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a, (3x2-2xy+y2) + (x2-xy+2y2) - (4x2-y2)
= 3x2-2xy+y2+x2-xy+2y2-4x2+y2
= 4y2-3xy
b, = x2-y2+2xy-x2-xy-2y2+4xy-1
= -3y2+5xy
c, M=5xy+x2-7y2+(2xy-4y)2 = 5xy+x2-7y2+4x2y2-16xy2+16y2 = 5xy+x2+9y2+4x2y2-16xy2
\(a)P\left(x\right)+Q\left(x\right)=4x^2y-2xy^2+x^2y-15+2xy+\left(2xy^2+3x^2y-4xy-x^2y^2\right)\)
\(=4x^2y^2-2xy^2+x^2y-15+2xy+2xy^2+3x^2y-4xy-x^2y^2\)
\(=\left(4x^2y^2-x^2y^2\right)+\left(-2xy^2+2xy^2\right)+\left(x^2y+3x^2y\right)-15+\left(2xy-4xy\right)\)
\(=3x^2y^2+4x^2y-2xy-15\)
\(P\left(x\right)-Q\left(x\right)=4x^2y^2-2xy^2+x^2y-15+2xy-\left(2xy^2+3x^2y-4xy-x^2y^2\right)\)
\(=4x^2y^2-2xy^2+x^2y-15+2xy-2xy^2-3x^2y+4xy+x^2y^2\)
\(=\left(4x^2y+x^2y^2\right)+\left(-2xy^2-2xy^2\right)+\left(x^2y-3x^2y\right)-15+\left(2xy+4xy\right)\)
\(=5x^2y-4xy^2-2x^2y+6xy-15\)
a) (-1/2xy^2)^2 . (-2x^2y)
= 1/4x^2y^4.(-2x^2y)
= -1/2x^4y^5
b) (-x^2y^3z)^3 . (2xy^2)^2
= -x^6y^9z^3 . 4x^2y^4
= -4x^8y^13z^3
a) \(a^3+a^2b-a^2c-abc=a^2\left(a+b\right)-ac\left(a+b\right)=a\left(a+b\right)\left(a-c\right)\)
b) mk chỉnh lại đề
\(x^2+2xy+y^2-xz-yz=\left(x+y\right)^2-z\left(x+y\right)=\left(x+y\right)\left(x+y-z\right)\)
c) \(4-x^2-2xy-y^2=4-\left(x+y\right)^2=\left(2-x-y\right)\left(2+x+y\right)\)
d) \(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
Ta có:
A+B=3x3+2xy3+y2+2x2y2+x3-2xy3-5
A+B=(3+1).x3+(2-2).xy3+y2-2xy3-5
A+B=4x3+y2-2xy3-5
câu khác tương tự
A + B = (3x3 + 2xy3 +y2 ) + ( 2x2y2 +x3 - 2xy3 - 5 )
= 3x3+2xy3 + y2 + 2x2y2 + x3 - 2xy3 - 5
= ( 3x3 +x3 ) + ( 2xy3 - 2xy3 ) + y2 +2x2y2 -5
= 4x3 + y2 + 2x2y2 -5
A - B = ( 3x3 + 2xy3 + y2 ) - ( 2x2y2 + x3 - 2xy3 - 5)
= 3x3 + 2xy3 + y2 - 2x2y2 - x3 + 2xy3 - 5
= ( 3x3 - x3 ) + ( 2xy3 + 2xy3 ) + y2 - 2x2y2 - 5
= 2x3 + 4xy3 + y2 - 2x2y2 - 5
x2+2xy+y2=9
=>(x2+xy)+(xy+y2)=9
=>x(x+y)+y(x+y)=9
=>(x+y)(x+y)=3.3
=>x+y=3
x2-2xy+y2=1
=>(x2-xy)+(y2-xy)=1
=>x(x-y)+y(y-x)=1
=>x(x-y)-y(x-y)=1
=>(x-y)(x-y)=1.1
=>x-y=1
x+y+x-y=3+1
=>2x=4
=>x=2
=>y=2-1
=>y=1
vậy x=2 và y=1
Nhân phân phối là ra thôi
a)
\(VT=\left(x-1\right)\left(x+1\right)=x.x+x.1-1.x+\left(-1\right).1\)
\(=\left(x^2-1\right)+\left(x-x\right)=x^2-1+0=x^2-1=VP\Rightarrow dccm\)
c) thay vì c/m A=B ta chứng Minh B=A
\(VP=\left(x+1\right)\left(x^2-x+1\right)=\left(x^3-x^2+x\right)+\left(x^2-x+1\right)\)
\(=\left(x^3+1\right)+\left(-x^2+x^2\right)+\left(x-x\right)=x^3+1+0+0=x^3+1=VT\Rightarrow VT=VP\Rightarrow dpcm\)\(=x^3+1+0+0=x^3+1=VT\Rightarrow VT=VP\Rightarrow dpcm\)
a: \(M=6x^2+9xy-y^2-5x^2+2xy=x^2+7xy-y^2\)
b: \(M=-7xyz-15x^2yz^2+2xy^3\)
c: \(M=25u^2v-13uv^2+u^3-11u^2v+2u^3=14u^2v-13uv^2+3u^3\)
d: \(M=x^2-7xy+8y^2+4xy-3y^2=x^2-3xy+5y^2\)
A=x2+y2-2xy-x2+y2+2xy
=x2-x2+y2+y2+2xy-2xy
=y4
vậy da thức A sau khithu gọn là: y4
a,A=(x2+y2-2xy)+(-x2+y2+2xy)
= x2+y2-2xy+-x2+y2+2xy
=(x2-x2)+(y2+y2)+(-2xy+2xy)
= 2y2