Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(x-2\right)\left(x^2+2x+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+7+2x+4-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+6\right)=0\)
\(\Leftrightarrow x-2=0\) (Vì: \(x^2+4x+6>0\) )
\(\Leftrightarrow x=2\)
b) \(2x^3+x^2-6x=0\)
\(\Leftrightarrow x\left(2x^2+x-6\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2+4x\right)-\left(3x+6\right)\right]=0\)
\(\Leftrightarrow x\left[2x\left(x+2\right)-3\left(x+2\right)\right]=0\)
\(\Leftrightarrow x\left(x+2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+2=0\\2x-3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\\x=\frac{3}{2}\end{array}\right.\)
c) \(4x^2+4xy+x^2-2x+1+y^2=0\)
\(\Leftrightarrow\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x+y\right)^2+\left(x-1\right)^2=0\)
\(\Leftrightarrow\begin{cases}2x+y=0\\x-1=0\end{cases}\)\(\Leftrightarrow\begin{cases}y=-2\\x=1\end{cases}\)
\(1.5x\left(x^2+2x-1\right)-3x^2\left(x-2\right)=5x^3+10x^2-5x-3x^3+6x^2\)
\(=2x^3+16x^2-5x\)
\(=\left(2x^3-x\right)+\left(16x^2-4x\right)\)
\(=x\left(2x^2-1\right)+4x\left(4x-1\right)\left(ĐCCM\right)\)
Ta có : x3 - 2x2 - 4xy2 + x
= x(x2 - 2x + 1 - 4y2)
= x[(x - 1)2 - 4y2]
= x(x - 1 - 2y)(x - 1+ 2y)
x3 - 2x2 - 4xy2 + x
= x ( x2 - 2x - 4y2 + 1 )
= x [ ( x2 - 2x + 1 ) - 4y2 ]
= x [ ( x - 1 )2 - 4y2 ]
= x ( x - 1 + 2y ) ( x - 1 + 2y )
........
a: =2x^3-14x^2-6x
b: =3x2-5x-9x+15
=3x^2-14x+15
c: =-8x^4y^2+4xy^4-28x^2y^3
d: =x^3-4x^2-2x^2+8x+3x-12
=x^3-6x^2+11x-12
a) = (x-3)(x+3) +(x-3((x-3)
= (x-3)(x+3+x-3)
= 2x(x-3)
làm cho 1 câu thui
Ta có:
a) 6x2y - 3y2 - 2x2 + y = (6x2y - 2x2) - (3y2 - y) = 2x2(3y - 1) - y(3y - 1) = (2x2 - y)(3y - 1)
b) 2x2 + x - 4xy - 2y + 2x + 1 = (x2 + x) - (4xy + 2y) + (x2 + 2x + 1) = x(x + 1) - 2y(2x + 1) + (x + 1)2
= (x + x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1) - 2y(2x + 1) = (2x + 1)(x + 1 - 2y)
c) 16x2y - 4xy2 - 4x3 + x2y = 4xy(4x - y) - x2(4x - y) = (4xy - x2)(4x - y)
d) 4x2 - 20x + 25 - 36y2 = (2x - 5)2 - (6y)2 = (2x - 5 - 6y)(2x - 5 + 6y)
e) x2 - 4y2 + 6x - 4y + 8 = (x2 + 6x + 9) - (4y2 + 4y + 1) = (x + 3)2 - (2y + 1)2 = (x + 3 - 2y - 1)(x + 3 + 2y + 1) = (x + 2 - 2y)(x + 4 + 2y)
g) Ta có : x10 + x5 + 1
= (x10 - x) + (x5 - x2) + (x2 + x + 1)
= x(x9 - 1) + x2(x3 - 1) + (x2 + x + 1)
= x(x3 - 1)(x6 + x3 + 1) + x2(x3 - 1) + (x2 + x + 1)
= (x7 + x4 + x)(x - 1)(x2 + x + 1) + x2(x - 1)(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x8 - x7 + x 5 - x4 + x2 - x + x4 + x3 + x2 + 1)
= (x2 + x + 1)(x8 - x7 + x5 + x3 - x + 1)
h) TT trên (dài dòng ktl)
a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
\(x^3+2x^2+x-4xy^2\)
\(=x\left(x^2+2x+1\right)-4xy^2\)
\(=x\left(x+1\right)^2-4xy^2\)
\(=x\left(\left(x+1\right)^2-4y^2\right)\)
\(=x\left(\left(x+1-2y\right)\left(x+1+2y\right)\right)\)
\(\text{x3+2x2+x−4xy2 =x(x2+2x+1)−4xy2 =x(x+1)2−4xy2 =x((x+1)2−4y2) =x((x+1−2y)(x+1+2y))}\)