\left(a-b\right)\left(a+b\right)}{\left(a-b\right)\left(a+b\right)-\left(a+b\right)}=\dfrac{a-b}{a-b-1}\)
c: \(=\dfrac{\left(a+1\right)\left(a^2-a+1\right)}{2\left(a+1\right)^2}=\dfrac{a^2-a+1}{2a+2}\)
Giải:
\(B=\left(4a^2-4ab+b^2\right)\left(2a+b\right)\)
\(\Leftrightarrow B=\left(2a-b\right)^2\left(2a+b\right)\)
Thay các giá trị của a và b, ta được:
\(B=\left(2.\dfrac{1}{2}-\dfrac{1}{3}\right)^2\left(2.\dfrac{1}{2}+\dfrac{1}{3}\right)\)
\(\Leftrightarrow B=\left(1-\dfrac{1}{3}\right)^2\left(1+\dfrac{1}{3}\right)\)
\(\Leftrightarrow B=\dfrac{4}{9}.\dfrac{4}{3}\)
\(\Leftrightarrow B=\dfrac{16}{27}\)
Vậy ...
B \(=\left[\left(2a\right)^2-2ab+b^2\right]\left(2a+b\right)\)
\(B=\left(2a-b\right)^2\left(2a+b\right)=\left(2a+b\right)\left(2a-b\right)\left(2a-b\right)=\left(4a^2-b^2\right)\left(2a-b\right)\)
Thế a = \(\dfrac{1}{2}\) ; b = \(\dfrac{1}{3}\)ta được:
\(B=\left[4\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{3}\right)^2\right]\left(2.\dfrac{1}{2}-\dfrac{1}{3}\right)\)
\(B=\dfrac{16}{27}\)
\(B=\left(\dfrac{a}{\left(a-4\right)\left(a+4\right)}-\dfrac{a-4}{a\left(a+4\right)}\right):\dfrac{2a-4}{a\left(a+4\right)}-\dfrac{a}{a-4}\)
\(=\dfrac{a^2-\left(a-4\right)^2}{a\left(a-4\right)\left(a+4\right)}\cdot\dfrac{a\left(a+4\right)}{2\left(a-2\right)}-\dfrac{a}{a-4}\)
\(=\dfrac{a^2-a^2+8a-16}{a-4}\cdot\dfrac{1}{2\left(a-2\right)}-\dfrac{a}{a-4}\)
\(=\dfrac{8\left(a-2\right)}{2\left(a-2\right)}\cdot\dfrac{1}{a-4}-\dfrac{a}{a-4}\)
\(=\dfrac{4}{a-4}-\dfrac{a}{a-4}=-1\)
\(=\left(\dfrac{2\left(2a+b\right)-6b-4\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\right):\dfrac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)
\(=\dfrac{4a+2b-6b-8a+4b}{8a^2}\)
\(=\dfrac{-4a}{8a^2}=\dfrac{-1}{2a}\)