Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
\(a,3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(3-x\right),\left(x+4\right)\)
\(a,4x^4-8x^3+4x^2\)
\(=4x^2\cdot\left(x^2-2x+1\right)\)
\(=4x^2\cdot\left(x-1\right)^2\)
\(b,x^2-y^2+5\cdot\left(y-x\right)\)
\(=\left(x-y\right)\cdot\left(x+y\right)-5\cdot\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(x+y-5\right)\)
\(c,3x^2-6xy+3y^2-12z^2\)
\(=3\cdot\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\cdot\left[\left(x-y\right)^2-\left(2x\right)^2\right]\)
\(=3\cdot\left(x-y-2x\right)\cdot\left(x-y+2x\right)\)
e) \(\left(x-y\right)^2+4\left(x-y\right)-12\)
\(=\left(x-y\right)\left[\left(x-y\right)+4\right]-12\)
\(=\left(x-y\right)\left(x-y+4\right)-12\)
f) \(x^2-2xy+y^2+3x-3y-10\)
\(=\left(x^2-2xy+y^2\right)+\left(3x-3y\right)-10\)
\(=\left(x-y\right)^2+3\left(x-y\right)-10\)
\(=\left(x-y\right)\left[\left(x-y\right)+3\right]-10\)
\(=\left(x-y\right)\left(x-y+3\right)-10\)
g) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left[\left(x^2+4x+8\right)+3x\right]+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+4x+8+3x\right)+2x^2\)
\(=\left(x^2+4x+8\right)\left(x^2+7x+8\right)+2x^2\)
\(3x\left(x-5\right)-x\left(4+3x\right)=43\)
\(\Leftrightarrow3x^2-15x-4x-3x^2=43\)
\(\Leftrightarrow-19x=43\)
\(\Leftrightarrow x=\frac{-43}{19}\)
Áp dụng HĐT a2 - b2 = ( a - b )( a + b )
và tính chất an.bn = ( a.b )n ( với n ∈ N* )
a) ( 3x + 1 )2 - ( x + 1 )2
= [ ( 3x + 1 ) - ( x + 1 ) ][ ( 3x + 1 ) + ( x + 1 ) ]
= ( 3x + 1 - x - 1 )( 3x + 1 + x + 1 )
= 2x( 4x + 2 )
= 2x.2( 2x + 1 )
= 4x( 2x + 1 )
b) ( x + y )2 - ( x - y )2
= [ ( x + y ) - ( x - y ) ][ ( x + y ) + ( x - y ) ]
= ( x + y - x + y )( x + y + x - y )
= 2y.2x = 4xy
c) ( 2xy + 1 )2 - ( 2x + y )2
= [ ( 2xy + 1 ) - ( 2x + y ) ][ ( 2xy + 1 ) + ( 2x + y ) ]
= ( 2xy + 1 - 2x - y )( 2xy + 1 + 2x + y )
= [ ( 2xy - 2x ) - ( y - 1 ) ][ ( 2xy + 2x ) + ( y + 1 ) ]
= [ 2x( y - 1 ) - ( y - 1 ) ][ 2x( y + 1 ) + ( y + 1 ) ]
= ( y - 1 )( 2x - 1 )9 y + 1 )( 2x + 1 )
d) 9( x - y )2 - 4( x + y )2
= 32( x - y )2 - 22( x + y )2
= [ 3( x - y ) ]2 - [ 2( x + y ) ]2
= ( 3x - 3y )2 - ( 2x + 2y )2
= [ ( 3x - 3y ) - ( 2x + 2y ) ][ ( 3x - 3y ) + ( 2x + 2y ) ]
= ( 3x - 3y - 2x - 2y )( 3x - 3y + 2x + 2y )
= ( x - 5y )( 5x - y )
e) ( 3x - 2y )2 - ( 2x - 3y )2
= [ ( 3x - 2y ) - ( 2x - 3y ) ][ ( 3x - 2y ) + ( 2x - 3y ) ]
= ( 3x - 2y - 2x + 3y )( 3x - 2y + 2x - 3y )
= ( x + y )( 5x - 5y )
= ( x + y )5( x - y )
f) ( 4x2 - 4x + 1 ) - ( x + 1 )2
= ( 2x - 1 )2 - ( x + 1 )2
= [ ( 2x - 1 ) - ( x + 1 ) ][ ( 2x - 1 ) + ( x + 1 ) ]
= ( 2x - 1 - x - 1 )( 2x - 1 + x + 1 )
= 3x( x - 2 )
a, x^5+x^4+x^3-x^3-x²-x+x²+x+1
= x^3(x²+x+1)-x(x²+x+1)+1(x²+x+1)
= (x²+x+1).(x³-x²+1)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
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)