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

2. \(x^2+5x+\dfrac{25}{4}=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)

3. \(16x^2-8x+1=\left(4x-1\right)^2\)

4. \(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x-y+1\right)\left(x+y\right)\)

5. \(\dfrac{1}{4}x^2-\dfrac{4}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)\left(\dfrac{1}{2}x+\dfrac{2}{3}y\right)\)

6. \(a^2-2ab+b^2-x^2=\left(a-b\right)^2-x^2=\left(a-b-x\right)\left(a-b+x\right)\)

7. \(4x^2-20x+25-y^2=\left(2x-5\right)^2-y^2=\left(2x-5-y\right)\left(2x-5+y\right)\)

Kết quả nè

1.(x^2-1)^2×(x^2+1)^2

2.(2x-5)^2÷4

3.(4x-1)^2

4.(x+y)^2×(x-y)

5.(1/4x-4/9y)×(1/4x+1/9y)

4. 4x² + 4x + 1 = ( 2x + 1)²

5. \(\dfrac{1}{4}x-\dfrac{2}{3}xy+\dfrac{4}{9}y^2\) \(=\left(\dfrac{1}{2}x\right)^2-2.\dfrac{1}{2}x.\dfrac{2}{3}+\left(\dfrac{2}{3}y\right)^2\)

\(=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)^2\)

6. \(4a^2-\dfrac{4}{3}ab+\dfrac{1}{9}b^2=\left(2a\right)^2-2.2a.\dfrac{1}{3}+\left(\dfrac{1}{3}b\right)^2=\left(2a-\dfrac{1}{3}b\right)^2\)

7.

\(9x^2+4xy+\dfrac{4}{9}y^2-25z^2=\left(3x+\dfrac{2}{3}y\right)^2-\left(5z\right)^2=\left(3x+\dfrac{2}{3}y-5z\right)\left(3x+\dfrac{2}{3}y+5z\right)\)

x² - 6x + 9 = (x-3)²

x² - 2x + 1 - y² = ( x-1)² - y² = ( x-1-y)(x-1+y)

4x² +12xy +9y² = ( 2x+3y)²

\(4.4x^2+4x+1\)

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

\(5.\dfrac{1}{4}x^2-\dfrac{2}{3}xy+\dfrac{4}{9}y^2\)

\(=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)^2\)

\(6.4a^2-\dfrac{4}{3}ab+\dfrac{1}{9}b^2\)

\(=\left(2a-\dfrac{1}{3}b\right)^2\)

\(1,\)

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

\(=\left(x-3y\right)\left(x+3y\right)-4\left(x+3y\right)\)

\(=\left(x+3y\right)\left(x-3y-4\right)\)

\(3,\)

\(-x^2+2xy-y^2+25\)

\(=-\left(x^2-2xy+y^2\right)+25\)

\(=25-\left(x-y\right)^2\)

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

\(=\left(5-x+y\right)\left(5+x-y\right)\)

\(5,\)

\(x^3-6x^2+9\)

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

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

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

Dấu '=' xảy ra khi x=-1

2: \(=-\left(4x^2-12x-10\right)\)

\(=-\left(4x^2-12x+9-19\right)\)

\(=-\left(2x-3\right)^2+19< =19\)

Dấu '=' xảy ra khi x=3/2

3: \(=-\left(x^2+4x+4-4\right)=-\left(x+2\right)^2+4< =4\)

Dấu '=' xảy ra khi x=-2

giúp với trí ơi

giúp nốt trí ui

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

\(=\left(2x+y\right)\left(2x-y\right)-\left(2x+y\right)=\left(2x+y\right)\left(2x-y-1\right)\)

b) \(9x^2-25y^2-6x+10y=\left(3x-5y\right)\left(3x-5y\right)-2\left(3x-5y\right)\)

\(=\left(3x-5y\right)\left(3x+5y-2\right)\)

c) \(x^3-2x^2+2x-1=x^3-1-2x^2+2x\)

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

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

d) \(x^4+2x^3-4x-4=x^4-4+2x^3-4x\)

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

6, x mũ 4 - 4x mũ 3 - 8x mũ 2 + 8x =x (x+2) (x^2-6x+4)

8, x mũ 4 + 2x mũ 3 + x mũ 2 - y mũ 2  = -(y-x^2-x) (y+x^2+x)

10, 4x mũ 2 ( x + y ) -x - y  = (2x-1) (2x+1) (y+x)

a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)

b, \(x^2-4x+4-y^2=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2\)

\(=\left(x-2-y\right)\left(x-2+y\right)\)

c, \(3x^2-2x-5=3x^2-5x+3x-5=x\left(3x-5\right)+3x-5\)

\(=\left(3x-5\right)\left(x+1\right)\)