$a^4-8\\=(a^2-\sqrt 8)(a^2+\sqrt 8)$

√

\(a,4x^2-y^2+2y-1\)

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

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

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

a) \(19+8\sqrt{3}=3+2\sqrt{3}\cdot4+16=\left(\sqrt{3}+4\right)^2\)

b) \(11-4\sqrt{6}=3-2\sqrt{3}\cdot2\sqrt{2}+8=\left(\sqrt{3}-2\sqrt{2}\right)^2\)

c) \(9-4\sqrt{2}=8-2\cdot2\sqrt{2}+1=\left(2\sqrt{2}-1\right)^2\)

d) \(21+6\sqrt{10}=18+2\cdot3\sqrt{2}\cdot\sqrt{5}+5-2=\left(3\sqrt{2}+\sqrt{5}\right)^2-\left(\sqrt{2}\right)^2\)

e) \(23+6\sqrt{10}=18+2\cdot3\sqrt{2}\cdot\sqrt{5}+5=\left(3\sqrt{2}+\sqrt{5}\right)^2\)

f) \(49-20\sqrt{6}=\left(5\sqrt{2}\right)^2-2\cdot5\sqrt{2}\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-13=\left(5\sqrt{2}-2\sqrt{3}\right)^2-\left(\sqrt{13}\right)^2\)

nó dễ ợt mà -_- 

x²y²+4xy+4=(xy+2)² xong :)) 

`1)(a^[1/4]-b^[1/4])(a^[1/4]+b^[1/4])(a^[1/2]+b^[1/2])`

`=[(a^[1/4])^2-(b^[1/4])^2](a^[1/2]+b^[1/2])`

`=(a^[1/2]-b^[1/2])(a^[1/2]+b^[1/2])`

`=a-b`

`2)(a^[1/3]-b^[2/3])(a^[2/3]+a^[1/3]b^[2/3]+b^[4/3])`

`=(a^[1/3]-b^[2/3])[(a^[1/3])^2+a^[1/3]b^[2/3]+(b^[2/3])^2]`

`=(a^[1/3])^3-(b^[2/3])^3`

`=a-b^2`

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

cam on ban

a)8x3 + * + * + 27y3 = (* + *)3

=>A=(2x+3y)^3

b) (2x+1)^3

c)(x-2y)^3

d)(3x-2)(3x+2)

e)(3x-1)(9x^2+3x+1)

f)....................

6: \(27x^3+1=\left(3x+1\right)\left(9x^2-3x+1\right)\)

7: \(\left(2x+1\right)^2=4x^2+4x+1\)

8: \(\left(2x-1\right)^2=4x^2-4x+1\)

9: \(9-16x^2=\left(3-4x\right)\left(3+4x\right)\)

nhân tung

(xy² - 1/2)(2 + 4xy²)

= 4(xy² - 1/2)(xy² + 1/2)

= 4[(xy²)² - (1/2)²]

= 4(x²y⁴ - 1/4)