Question

Câu 1 : Phân tích thành nhân tử

A= (x²+x)² + 2(x²+x) +1

B= ( x-a)⁴-(x+a)⁴

C= (x²-10x+25)-4y²

D= 49(y-4)² -9(y+2)²

E= x²-6xy+9y²

F=x³+6x²y+12xy²+8y³

G= x³-64

H= 125x³+y⁶

Câu 2 : Tìm x biết

a) x³-3x²+3x-1=0

b) 4x³-36x=0

c) x⁶-1=0

d) x³-6x²+12x-8

Mn giúp mình với ạ !

Answer 1

a) x³-3x²+3x-1=0

⇔ ( x - 1 )\(^3\) = 0

⇔ x - 1 = 0

⇔ x = 1

b) 4x³-36x=0

⇔ 4x ( x\(^2\) - 9 ) = 0

⇔ 4x ( x - 3 ) ( x + 3 ) = 0

\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)

c) x⁶-1=0

⇔ x\(^6\) = 1

⇔ x = \(\pm\)1

d) x³-6x²+12x-8 = 0

⇔ ( x - 2 )\(^3\) = 0

⇔ x - 2 = 0

⇔ x = 2

Answer 2

C= (x²-10x+25)-4y²

= ( x - 5 )\(^2\) - 4y\(^2\) = ( x - 5 - 4y ) ( x - 5 + 4y )

E= x²-6xy+9y² = ( x - 3y )\(^2\)

F=x³+6x²y+12xy²+8y³ = ( x + 2 )\(^3\)

G= x³-64 = ( x - 4 ) ( x\(^2\) + 4x +16 )

H= 125x³+y⁶ = ( 5x )\(^3\) + ( y\(^2\) )\(^3\) = ( 5x + y\(^2\) ) ( 25x\(^2\) - 5xy\(^2\) + y\(^4\) )

Answer 3

Mình cần gấp lắm ạ !!! Help mee