Giải hệ pt sau:
a, ( x + 2y - 2011 ).( x2 +2012 ) = 2y
b, x2 + 4y2 - 2xy +2012 = 4026y
Cho (x + 2y)(x2 - 2xy + 4y2) =0 và (x - 2y)(x2 + 2xy + 4y2) = 16. Tìm x và y
\(\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\)
\(\Leftrightarrow x^3+8y^3=0\)
\(\Leftrightarrow x^3=-8y^3\)
\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\)
\(\Leftrightarrow x^3-8y^3=16\)
\(\Leftrightarrow-8y^3-8y^3=16\)
\(\Leftrightarrow y^3=-1\Rightarrow y=-1\Rightarrow x=2\)
(x-2y).(x2+2xy+4y2)-(x-y).(x2+8y2)
mọi người giúp em giải bài này với ạ
\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x-y\right)\left(x^2+8y^2\right)\)
\(=x^3-8y^3-\left(x^3-x^2y+8xy^2-8y^3\right)\)
\(=x^3-8y^3-x^3+x^2y-8xy^2+8y^3\)
\(=x^2y-8xy^2\)
\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x-y\right)\left(x^2+8y^2\right)\\ =x^3-8y^3-\left(x^3+8y^3-x^2y-8y^3\right)\\ =x^3-8y^3-x^3-8y^3+x^2y+8y^3\\ =-8y^3+x^2y\)
phân tích đa thức thành nhân tử bằng phương pháp nhóm hạng tử:
a) x2 ( x+ 2y) -x -2y
b)3x2- 3y2 -2 (x-y)2
c) x^2- 2x-4y2 - 4y
d) x3 - 4x2 - 9x +36
các bạn giải giúp mình với. Mình đang cần gấp
a) x2 ( x+ 2y) -x -2y
= x2 ( x+ 2y) -(x+2y)
= (x2-1)(x+2y)
= (x-1)(x+1)(x+2y)
b)3x2- 3y2 -2 (x-y)2
= 3(x2-y2) -2 (x-y)2
= 3(x-y)(x+y)-2(x-y)(x-y)
\(=\left(x-y\right)\left[3\left(x+y\right)-2\left(x-y\right)\right]\\ =\left(x-y\right)\left(3x+3y-2x+2y\right)\\ =\left(x-y\right)\left(x+5y\right)\)
c) x2- 2x-4y2 - 4y
= (x2-4y2)-(2x+4y)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\\ =\left(x+2y\right)\left(x-2y-2\right)\)
d) x3 - 4x2 - 9x +36
= (x3+3x2)-(7x2+21x)+(12x+36)
= x2(x+3)-7x(x+3)+12(x+3)
=(x2-7x+12)(x+3)
\(=\left[\left(x^2-3x\right)-\left(4x-12\right)\right]\left(x+3\right)\\ =\left[x\left(x-3\right)-4\left(x-3\right)\right]\left(x+3\right)=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
a) = x2 ( x+ 2y) -(x+2y)
= (x2-1)(x+2y)
= (x-1)(x+1)(x+2y)
b)= 3(x2-y2) -2 (x-y)2
= 3(x-y)(x+y)-2(x-y)(x-y)
=(x−y)[3(x+y)−2(x−y)]
=(x−y)(3x+3y−2x+2y)
=(x−y)(x+5y)
=(x−y)[3(x+y)−2(x−y)]
=(x−y)(3x+3y−2x+2y)
=(x−y)(x+5y)
c)= (x2-4y2)-(2x+4y)
=(x−2y)(x+2y)−2(x+2y)
=(x+2y)(x−2y−2)
=(x−2y)(x+2y)−2(x+2y)
=(x+2y)(x−2y−2)
d)= (x3+3x2)-(7x2+21x)+(12x+36)
= x2(x+3)-7x(x+3)+12(x+3)
=(x2-7x+12)(x+3)
=[(x2−3x)−(4x−12)](x+3)
=[x(x−3)−4(x−3)](x+3)
=(x−4)(x−3)(x+3)
a: \(x^2\left(x+2y\right)-x-2y\)
\(=\left(x+2y\right)\left(x^2-1\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
b: \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
c: Ta có: \(x^2-2x-4y^2-4y\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
d: Ta có: \(x^3-4x^2-9x+36\)
\(=x^2\left(x-4\right)-9\left(x-4\right)\)
\(=\left(x-4\right)\left(x^2-9\right)\)
\(=\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
Rút gọn biểu thức:
( x - 2y)3 - ( x + 2y) (x2 - 2xy + 4y2) + 6x2y
\(\left(x-2y\right)^3-\left(x+2y\right)\left(x^2-2xy+4y^2\right)+6x^2y\\ x^3-6x^2+12xy^2-8y^3-\left(x^3+8y^3\right)+6x^2y\\ x^3-6x^2+12xy^2-8y^3-x^3-8y^3+6x^2y\\ =12xy^2-16y^3\)
=x^3-6x^2y+12xy^2-8y^3-x^3-8y^3+6x^2y
=-16y^3+12xy^2
thực hiện phép tính
a)1/2x2.2x3-4x2+3
b)2y(xy-1)(xy+1)
c)(x+2)x2-x+1
d)(x-2y)x2+2xy+4y2
a: \(\dfrac{1}{2}x^2\cdot2x^3-4x^2+3=x^5-4x^2+3\)
b: \(2y\left(xy-1\right)\left(xy+1\right)=2y\left(x^2y^2-1\right)=2x^2y^3-2y\)
Tính giá trị biểu thức
M=(x+3)(x2-3x+9)-(3-2x)(4x2+6x+9) tại x = 20
N=(x-2y)(x2+2xy+4y2)+16y3 biết x+2y=0
\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(3-2x\right)\left(4x^2+6x+9\right)\)
\(M=\left(x^3+3^3\right)-\left[3^3-\left(2x\right)^3\right]\)
\(M=x^3+27-27+8x^3\)
\(M=9x^3\)
Thay x=20 vào M ta có:
\(M=9\cdot20^3=72000\)
Vậy: ...
\(N=\left(x-2y\right)\left(x^2+2xy+4y^2\right)+16y^3\)
\(N=x^3-\left(2y\right)^3+16y^3\)
\(N=x^3-8y^3+16y^3\)
\(N=x^3+8y^3\)
\(N=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
Thay \(x+2y=0\) vào N ta có:
\(N=0\cdot\left(x^2-2xy+4y^2\right)=0\)
Vậy: ...
Biểu thức nào dưới đây là bình phương của hiệu x - 2y:
A. x2 + 2xy + 4y2.
B. x2 – 2xy + 4y2 .
C. x2 – 4xy + 4y2 .
D. x2 + 4xy + 4y2
Bài 1. Rút gọn các biểu thức sau.
a) (x + 2y)(x2 - 2xy + 4y2) – (x - y)(x2 + xy + y2)
b) (x + 1)(x - 1)2 – (x + 2)(x2 - 2x + 4)
a) Ta có: \(\left(x+2y\right)\left(x^2-2xy+4y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3+\left(2y\right)^3-\left(x^3-y^3\right)\)
\(=x^3+8y^3-x^3+y^3\)
\(=9y^3\)
b) Ta có: \(\left(x+1\right)\left(x-1\right)^2-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)-\left(x+2\right)\left(x^2-2x+4\right)\)
\(=x^3-2x^2+x+x^2-2x+1-\left(x^3+8\right)\)
\(=x^3-x^2-x+1-x^3-8\)
\(=-x^2-x-7\)
chứng tỏ
a) x2 + 8y2 =( x +2y ) ( x2- 2xy +4y2)
b) (x-y) (x2+xy+y2 ) -3xy (x-y) =( x-y)3
c) (x-3y) (x2 +3xy +9y2 ) - ( 3y +x ) ( 9y2 -3xy + x2) = -54y3
cíu em vớii
\(a,VP=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ =\left(x+2y\right)\left[x^2-x.2y+\left(2y\right)^2\right]\\ =x^3+\left(2y\right)^3=x^3+8y^3=VT\left(đpcm\right)\\ b,VT=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\left(x-y\right)\\ =x^3-y^3-3xy\left(x-y\right)\\ =x^3-3x^2y+3xy^2-y^3\\ =\left(x-y\right)^3=VP\left(đpcm\right)\)
\(c,VT=\left(x-3y\right)\left(x^2+3xy+9y^2\right)-\left(3y+x\right)\left(9y^2-3xy+x^2\right)\\ =\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]-\left(x+3y\right).\left[x^2-x.3y+\left(3y\right)^2\right]\\ =x^3-27y^3-\left(x^3+27y^3\right)\\ =-54y^3=VP\left(đpcm\right)\)