1) ( x - y)2 - ( x + y)2 = -4xy
\(\Leftrightarrow\)( x - y - x + y ) ( x - y + x + y ) = -4xy
\(\Leftrightarrow\)2x + 4xy = 0
\(\Leftrightarrow\)2x ( 1 + 2y ) = 0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=0\\1+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}0\\-\dfrac{1}{2}\end{matrix}\right.\)
2) ( 7n -2)2 - ( 2n - 7)2
= ( 7n - 2 - 2n - 7 )( 7n - 2 + 2n - 7 )
= ( 5n - 9 )( 9n - 9 )
Ta có: 9n \(⋮\) 9 với mọi n
9 \(⋮\) 9 với mọi n
\(\Rightarrow\)9n - 9 \(⋮\) 9 với mọi n
\(\Rightarrow\) đpcm
3) F = x2 + 6x + 1
F = x2 + 2.x.3 + 9 - 8
F = ( x + 3 )2 - 8
Vì ( x + 3)2 \(\ge\) 0 với mọi x
\(\Rightarrow\) ( x + 3 )2 - 8 \(\ge\) -8 với mọi x
\(\Rightarrow\) F \(\ge\) -8 với mọi x
Vậy min F = -8 \(\Leftrightarrow\) ( x + 3 )2 = 0
\(\Leftrightarrow\) x = -3
1. Ta có: \(\left(x-y\right)^2-\left(x+y\right)^2=\left(x-y+x+y\right)\left(x-y-x-y\right)=2x.\left(-2y\right)=-4xy\)
2. Ta có: \(\left(7n-2\right)^2-\left(2n-7\right)^2=\left(7n-2-2n+7\right)\left(7n-2+2n-7\right)=\left(5n+5\right)\left(9n-9\right)=9\left(n-1\right)\left(5n+5\right)\)
\(\Rightarrow\left(7n-2\right)^2-\left(2n-7\right)^2\) chia hết cho 9 với mọi giá trị nguyên của n.
3. Ta có: \(F=-x^2+6x+1=-\left(x^2-6x-1\right)=-\left(x^2-6x+9-10\right)=-\left(x-3\right)^2+10\)
Vì \(-\left(x-3\right)^2\le0\Rightarrow-\left(x-3\right)^2+10\le10\)
=> MaxF=10 <=> \(-\left(x-3\right)^2+10=10\Leftrightarrow-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy MaxF=10 khi x=3.
4. Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2-a^2x^2-2abxy-b^2y^2=0\Leftrightarrow a^2y^2+b^2x^2-2abxy=0\Leftrightarrow\left(ay-bx\right)^2=0\Leftrightarrow ay-bx=0\)
=> đpcm.
