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\(M=\frac{1}{ab}+\frac{1}{a^2+ab}+\frac{1}{b^2+ab}+\frac{1}{a^2+b^2}\)
\(=\left(\frac{1}{2ab}+\frac{1}{a^2+b^2}\right)+\left(\frac{1}{a^2+ab}+\frac{1}{b^2+ab}\right)+\frac{1}{2ab}\)
\(\ge\frac{\left(1+1\right)^2}{a^2+2ab+b^2}+\frac{\left(1+1\right)^2}{a^2+ab+b^2+ab}+\frac{2}{\left(a+b\right)^2}\)
\(=\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{2}{\left(a+b\right)^2}\)
\(\ge\frac{4}{1}+\frac{4}{1}+\frac{2}{1}=10\)
Dấu = xảy ra khi a = b = \(\frac{1}{2}\)
a) \(6x^2+6\)
\(=6\left(x^2+1\right)\)
b) \(2x^2-18\)
\(=2\left(x^2-9\right)\)
\(=2\left(x-3\right)\left(x+3\right)\)
c) \(3x^2-3xy+4x-4y\)
\(=\left(3x^2-3xy\right)+\left(4x-4y\right)\)
\(=3x\left(x-y\right)+4\left(x-y\right)\)
\(=\left(3x-4\right)\left(x-y\right)\)
a) \(\left(x^3-9x^2+27x-27\right)\)\(:\)\(\left(x-3\right)\)
\(=\left(x-3\right)^3\)\(:\)\(\left(x-3\right)\)
\(=\left(x-3\right)^2\)
c) \(\frac{x^2-4}{2x}:\frac{3x-6}{6}\)
\(=\frac{\left(x-2\right)\left(x+2\right)}{2x}.\frac{6}{3\left(x-2\right)}\)
\(=\frac{\left(x+2\right)}{x}\)