Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Phân tích các đa thức sau thành nhân tử :
a) x - y + 5x - 5y
= ( x + 5x ) - ( y + 5y )
= x . ( 1 + 6 ) - y . ( 1 + 6 )
= ( 1 + 6 ) . ( x - y )
\(a,x-y+5x-5y=\left(x-y\right)+5\left(x-y\right)=6\left(x-y\right)\)
a) 2x3 + 6xy - x2z - 3yz
= ( 2x3 + 6xy ) - ( x2z + 3yz )
= 2x( x2 + 3y ) - z( x2 + 3y )
= ( x2 + 3y )( 2x - z )
b) x2 - 6xy + 9y2 - 49
= ( x2 - 6xy + 9y2 ) - 49
= ( x - 3y )2 - 72
= ( x - 3y - 7 )( x - 3y + 7 )
c) x3 + 4x2 + 16x + 64
= ( x3 + 4x2 ) + ( 16x + 64 )
= x2( x + 4 ) + 16( x + 4 )
= ( x + 4 )( x2 + 16 )
a) =(2x^3-x^2z)+(6xy-3yz)
=x^2(2x-z)+3y(2x-z)
=(x^2+3y)(2x-z)
b) =(x^2-6xy+9y^2)-7^2
=(x-3y)^2-7^2
=(x-3y+7)(x-3y-7)
c) =(x^3+4x^2)+(16x+64)
=x^2(x+4)+16(x+4)
=(x^2+16)(x+4)
a) A = (x + 1)(y - 2) - (2 - y)2
= -[(x + 1)(2 - y) + (2 - y)2]
= -[(x + 1 - 2 + y)(2 - y)]
= -[(x - 1 + y)(2 - y)]
= (x - 1 + y)(y - 2)
Bài 2:
a) \(A=\left(x+1\right)\left(y-2\right)-\left(2-y\right)^2\)
\(A=\left(x+1\right)\left(y-2\right)-\left(y-2\right)^2\)
\(A=\left(y-2\right)\left(x+1-y+2\right)\)
\(A=\left(y-2\right)\left(x-y+3\right)\)
b) \(B=x^2-6xy+9y^2+4x-12y\)
\(B=\left[x^2-2\cdot x\cdot3y+\left(3y\right)^2\right]+4\left(x-3y\right)\)
\(B=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(B=\left(x-3y\right)\left(x-3y+4\right)\)
Bài 3:
a) \(3\left(x-2\right)\left(x+3\right)-x\left(3x+1\right)=2\)
\(\left(3x^2+3x-18\right)-\left(3x^2+x\right)-2=0\)
\(3x^2+3x-18-3x^2-x-2=0\)
\(2x-20=0\)
\(x=10\)
b) \(6x^2+13x+5=0\)
\(6x^2+10x+3x+5=0\)
\(2x\left(3x+5\right)+\left(3x+5\right)=0\)
\(\left(3x+5\right)\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+5=0\\2x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-5}{3}\\x=\frac{-1}{2}\end{cases}}}\)
\(1.\)
\(a.\)
\(x^2-2x=x\left(x-2\right)\)
b.
\(3y^3+6xy^2+3x^2y\)
\(=3y\left(y^2+2xy+x^2\right)\)
\(=3y\left(x+y\right)^2\)
\(c.\)
\(x^2-2xy-xy+2y^2\)
\(=x\left(x-2y\right)-y\left(x-2y\right)\)
\(=\left(x-y\right)\left(x-2y\right)\)
\(2.\)
\(a.\)
\(x^2-y^2+5x-5y\)
\(=\left(x-y\right)\left(x+y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+5\right)\)
\(b.\)
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(c.\)
\(x^2-6xy+9y^2-16\)
\(=\left(x^2-6xy+9y^2\right)-4^2\)
\(=\left(x-3\right)^2-4^2\)
\(=\left(x-3-4\right)\left(x-3+4\right)\)
\(=\left(x-7\right)\left(x+1\right)\)
Tương tự câu \(d,e,g\)
\(3.\)
\(a.\)
\(x^3-2x=0\)
\(\Rightarrow x\left(x^2-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{2}\end{matrix}\right.\)
\(b.\)
\(x\left(x-4\right)+\left(x-4\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
\(c.\)
\(x\left(x-3\right)+4x-12=0\)
\(\Rightarrow x\left(x-3\right)+3\left(x-3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Tương tự \(d,e,g\)
a, ( 2x - 1)^2 - (4x + 2) ^2 = ( 2x - 1 - 4x- 2) ( 2x - 1 + 4x + 2) = (-2x-3)(6x+1) = - (2x+3)(6x+1)
b, 8x^3 + 12x^2y + 6xy^2 + y^3
= (2x)^2 + 3.(2x)^2 . y + 3.2x.y^2 + y^3
= (2x + y)^3
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)
\(a,4x^4-8x^3+4x^2\)
\(=4x^2\cdot\left(x^2-2x+1\right)\)
\(=4x^2\cdot\left(x-1\right)^2\)
\(b,x^2-y^2+5\cdot\left(y-x\right)\)
\(=\left(x-y\right)\cdot\left(x+y\right)-5\cdot\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(x+y-5\right)\)
\(c,3x^2-6xy+3y^2-12z^2\)
\(=3\cdot\left(x^2-2xy+y^2-4x^2\right)\)
\(=3\cdot\left[\left(x-y\right)^2-\left(2x\right)^2\right]\)
\(=3\cdot\left(x-y-2x\right)\cdot\left(x-y+2x\right)\)
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)