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1) x2-4x+5+y2+2y=0
<=>x2-4x+4+y2+2y+1=0
<=>(x-2)2+(x+1)2=0
<=>x-2=0 và x+1=0
<=>x=2 và x=-1
2)2p.p2-(p3-1)+(p+3)2p2-3p5
<=>2p3-p3+1+2p3+6p2-3p5
<=>3p3+6p2-3p5+1
3)(0.2a3)2-0.01a4(4a2-100)=0,04a6-0,04a6+1
=1
4)a) x(2x+1)-x2(x+20)+(x3-x+3)=2x2+x-x3-20x2+x3-x+3
=-18x2+3(đề sai)
b) x(3x2-x+5)-(2x3+3x-16)-x(x2-x+2)=3x3-x2+5x-2x3-3x+16-x3+x2-2x
=16
Vậy x(3x2-x+5)-(2x3+3x-16)-x(x2-x+2) không phụ thuộc vào x
5)a) x(y-z)+y(z-x)+z(x-y)=xy-xz+yz-xy+xz-yz=0
b) x(y+z-yz)-y(z+x-xz)+z(y-x)=xy+xz-xyz-yz-xy+xyz+yz-xz=0
6)M+(12x4-15x2y+2xy2+7)=0
<=>M =-(12x4-15x2y+2xy2+7)
<=>M =-12x4+15x2y-2xy2-7
1a : x = -1
2a : x = 10
còn mấy bài khác mình không biết giải nha
giải hộ câu c, d và f thôi nhá, mấy câu kia biết là rồi
mik cần gấp
Bài 1:
\(a,\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(x^2-xy+1\right)=\left(x^3y^2+\dfrac{1}{2}x^2y^3\right)\left(x^2-xy+1\right)=x^5y^2-x^4y^3+x^3y^2+\dfrac{1}{2}x^3y^3-\dfrac{1}{2}x^3y^4+\dfrac{1}{2}x^2y^3\)
\(b,\left(\dfrac{1}{2}x-1\right)\left(2x-3\right)=x^2-\dfrac{3}{2}x-2x+3=x^2-\dfrac{7}{2}x+3\)\(c,\left(x-7\right)\left(x-5\right)=x^2-5x-7x+35=x^2-12x+35\)\(f,\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x-1\right)=\left(x^2-\dfrac{1}{4}\right)\left(4x-1\right)=4x^3-x^2-x+\dfrac{1}{4}\)Bài 2 ,
\(\left(x-1\right)\left(x^2+x+1\right)=x^3+x^2+x-x^2-x-1=x^3-1\Rightarrowđpcm\)\(b,\left(x^3+x^2y+xy^2+y^3\right)\left(x-y\right)=x^4+x^3y+x^2y^2+y^3x+x^3y-x^2y^2-xy^3-y^4=x^4-y^4\)
a ) \(\left(x+1\right)\left(x-2\right)=x^2-2x+x-2=x^2-x-2\)
b ) \(\left(4x^4y^4-12x^2y^2\right):4x^2y^2=x^2y^2-3\)
c ) \(\frac{3x^2-1}{2x}+\frac{x^2+1}{2x}=\frac{3x^2-1+x^2+1}{2x}=\frac{4x^2}{2x}=2x\)
d ) \(\frac{x^2}{x-1}+\frac{2x}{1-x}+\frac{1}{x-1}=\left(\frac{x^2}{x-1}+\frac{1}{x-1}\right)+\frac{2x}{1-x}\)
\(=\frac{x^2+1}{x-1}+\frac{2x}{1-x}=\frac{x^2+1}{x-1}+\frac{-2x}{x-1}=\frac{x^2+1-2x}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
a) .......=x2-x-2
b) .........=x2y2-3
c) .......=(3x2-1+x2+1)/2x=4x2/2x=2x
d) x2 /(x-1)+(-2x)/(x-1)+1/(x-1)=(x2-2x+1)/(x-1)=(x-1)2/(x-1)=x-1
e)...
x-y=4
=> x2-2xy+y2=16
<=> 106-2xy =16 (vì x2+y2 =106)
=>xy=(106-16)/2=45
ta có x3 -y3 =(x-y)(x2+xy+y2 )
=4(106+45)=604
a.\(2x\left(7x^2-5x-1\right)=14x^3-10x^2-2x\)
b.\(-2x^3\left(2x^2-3y+5yz\right)=-4x^5+6x^3y-10x^3yz\)
c.\(\left(2x-y\right)\left(4x^2-2xy+y^2\right)=2x\left(4x^2-2xy+y^2\right)-y\left(4x^2-2xy+y^2\right)\)
\(=8x^2-4x^2y+4xy^2-4x^2y+2xy^2-y^3\)
a.2x(7x2−5x−1)=14x3−10x2−2x
b.−2x3(2x2−3y+5yz)=−4x5+6x3y−10x3yz
c.(2x−y)(4x2−2xy+y2)=2x(4x2−2xy+y2)−y(4x2−2xy+y2)
=8x2−4x2y+4xy2−4x2y+2xy2−y3
b) \(=y^3-1+\frac{2}{3}x^3y-2xy+\frac{1}{3}x^2y^3-y^3\)
\(=\frac{2}{3}x^3y+\frac{1}{3}x^2y^3-2xy-1\)
Cách làm