LG
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
20 tháng 5 2018
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
9 tháng 12 2018
a) \(2x\left(x-3\right)^2+5x\left(3-x\right)\)
\(=2x\left(x-3\right)^2-5x\left(x-3\right)\)
\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)
\(=\left(x-3\right)\left(2x^2-6x-5x\right)\)
\(=\left(x-3\right)\left(2x^2-11x\right)\)
\(=x\left(x-3\right)\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)
\(=\left(x+3\right)^2-2^2\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2\left(y-1\right)\right]^2\)
\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)
\(=\left(x+3-2y+2\right)\left(x+3+2y-2\right)\)
\(=\left(x-2y+5\right)\left(x+2y+1\right)\)
N
9 tháng 12 2018
a) \(2x.\left(x-3\right)^2+5x.\left(-x+3\right)=2x.\left(x-3\right)^2-5x.\left(x-3\right)\)
\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)
17 tháng 6 2015
c, =(5x)^3 + (y^2)^ 3 = (5x+y^2)(25x^2 - 5xy^2 + y^4)
d, = (0,5.(a+1))^3-1^3 = ( 0,5(a+1) - 1 ) ( 0,25(a+1) ^2 +a,5(a+1) + 1)
e,2x( x+ 1 ) + 2(x+ 1 ) = 2(x+1)(x+1) = 2(x+1)^2
g, y^2 (x^2 + y) - zx^2 - zy = x^2.y^2 - z.x^2 + y^3 - zy = x^2 (y^2 - z) + y (y^2 -z) = (x^2 +y) (y^2 -z)
h,4.x(x-2y) + 8.y(2y -x) = 4x( x- 2 y ) -8 (x - 2y) = (4x - 8) (x-2y)=4(x-2)(x-2y)
k,=(x+1)(3x(x+1)-5x+7) =(x+1) (3x^2 +3x - 5x + 7)
VP
0
HT
1 tháng 9 2020
B1:
a) \(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5\left(x^2-4x^2y^2+y^2\right)\)
b) \(=2\left(x^8-16\right)=2\left(x^4-4\right)\left(x^4+4\right)=2\left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\)
HT
1 tháng 9 2020
B2:
a) Đặt \(x^2-3x+1=y\)
=> \(y^2-12y+27\)
\(=\left(y^2-12y+36\right)-9\)
\(=\left(y-6\right)^2-3^2\)
\(=\left(y-9\right)\left(y-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x+1\right)\left(x-4\right)\left(x^2-3x-10\right)\)
b) Đặt \(x^2+7x+11=t\)
Ta có: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
S
0
5 tháng 9 2020
a) x3 - 1 + 5x2 - 5 + 3x - 3
= x3 + 5x2 + 3x - 9
= x3 + 6x2 - x2 + 9x - 6x - 9
= ( x3 + 6x2 + 9x ) - ( x2 + 6x + 9 )
= x( x2 + 6x + 9 ) - ( x2 + 6x + 9 )
= ( x2 + 6x + 9 )( x - 1 )
= ( x + 3 )2( x - 1 )
b) a5 + a4 + a3 + a2 + a + 1
= ( a5 + a4 + a3 ) + ( a2 + a + 1 )
= a3( a2 + a + 1 ) + 1( a2 + a + 1 )
= ( a2 + a + 1 )( a3 + 1 )
= ( a2 + a + 1 )( a + 1 )( a2 - a + 1 )
c) x3 - 3x2 + 3x - 1 - y3
= ( x3 - 3x2 + 3x - 1 ) - y3
= ( x - 1 )3 - y3
= ( x - 1 - y )[ ( x - 1 )2 + ( x - 1 )y + y2 ]
= ( x - 1 - y )( x2 - 2x + 1 + xy - y + y2 )
d) 5x3 - 3x2y - 45xy2 + 27y3
= ( 5x3 - 45xy2 ) - ( 3x2y - 27y3 )
= 5x( x2 - 9y2 ) - 3y( x2 - 9y2 )
= ( 5x - 3y )( x2 - 9y2 )
= ( 5x - 3y )[ x2 - ( 3y )2 ]
= ( 5x - 3y )( x - 3y )( x + 3y )
PV
1
TN
25 tháng 9 2016
a, x3+x+2
=x3-x2+2x+x2-x+2
=x(x2-x+2)+(x2-x+2)
=(x+1)(x2-x+2)
b, x3-2x-1
=x3-x2-x+x2-x-1
=x(x2-x-1)+(x2-x-1)
=(x+1)(x2-x-1)
c, x3+3x2-4
=x2(x+3)-4
=(x-1)(x+2)2
d, x3+3x2y-9xy2+5y3
=(x3-3x2y+3xy2-y3)+(6y3-12xy2+6x2y)
=(x-y)3+6y(x-y)2
=(x-y)2(x+5y)
VX
3
19 tháng 6 2016
a) \(x^3-3x^2-3x+1=x^3-3x-\left(3x^2-1\right)\)
\(=x\left(x^2-3\right)-\left(\sqrt{3x^2}-1\right)\left(\sqrt{3x^2}+1\right)\)
a/\(x^3-3x^2+3x-1=\left(x-1\right)^3\)
b/\(1-x^2y^4=1^2-\left(xy^2\right)^2=\left(1-xy^2\right)\left(1+xy^2\right)\)
c/\(8-125x^3=2^3-\left(5x\right)^3=\left(2-5x\right)\left(4+10x+25x^2\right)\)