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Bài 1:
a: \(P=\left(\dfrac{x-2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+2}{\left(x+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{x^2-x-2-x^2-x+2}{\left(x-1\right)\left(x+1\right)^2}\cdot\dfrac{\left(x-1\right)^2\cdot\left(x+1\right)^2}{4}\)
\(=\dfrac{-2x}{1}\cdot\dfrac{x-1}{4}=-\dfrac{x\left(x-1\right)}{2}\)
b: Để \(\dfrac{P-4}{5}=x\) thì P-4=5x
=>P=5x+4
\(\Leftrightarrow-\dfrac{x\left(x-1\right)}{2}=5x+4\)
=>-x2+x=10x+8
=>x2-x=-10x-8
=>x2+9x+8=0
=>x=-8(nhận) hoặc x=-1(loại)
a) \(\left(a+b\right)^3-\left(a-b\right)^3-6a^2b\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)-6a^2b\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6a^2b\)
\(\Leftrightarrow2b^3\)
b) \(\left(a+b\right)^3-\left(a-b\right)^3-6ab^2\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)-6ab^2\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6ab^2\)
\(\Leftrightarrow2b^3+6a^2b-6ab^2\)
\(\left(x+2\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=\left(x+2\right)\left(x-2-x-2\right)\)
\(=\left(-4\right)\left(x+2\right)\)
\(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(=a^2+b^2+c^2+2ab-2ac-2bc-a^2+2ac-c^2-2ab+2bc\)
\(=b^2\)