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B1:
a) \(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5\left(x^2-4x^2y^2+y^2\right)\)
b) \(=2\left(x^8-16\right)=2\left(x^4-4\right)\left(x^4+4\right)=2\left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\)
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: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
1/a ) = (x+y)3 -(x+y)
= (x+y)[(x+y)2+1]
c) = 5(x2-xy+y2)-20z2
=5(x-y)2-20z2
= 5 [ (x-y)2- 4z2 ]
=5(x-y-4z)(x-y+4z)
Bài 1:
a) x3-x+3x2y+3xy2+y3-y
=x3+2x2y-x2+xy2-xy+x2y+2xy2-xy+y3-y2+x2+2xy-x+y2-y
=x(x2+2xy-x+y2-y)+y(x2+2xy-x+y2-y)+(x2+2xy-x+y2-y)
=(x2+2xy-x+y2-y)(x+y+1)
=[x(x+y-1)+y(x+y-1)](x+y+1)
=(x+y-1)(x+y)(x+y+1)
c) 5x2-10xy+5y2-20z2
=-5(2xy-y2+4z2-2)
Bài 2:
5x(x-1)=x-1
=>5x2-6x+1=0
=>5x2-x-5x+1
=>x(5x-1)-(5x-1)
=>(x-1)(5x-1)=0
=>x=1 hoặc x=1/5
b) 2(x+5)-x2-5x=0
=>2(x+5)-x(x+5)=0
=>(2-x)(x+5)=0
=>x=2 hoặc x=-5
a,\(\frac{1}{5}x^2y\left(15xy^2-5y+3xy\right)=3x^3y^3-x^2y^2+\frac{3}{5}x^3y^2\)
b,\(5x^3-5x=5x\left(x^2-1\right)=5x\left(x-1\right)\left(x+1\right)\)
c, \(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(3x-5\right)\left(x-y\right)\)
1) 1/5x2y( 15xy2 - 5y + 3xy ) = 3x3y3 - x2y2 + 3/5x3y2
2) a) 5x3 - 5x = 5x( x2 - 1 ) = 5x( x2 - 12 ) = 5x( x - 1 )( x + 1 )
b) 3x2 + 5y - 3xy - 5x = ( 3x2 - 3xy ) + ( 5y - 5x )
= 3x( x - y ) + 5( y - x )
= 3x( x - y ) + 5[ -( x - y ) ]
= 3x( x - y ) - 5( x - y )
= ( 3x - 5 )( x - y )
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)
\(=\left[\left(x+y\right)^3-1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+1+2\left(x+y\right)\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+y^2+2xy+1+2x+2y-3xy\right)\)
\(=\left(x+y+1\right)\left(x^2+y^2-xy+1+2x+2y\right)\)
\(=\left(x+y-1\right)\left[\left(x^2+1+2x\right)\left(y^2-xy+2y\right)\right]\)
\(=\left(x+y-1\right)\left(x+1\right)^2\left(y-x+2\right)y\)
Mấy câu trên dễ
\(M=4a^2-6a+12\)
\(M=\left(2a\right)^2-2\cdot2a\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2+\frac{39}{4}\)
\(M=\left(2a-\frac{3}{2}\right)^2+\frac{39}{4}\ge\frac{39}{4}\forall x\left(đpcm\right)\)
1. a) 2x2y - 3xy2 - 6x + 9y = 2x( xy - 3 ) - 3y ( xy - 3) = ( 2x - 3y)(xy - 3)
b) x2 - 2x + 8 = x2 - 2x + 12 - 1 + 9 = ( x - 1 )2 + 32 ( xem lại đề bài )
2. a) ( 2x - 1) 2 - (2x-1)(2x+3) = 5
(2x-1)(2x-1-2x-3) = 5
-4(2x-1) = 5
2x - 1 = -1,25
2x = -0,25
x= -0,125
b) x(x-9 ) = 0
x= 0 hoặc x = 9
c, ko hiểu
3, M = (2a)2 - 2.2a.1,5 + ( 1,5)2 + 9,75
M= ( 2a - 1,5)2 + 9,75
Vì ( 2a - 1,5 )2 \(\ge\)0 \(\forall x\)
\(\Rightarrow\)( 2a - 1,5)2 + 9,75 \(\ge9,75\forall x\)
Vậy biểu thức trên luôn dương
b: \(=x\left(x^4-y^4\right)+y\left(x^4-y^4\right)\)
\(=\left(x+y\right)\left(x^4-y^4\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)\left(x+y\right)^2\cdot\left(x-y\right)\)
a) x3 + y3 - 3xy + 1
= ( x + y )3 - 3xy( x + y ) - 3xy + 1
= [ ( x + y )3 + 1 ] - [ 3xy( x + y ) + 3xy ]
= ( x + y + 1 )( x2 + 2xy + y2 - x - y + 1 ) - 3xy( x + y + 1 )
= ( x + y + 1 )( x2 - xy + y2 - x - y + 1 )
b) ( 4 - x )5 + ( x - 2 )5 - 32
= [ -( x - 4 ) ]5 + ( x - 2 )5 - 32
Đặt t = x - 3
đthức <=> ( 1 - t )5 + ( 1 + t )5 - 32 ( chỗ này bạn dùng nhị thức Newton để khai triển nhé )
= 10t4 + 20t2 - 30
Đặt y = t2
đthức = 10y2 + 20y - 30
= 10y2 - 10y + 30y - 30
= 10y( y - 1 ) + 30( y - 1 )
= 10( y - 1 )( y + 3 )
= 10( t2 - 1 )( t2 + 3 )
= 10( t - 1 )( t + 1 )( t2 + 3 )
= 10( x - 3 - 1 )( x - 3 + 1 )[ ( x - 3 )2 + 3 ]
= 10( x - 4 )( x - 2 )( x2 - 6x + 12 )
a,\(x^3+y^3-3xy+1\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)+1-3x^2y-3xy^2-3xy\)
\(=\left[\left(x+y\right)^3+1\right]-3xy\left(x+y+1\right)\)
\(=\left(x+y+1\right)\left[\left(x+y\right)^2-\left(x+y\right)+1\right]-3xy\left(x+y+1\right)\)
\(=\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1-3xy\right)\)
\(=\left(x+y+1\right)\left(x^2+y^2-xy-x-y+1\right)\)