\(x^2+2\left(x+1\right)^2+3\left(x-2\right)^2+4\left(x+3\right)^2\)

\(=x^2+2\left(x^2+2x+1\right)+3\left(x^2-4x+4\right)+4\left(x^2+6x+9\right)\)

\(=x^2+2x^2+4x+2+3x^2-12x+12+4x^2+24x+36\)

\(=10x^2+16x+50\)

dưới dạng tổng các bình phương mà

Câu 7:

a: \(A=x^2-2x+7\)

\(=x^2-2x+1+6\)

\(=\left(x-1\right)^2+6\ge6\forall x\)

Dấu '=' xảy ra khi x-1=0

=>x=1

b: \(B=5x^2-20x\)

\(=5\left(x^2-4x\right)\)

\(=5\left(x^2-4x+4-4\right)=5\left(x-2\right)^2-20\ge-20\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

Câu 4:

a: \(A=\left(x-y\right)^2+\left(x+y\right)^2\)

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

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

b: \(B=\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+5-8x^2+24x-18=-4x^2+20x-13\)

Câu 2:

a: \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)

b: \(x^2+10x+25=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\)

c: \(\frac{x^2}{4}-x+1=\left(\frac12x\right)^2-2\cdot\frac12x\cdot1+1^2=\left(\frac12x-1\right)^2\)

d: \(9\left(x+1\right)^2-6\left(x+1\right)+1\)

\(=\left(3x+3\right)^2-2\cdot\left(3x+3\right)\cdot1+1^2\)

\(=\left(3x+3-1\right)^2=\left(3x+2\right)^2\)

e: \(\left(x-2y\right)^2-8\left(x^2-2xy\right)+16x^2\)

\(=\left(x-2y\right)^2-2\cdot\left(x-2y\right)\cdot4x+\left(4x\right)^2\)

\(=\left(x-2y-4x\right)^2=\left(-3x-2y\right)^2\)

a. (x + y)² = x² + 2xy + y²

b. (x - 2y)² = x² - 4xy - 4x²

c. (xy² + 1)(xy² - 1) = x²y⁴ - 1

d. (x + y)²(x - y)² = (x² + 2xy + y²)(x² - 2xy + y²) = x⁴ - (2xy + y²)² = x⁴ - (4x²y² + y⁴) = x⁴ - 4x²y² - y⁴

^{Chucs hocj toots}

Câu 2: 

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(x^2+10x+25=\left(x+5\right)^2\)

d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)

e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)

A=x^2+2(x^2+2x+1)+3(x^2+4x+4)+4(x^2+6x+9)

  =x^2+2x^2+4x+2+3x^2+12x+12+4x^2+24x+36

  =10x^2+40x+50

  =(9x^2+30x+25)+(x^2+10x+25)

  =(3x+5)^2+(x+5)^2

a=(x+3)

b=(x+1/2)

c=(xy^2+1)

Good luck!

\(a,\left(x+3\right)^2\)

\(b,\left(x+\frac{1}{2}\right)^2\)

\(c,\left(xy^2+1\right)^2\)

\(x^2+\left(x+1\right)^2+3\left(x+2\right)^2+4\left(x+3\right)^2\)

\(=x^2+x^2+1+3x^2+4+4x^2+9\)

\(=x^2+x^2+1+3x^2+3+4x^2+9+1\)

\(=2x^2+1+3x^2+3+4x^2+9+1\)

Từ đây ghép x vào rồi tính nốt đẳng thức thôi nhé

\(x^2+2\left(x+1\right)^2+3\left(x+1\right)^2+4\left(x+1\right)^2\)

\(=x^2+9\left(x+1\right)^2\)

\(=x^2+3^2.\left(x+1\right)^2\)

\(=x^2+\left(3x+3\right)^2\)