Bài 1:
Ta có:
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x^2+2xy+y^2\right)-\left(x^2-2xy+y^2\right)\)
\(=x^2+2xy+y^2-x^2+2xy-y^2\)
\(=4xy\)
Bài 2:
a) \(A=x^2+8x+2017\)
\(A=x^2+2.x.4+16+2001\)
\(A=\left(x+4\right)^2+2001\)
Vì \(\left(x+4\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x+4\right)^2+2001\ge2001\) với mọi x
\(\Rightarrow Amin=2001\Leftrightarrow x=-4\)
Bài 2:
b) Ta có:
\(a^3-3ab-b^3\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)-3ab\)
Thay a - b = 1 vào, ta được
\(=1.\left(a^2+ab+b^2\right)-3ab\)
\(=a^2+ab+b^2-3ab\)
\(=a^2-2ab+b^2\)
\(=\left(a-b\right)^2\)
\(=1\)
Bài 2:
c) Ta có:
\(x+\dfrac{1}{x}=2\)
\(\Rightarrow\left(x+\dfrac{1}{x}\right)^2=4\)
\(\Rightarrow x^2+2.x.\dfrac{1}{x}+\dfrac{1}{x^2}=4\)
\(\Rightarrow x^2+\dfrac{1}{x^2}+2=4\)
\(\Rightarrow x^2+\dfrac{1}{x^2}=4-2=2\)
1.
Biến đổi vế trái , ta có :
(x+y)2 -( x-y)2
= x2 +2xy+ y2 - ( x2 -2xy +y2)
= x2+2xy+y2-x2+2xy-y2
= 4xy
=> đpcm