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Bài 1:
\(P=2a^2-2b^2-a^2+2ab-b^2+a^2+2ab+b^2+b^2=2a^2-b^2+4ab\\ Q=\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(2x-3\right)\left(2x+3\right)\\ Q=\left(2x+3-2x+3\right)^2=9^2=81\)
Bài 2:
\(Sửa:A=x^2+2xy+y^2-4x-4y+2=\left(x+y\right)^2-4\left(x+y\right)+4-2\\ A=\left(x+y-2\right)^2-2=\left(3-2\right)^2-2=1-2=-1\)
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
Bài 8:
Ta có: \(A=-x^2+2x+4\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=1
\(a,\left(a+b\right)^2-\left(a-b\right)^2\)
\(=a^2+2ab+b^2-a^2+2ab-b^2\)
\(=4ab\)
\(b,\left(a+b\right)^3-\left(a-b\right)-\left(2b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-a+b-8b^3\)
a) \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(\left(2b\right)\left(2a\right)\)
\(4ab\)
b) \(\left(a+b\right)^3-\left(a-b\right)-\left(2b\right)^3\)
\(a^3+3a^2b+3ab^2+b^3-a+b-8b^3\)
\(a\left(a^2-1\right)+3\left(a^2b+ab^2\right)+b\left(b^2+1-8b^2\right)\)
\(a\left(a-1\right)\left(a+1\right)+3\left[ab\left(a+b\right)\right]+b\left(-7b^2+1\right)\)