C1: \(\left(x+y\right)\left(x-y\right)=x\left(x-y\right)+y\left(x-y\right)=x^2-xy+xy-y^2=x^2-y^2\)

C2: x²-y²=(x-y)(x+y)

  <=> x²-y²-(x-y)(x+y)=0

   <=> x²-y²-[x(x+y)-y(x+y)] = 0

   <=> x²-y²-(x²+xy-xy-y²) = 0

    <=> x²-y²-(x²-y²) = 0

    <=> x²-y²-x²+y² = 0

    <=> 0 =0 (đúng)

Vậy .....

x^2 - y^2 = ( x + y )( x - y )

Co ( x + y )( x - y ) = x^2 - xy + xy - y^2 = x^2 - y^2

Ma x^2 - y^2 = x^2 - y^2

=> x^2 - y^2 = ( x + y )( x - y ) 

\(3x^2-2x.\left(5+1,5x\right)+10\)

\(=3x^2-2x.5-2x.1,5x+10\)

\(=3x^2-3x^2-10x+10\)

\(=10-10x\)

\(=10.\left(1-x\right)\)

cam on ban nha

Đặt a=x+y ; b=y+t ; c=t+x

Khi P=Q tức là: a²+b²+c²=ab+bc+ac

                 <=> 2(a²+b²+c²)=2(ab+bc+ac)

                 <=> 2a²+2b²+2c²=2ab+2bc+2ac

                 <=> (a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²) = 0

                  <=> (a-b)²+(b-c)²+(c-a)² = 0

Dấu "=" xảy ra <=> a=b=c (đpcm)

Vậy .....

a) \(3x+1=7x-11\)

\(\Leftrightarrow3x-7x=-11-1\)

\(\Leftrightarrow-4x=-12\Leftrightarrow x=3\)

b) \(5-3x=6x+7\)

\(\Leftrightarrow-3x-6x=7-5\)

\(\Leftrightarrow-9x=2\Leftrightarrow x=-\frac{2}{9}\)

(3x+1)\(^2\)-(x+1)\(^2\) =(3x+1+x+1)(3x+1-x-1) = 2x(4x+2) =4x(2x+1)