1. x² - 6x + 9=(x-3)²

2. 25 +  10x + x²=(x+5)²

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

4.\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)

5.x³ + 8y³=(x+8y)(x²-8xy+64y²)

6.8y³ -125=(2y-5)(4y²+10y+25)

7.a⁶-b³=(a²-b)(a⁴+a²b+b²)

8 x² - 10x + 25=(x-2)²

1) \(x^2-6x+9=\left(x-3\right)^2\)

2) \(25+10x+x^2=\left(5+x\right)^2\)

3) \(\dfrac{1}{4}a^2+2ab+4b^4=\left(\dfrac{1}{2}a+2b^2\right)^2\)

4) \(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}-y^4\right)^2\)

5) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

6) \(8y^3-125=\left(2y-5\right)\left(4y^2+10y+25\right)\)

7) \(a^6-b^3=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)

8) \(x^2-10x+25=\left(x-5\right)^2\)

9) \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)

b) 8a3 + 4a2b - 2ab2 – b3 = (8a3 – b3 ) + (4a2b - 2ab2 )

= (2a – b)(4a2 + 2ab + b2) + 2ab(2a – b)

= (2a – b)( 4a2 + 2ab + b2 + 2ab) = (2a – b)(2a + b)2

\(x^2-6x+7=x^2-6x+9-2\\ =\left(x-3\right)^2-2=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\\ x^4+64=x^4+16x^2+64-16x^2\\ =\left(x^2+8\right)^2-16x^2=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\\ a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2\\ =\left(a^2+2b^2\right)^2-4a^2b^2\\ =\left(a^2-2ab+2b^2\right)\left(a^2+2ab+2b^2\right)\)

= x² -7x -x +7

= x. (x-7) - (x-7)

= (x-1)(x-7)

Chúc bạn học tốt nha!

\(\dfrac{1}{9}-\dfrac{2}{3}y^4+y^8=\left(\dfrac{1}{3}\right)^2-2.\dfrac{1}{3}.y^4+\left(y^4\right)^2=\left(\dfrac{1}{3}+y^4\right)^2\)

\(y^8-\dfrac{2}{3}y^4+\dfrac{1}{9}=\left(y^4-\dfrac{1}{3}\right)^2\)

a2 – b2 – 4a + 4

= a2 – 4a + 4 – b2

= (a – 2)2 – b2

= (a – 2 + b)(a – 2 – b)

= (a + b – 2)(a – b – 2)

a, 4\(x^3\).y + \(\dfrac{1}{2}\)yz

  =y.(4\(x^3\) + \(\dfrac{1}{2}\)z)

b, (a² + b² - 5)² - 2.(ab + 2)²

 = [a² + b² - 5  - \(\sqrt{2}\)(ab + 2) ].[ a² + b² - 5 + \(\sqrt{2}\)(ab +2)]

a) \(4x^3y+\dfrac{1}{2}yz=y\left(4x^3+\dfrac{1}{2}z\right)\)

b) \(\left(a^2+b^2-5\right)^2-2.\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[a^2+b^2+2ab-1\right]\left[a^2+b^2-2ab-9\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

\(=\left[\left(a+b+1\right)\left(a+b-1\right)\right]\left[\left(a-b+3\right)\left(a-b-3\right)\right]\)

Mình đã làm sai, cô Hoài Thương đúng rồi.

(5+x)<2

Năm cộng x tất cả mũ 2

(5+x)²

\(25+10x+x^2\\ =x^2+10x+25\\ =x^2+2.5.x+5^2\\ =\left(x+5\right)^2\)

\(a,=\left(a-5\right)^2-4b^2=\left(a-2b-5\right)\left(a+2b-5\right)\\ b,=ax^2+a-a^2x-x=ax\left(a-x\right)+\left(a-x\right)=\left(ax+1\right)\left(a-x\right)\)

a: \(=\left(a-5-2b\right)\left(a-5+2b\right)\)

b: \(ax^2+a-a^2x-x\)

\(=ax\left(x-a\right)-\left(x-a\right)\)

\(=\left(x-a\right)\left(ax-1\right)\)

a) \(a^2+ab-7a-7b=a\left(a+b\right)-7\left(a+b\right)=\left(a+b\right)\left(a-7\right)\)

b) \(5ab+4c+20b+ac=5b\left(a+4\right)+c\left(a+4\right)=\left(a+4\right)\left(5b+c\right)\)

c) \(a^2+6a-b^2+9=\left(a+3\right)^2-b^2=\left(a+b-b\right)\left(a+3+b\right)\)

d) \(a^2-16=\left(a-4\right)\left(a+4\right)\)