\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow\) \(\dfrac{a}{x}=\dfrac{b}{y}\)

\(\Leftrightarrow ay=bx\)

\(\Leftrightarrow ay-bx=0\)

( Bất đẳng thức Bu - nhi - a - cốp - xki )

Áp dụng BĐT Bunhiacopxki :

\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)

Dấu đẳng thức xảy ra \(\Leftrightarrow\frac{a}{x}=\frac{b}{y}\)

\(\Leftrightarrow ay=bx\)

\(\Leftrightarrow ay-bx=0\)

Ta có đpcm.

Ta có:

VT = (x² + y²)(a² + b²)

= x²a² + x²b² + y²a² + y²b²

= (a²x² + b²y² + 2axby) + (a²y² - 2aybx + b²x²)

= (ax + by)² + (ay - bx)²

=> VT = VP => đpcm

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=x^2\left(a^2+b^2\right)+y^2\left(a^2+b^2\right)\)

\(=a^2x^2+b^2x^2+a^2y^2+b^2y^2\)

\(\left(ax+by\right)^2=a^2x^2+2abxy+b^2y^2\)

\(\Rightarrow\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+b^2x^2+a^2y^2+b^2y^2=a^2y^2+2abxy+b^2y^2\)

\(\Leftrightarrow a^2x^2+b^2x^2=2abxy\)

\(\Leftrightarrow a^2x^2+b^2x^2-2abxy=0\)

\(\Leftrightarrow\left(ax-bx\right)^2=0\)

\(\Leftrightarrow ax-bx=0\left(đpcm\right)\)

a) Ta có: \(\left(a+b\right)^2=4ab\)<=> \(a^2+b^2+2ab=4ab\)

                                               <=> \(a^2-2ab+b^2=0\)

                                                <=> \(\left(a-b\right)^2=0\)=> a=b (đpcm)

b) Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

<=> \(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)

<=> \(a^2y^2+b^2x^2-2axby=0\)

<=>\(\left(ay-bx\right)^2=0\)

<=>ay=bx(đpcm)

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 \(-\left(x-3\right)^2\le0\Rightarrow-\left(x-3\right)^2+10\le10\)

=> Max_F=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 Max_F=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.

Bạn viết đề sai tứ tung luôn :v

Điều cần phải chứng minh:

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(ay+bx\right)^2\)

\(VT=\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(VP=\left(ax-by\right)^2+\left(ay+bx\right)^2\)

\(=a^2x^2-2axby+b^2y^2+a^2y^2+2axby+b^2x^2\)

\(=a^2x^2+b^2y^2+a^2y^2+b^2x^2\)

\(VT=VP\rightarrowđpcm\)