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2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
Bài 7: Phân tích đa thức thành nhân tử
a) Ta có: \(a^2-b^2-2a+2b\)
\(=\left(a-b\right)\left(a+b\right)-2\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b-2\right)\)
b) Ta có: \(3x-3y-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(3+5x\right)\)
c) Ta có: \(16-x^2+4xy-4y^2\)
\(=16-\left(x^2-4xy+4y^2\right)\)
\(=16-\left(x-2y\right)^2\)
\(=\left(4-x+2y\right)\left(4+x-2y\right)\)
d) Ta có: \(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4-2x-3y+1\right)\left(x-y+4+2x+3y-1\right)\)
\(=\left(5-x-4y\right)\left(3x+2y+3\right)\)
e) Ta có: \(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+x\right)\)
f) Ta có: \(\left(x+3\right)^3+\left(x-3\right)^3\)
\(=\left(x+3+x-3\right)\left[\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\right]\)
\(=2x\cdot\left[x^2+6x+9-\left(x^2-9\right)+x^2-6x+9\right]\)
\(=2x\cdot\left(2x^2+18-x^2+9\right)\)
\(=2x\cdot\left(x^2+27\right)\)
g) Ta có: \(9x^2-3xy+y-6x+1\)
\(=\left(9x^2-6x+1\right)-y\left(3x-1\right)\)
\(=\left(3x-1\right)^2-y\left(3x-1\right)\)
\(=\left(3x-1\right)\left(3x-1-y\right)\)
h) Ta có: \(x^3-4x^2+12x-27\)
\(=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
13.
M \(=\)\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)\)\(+16\)
\(=\)\(\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)\) \(+16\)
\(=\left(x^2+10x+20\right)^2-16+16\)
\(=\left(x^2+10x+20\right)^2\) là một số chính phương
Nhiều quá, nhìn đã thấy ớn lạnh :(
Bạn nên chia nhỏ ra , post 1 hoặc 2 bài 1 lần thôi, đăng 1 lần 1 nùi thế này không ai dám làm đâu, bội thực chữ viết.
a) \(2x^2\left\{x^2+5x+6\right\}\)=\(2x^4+10x^3+12x^2\)
b) \(15x^2y^4:10x^2y\)=\(\frac{3}{2}y^3\)
c) \(\left\{16x^3y^2+20x^2y^3-8xy\right\}:4xy\)=\(4x^2y+5xy^2-2\)
Bài 1.
A = x2 + 2xy + y2 = ( x + y )2 = ( -1 )2 = 1
B = x2 + y2 = ( x2 + 2xy + y2 ) - 2xy = ( x + y )2 - 2xy = (-1)2 - 2.(-12) = 1 + 24 = 25
C = x3 + 3xy( x + y ) + y3 = ( x3 + y3 ) + 3xy( x + y ) = ( x + y )( x2 - xy + y2 ) + 3xy( x + y )
= -1( 25 + 12 ) + 3.(-12).(-1)
= -37 + 36
= -1
D = x3 + y3 = ( x3 + 3x2y + 3xy2 + y3 ) - 3x2y - 3xy2 = ( x + y )3 - 3xy( x + y ) = (-1)3 - 3.(-12).(-1) = -1 - 36 = -37
Bài 2.
M = 3( x2 + y2 ) - 2( x3 + y3 )
= 3( x2 + y2 ) - 2( x + y )( x2 - xy + y2 )
= 3( x2 + y2 ) - 2( x2 - xy + y2 )
= 3x2 + 3y2 - 2x2 + 2xy - 2y2
= x2 + 2xy + y2
= ( x + y )2 = 12 = 1
1, 2x2 - 8xy - 5x + 20y
= (2x2 - 5x) - (8xy - 20y)
= x(2x - 5) - 4y(2x - 5)
= (2x - 5) (x - 4y)
2, x3 - x2y - xy + y2
= (x3 - xy) - (x2y - y2)
= x(x2 - y) - y(x2 - y)
= (x2 - y) (x - y)
3, x2 - 2xy - 4z2 + y2
= (x2 - 2xy + y2) - 4z2
= (x - y)2 - (2z)2
= (x - y - 2z) (x - y + 2z)
4, a3 + a2b - a2c - abc
= (a3 - a2c) + (a2b - abc)
= a2(a - c) + ab(a - c)
= (a - c) (a2 + ab)
5, x3 + y3 + 3x2y + 3xy2 - x - y
= (x3 + 3x2y + 3xy2 + y3) - (x + y)
= (x + y) 3 - (x + y)
= (x + y) [(x + y)2 - 1]
= (x + y) (x + y - 1) (x + y + 1)
a: \(A=4\cdot15^2-70^2=-4000\)
b: \(B=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+y+1\right)^2\)
\(=100^2=10000\)
c: \(C=b^2-3b+a^2+3a-2ab\)
\(=\left(a-b\right)^2+3\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+3\right)\)
\(=\left(-5\right)\cdot\left(-5+3\right)=\left(-5\right)\cdot\left(-2\right)=10\)
d: \(D=\left(x-y\right)^3+3xy\left(x-y\right)+3xy\)
\(=\left(-1\right)^3-3xy+3xy\)
=-1
a: \(\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(y-x\right)^2}\)
\(=\dfrac{3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2}{\left(x-y\right)^2}\)
\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)
b: \(\dfrac{\left(x-2y\right)^3}{x^2-4xy+4y^2}\)
\(=\dfrac{\left(x-2y\right)^3}{\left(x-2y\right)^2}\)
=x-2y
c: \(\dfrac{x^3+y^3}{x+y}\)
\(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x+y}\)
\(=x^2-xy+y^2\)