phân tích thành nhân tử
a)x^2a - x^2b - y^2a + y^2b
k) 5x^2 - 5xy - 10x + 10y
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
18 tháng 10 2020
Mấy câu dễ mình làm trước nhé. Mấy câu khó hơn mình trình bày sau :)
1) 2x2 - 5xy - 3y2 = 2x2 + xy - 6xy - 3y2 = x( 2x + y ) - 3y( 2x + y ) = ( 2x + y )( x - 3y )
2) 7x2 + 3xy - 10y2 = 7x2 - 7xy + 10xy - 10y2 = 7x( x - y ) + 10y( x - y ) = ( x - y )( 7x + 10y )
3) x2 + 5x - 2 = ( x2 + 5x + 25/4 ) - 33/4 = ( x + 5/2 )2 - \(\left(\frac{\sqrt{33}}{2}\right)^2\)= \(\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)
6) x4 + 324 = ( x4 + 36x2 + 324 ) - 36x2 = ( x2 + 18 )2 - ( 6x )2 = ( x2 - 6x + 18 )( x2 + 6x + 18 )
18 tháng 10 2020
4) x8 + x7 + 1
= x8 + x7 + x6 - x6 + 1
= x6( x2 + x + 1 ) - ( x6 - 1 )
= x6( x2 + x + 1 ) - ( x3 - 1 )( x3 + 1 )
= x6( x2 + x + 1 ) - ( x - 1 )( x2 + x + 1 )( x3 + 1 )
= ( x2 + x + 1 )( x6 - ( x - 1 )( x3 + 1 ) ]
= ( x2 + x + 1 )( x6 - x4 + x3 - x + 1 )
5) x7 + x5 + 1
= x7 + x6 - x6 + x5 + 1
= x5( x2 + x + 1 ) - ( x6 - 1 )
= x5( x2 + x + 1 ) - ( x3 - 1 )( x3 + 1 )
= x5( x2 + x + 1 ) - ( x - 1 )( x2 + x + 1 )( x3 + 1 )
= ( x2 + x + 1 )[ x5 - ( x - 1 )( x3 + 1 ) ]
= ( x2 + x + 1 )( x5 - x4 + x3 - x + 1 )
7) x5 - 5x3 + 4x
= x5 - x3 - 4x3 + 4x
= x3( x2 - 1 ) - 4x( x2 - 1 )
= ( x2 - 1 )( x3 - 4x )
= ( x - 1 )( x + 1 )x( x2 - 4 )
= x( x - 1 )( x + 1 )( x - 2 )( x + 2 )
8) Xin hàng :)
12 tháng 11 2021
c: \(=\left(5x-y\right)\left(5x+y\right)\)
e: \(=\left(x-2\right)\left(x-3\right)\)
12 tháng 11 2021
a) x(4y-10x)
b)3(x+2y)+(x+1)
c)(5x-y)(5x+y)
d)5x(y-z)2
e)(x-3)(x-2)
f)(2x+y)3
JP
1
KN
4 tháng 3 2020
\(x^5-4x^3-5x\)
\(=x\left(x^4-4x^2-5\right)\)
\(=x\left(x^4-5x^2+x^2-5\right)\)
\(=x\left[x^2\left(x^2-5\right)+\left(x^2-5\right)\right]\)
\(=x\left(x^2+1\right)\left(x+\sqrt{5}\right)\left(x-\sqrt{5}\right)\)
I
4 tháng 3 2020
a/
\(a^4+b^4+c^4-2a^2b^2-2b^2c^2-2c^2a^2.\)
=>\(a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2-2\left(ac\right)^2\)
=>\(a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2+2\left(ac\right)^2-4\left(ca\right)^2\)
áp dụng hằng đẳng thức \(a^2-b^2-c^2=a^4+b^4+c^4-2\left(ab\right)^2-2\left(bc\right)^2+2\left(ac\right)^2\) ta đc
\(\left(a^2-b^2+c^2\right)-4\left(ac\right)^2\)
=> \(\left(a^2-b^2+c^2-2ac\right)\left(a^2-b^2+c^2+2ac\right)\)
13 tháng 12 2020
a, \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2=\left[\left(x+1\right)-\left(y-3\right)\right]^2\)
\(=\left(x+1-y+3\right)^2=\left(x-y+4\right)^2\)
b, \(a^2+b^2+2a-2b-2ab=\left(a^2-2ab+b^2\right)+\left(2a-2b\right)\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left[\left(a-b\right)+2\right]=\left(a-b\right)\left(a-b+2\right)\)
1 tháng 11 2023
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
1 tháng 11 2023
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
22 tháng 10 2021
\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)
\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
5x^2-5xy-10x+10y
(5x^2-5xy )-(10x-10y)
5x(x-y)-10(x-y)
(x-y)(5x-10)
a)\(x^2a-x^2b-y^2a+y^2b=x^2\left(a-b\right)-y^2\left(a-b\right)=\left(a-b\right)\left(x^2-y^2\right)=\left(a-b\right)\left(x-y\right)\left(x+y\right)\)
k) \(5x^2-5xy-10x+10y=5x\left(x-y\right)-10\left(x-y\right)=\left(x-y\right)\left(5x-10\right)\)