Giúp mình vs aaaaaaaaaaaaaaaaaaaa
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Bài 3:
a: \(P=x\left(x^2-y\right)+y\left(x-y^2\right)\)
\(=x^3-xy+xy-y^3\)
\(=x^3-y^3\)
Thay \(x=-\frac12;y=-\frac12\) vào P, ta được:
\(P=\left(-\frac12\right)^3-\left(-\frac12\right)^3=\left(-\frac18\right)-\left(-\frac18\right)=-\frac18+\frac18=0\)
b: \(Q=x^2\left(y^3-xy^2\right)+x^2y^2\left(x-y+1\right)\)
\(=x^2y^3-x^3y^2+x^3y^2-x^2y^3+x^2y^2=x^2y^2\)
Thay x=-10; y=-10 vào Q, ta được:
\(Q=\left(-10\right)^2\cdot\left(-10\right)^2=100\cdot100=10000\)
c: \(A=x^3+2xy-2x^3+2y^3+2x^3-y^3\)
\(=\left(x^3-2x^3+2x^3\right)+2xy+\left(2y^3-y^3\right)\)
\(=x^3+2xy+y^3\)
Thay x=2; y=-3 vào A, ta được:
\(A=2^3+2\cdot2\cdot\left(-3\right)+\left(-3\right)^3\)
=8-12-27
=-4-27
=-31
d:
x=1; y=-1
=>\(xy=1\cdot\left(-1\right)=-1\)
\(B=xy+x^2y^2-x^4y^4+x^6y^6-x^8y^8\)
\(=\left(xy\right)+\left(xy\right)^2-\left(xy\right)^4+\left(xy\right)^6-\left(xy\right)^8\)
\(=\left(-1\right)+\left(-1\right)^2-\left(-1\right)^4+\left(-1\right)^6-\left(-1\right)^8\)
=-1+1-1+1-1
=-1
e: x=-1; y=1
=>xy=-1
\(C=xy+x^2y^2+x^3y^3+\cdots+x^{10}y^{10}\)
\(=xy+\left(xy\right)^2+\left(xy\right)^3+\cdots+\left(xy\right)^{10}\)
\(=\left(-1\right)+\left(-1\right)^2+\left(-1\right)^3+\cdots+\left(-1\right)^{10}\)
=-1+1+(-1)+1+...+(-1)+1
=0
f: \(M=2x^2\left(x^2-5\right)+x\left(-2x^3+4x\right)+x^2\left(x+6\right)\)
\(=2x^4-10x^2-2x^4+4x^2+x^3+6x^2\)
\(=x^3\)
Khi x=-4 thì \(M=\left(-4\right)^3=-64\)
g: \(N=x^3\left(y+1\right)-xy\left(x^2-2x+1\right)-x\left(x^2+2xy-3y\right)\)
\(=x^3y+x^3-x^3y+2x^2y-xy-x^3-2x^2y+3xy\)
=2xy
Thay x=8; y=-5 vào N, ta được:
\(N=2\cdot8\cdot\left(-5\right)=-80\)
Bài 1:
d: \(x^2+2xy-3\cdot\left(-xy\right)\)
\(=x^2+2xy+3xy=x^2+5xy\)
e: \(\frac12x^2y\left(2x^3-\frac25xy^2-1\right)\)
\(=\frac12x^2y\cdot2x^3-\frac12x^2y\cdot\frac25xy^2-\frac12x^2y\)
\(=x^5y-\frac15x^3y^3-\frac12x^2y\)
f: \(\left(-xy^2\right)^2\left(x^2-2x+1\right)\)
\(=x^2y^4\left(x^2-2x+1\right)\)
\(=x^2y^4\cdot x^2-x^2y^4\cdot2x+x^2y^4\)
\(=x^4y^4-2x^3y^4+x^2y^4\)
g: (2xy+3)(x-2y)
\(=2xy\cdot x-2xy\cdot2y+3\cdot x-3\cdot2y\)
\(=2x^2y-4xy^2+3x-6y\)
h: \(\left(xy+2y\right)\left(x^2y-2xy+4\right)\)
\(=x^3y^2-2x^2y^2+4xy+2x^2y^2-4xy^2+8y\)
\(=x^3y^2+4xy-4xy^2+8y\)
i: \(4\left(x^2-\frac12y\right)\left(x^2+\frac12y\right)\)
\(=4\left(x^4-\frac14y^2\right)\)
\(=4\cdot x^4-4\cdot\frac14y^2=4x^4-y^2\)
k: \(2x^2\left(1-3x+2x^2\right)\)
\(=2x^2\cdot1-2x^2\cdot3x+2x^2\cdot2x^2\)
\(=2x^2-6x^3+4x^4\)
l: \(\left(2x^2-3x+4\right)\left(-\frac12x\right)\)
\(=-\frac12x\cdot2x^2+3x\cdot\frac12x-4\cdot\frac12x=-x^3+\frac32x^2-2x\)
m: \(\frac12xy\left(-x^3+2xy-4y^2\right)\)
\(=-\frac12xy\cdot x^3+\frac12xy\cdot2xy-\frac12xy\cdot4y^2\)
\(=-\frac12x^4y+x^2y^2-2xy^3\)
n: \(\frac12x^2y\left(2x^3-\frac25xy^2-1\right)\)
\(=\frac12x^2y\cdot2x^3-\frac12x^2y\cdot\frac25xy^2-\frac12x^2y\)
\(=x^5y-\frac15x^3y^3-\frac12x^2y\)
Bài 2:
a: \(\Leftrightarrow x:\dfrac{4}{5}=\dfrac{-4}{6}+\dfrac{3}{6}=\dfrac{-1}{6}\)
=>x=-1/6*4/5=-4/30=-2/15
b: =>1/6x=5/6-7/12+1/3=10/12-7/12+4/12=7/12
=>x=7/2
c: =(47/11-3x)*11/5=-21/5
=>47/11-3x=-21/5:11/5=-21/11
=>3x=68/11
=>x=68/33
Ta có: \(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\)
\(=\frac{x^2+y^2}{\left(x-y\right)^2}-\frac{2}{xy}\)
\(=\frac{xy\left(x^2+y^2\right)-2\left(x-y\right)^2}{xy\left(x-y\right)^2}=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}\)
TA có: \(\left(\frac{x^2+y^2}{x^2-2xy+y^2}-\frac{2}{xy}\right):\left(\frac{1}{x}-\frac{1}{y}\right)^2\)
\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}:\left(\frac{y-x}{xy}\right)^2\)
\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}:\frac{\left(x-y\right)^2}{x^2y^2}\)
\(=\frac{x^3y+xy^3-2x^2+4xy-2y^2}{xy\left(x-y\right)^2}\cdot\frac{x^2y^2}{\left(x-y\right)^2}=\frac{\left(x^3y+xy^3-2x^2+4xy-2y^2\right)\cdot xy}{\left(x-y\right)^4}\)
Ta có:
\(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\\ =\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{x^3-y^3}\\ =\dfrac{2\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\\ =\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
\(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\) \(=\dfrac{x^2+xy+y^2}{x^3-y^3}-\dfrac{3xy}{x^3-y^3}+\dfrac{\left(x-y\right)^2}{x^3-y^3}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{x^3-y^3}\)
\(=\dfrac{2x^2+2y^2-4xy}{x^3-y^3}\)
\(=\dfrac{2x^2-2xy-2xy+2y^2}{x^3-y^3}\)
\(=\dfrac{2x\left(x-y\right)-2y\left(x-y\right)}{x^3-y^3}\)
\(=\dfrac{\left(2x-2y\right)\left(x-y\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x-2y}{x^2+xy+y^2}\)
\(=\left[\left(-6x\right)+\left(x^2+9\right)\right]\left[\left(-6x\right)-\left(x^2+9\right)\right]\)
\(=\left(-6x\right)^2-\left(x^2+9\right)^2\)
\(=36x^2-\left(x^4+18x^2+81\right)\)
\(=-x^4+18x^2-81\)
\(=-\left(x^4-18x^2+81\right)\)
\(=-\left(x^2-9\right)^2\)
Ta có: \(\left(x^2-6x+9\right)\left(-x^2-6x-9\right)\)
\(=-\left(x^2-6x+9\right)\left(x^2+6x+9\right)\)
\(=-\left[\left(x-3\right)^2\cdot\left(x+3\right)^2\right]\)
\(=-\left(x^2-9\right)^2\)
\(a,\dfrac{-13}{5}-\dfrac{-7}{15}.\dfrac{120}{63}+\dfrac{4}{5}\\ =\dfrac{-13}{5}+\dfrac{8}{9}+\dfrac{4}{5}\\ =\dfrac{-77}{45}+\dfrac{4}{5}\\ =\dfrac{-41}{45}\\ b.\left(7\dfrac{5}{9}+2\dfrac{2}{3}\right)-5\dfrac{2}{9}\\ =\left(\dfrac{68}{9}+\dfrac{8}{3}\right)-\dfrac{47}{9}\\ =\dfrac{92}{9}-\dfrac{47}{9}\\ =5\)
2)\(-\dfrac{13}{5}--\dfrac{7}{15}.\dfrac{120}{63}+\dfrac{4}{5}\)
\(=-\dfrac{13}{5}+\dfrac{7.15.8}{15.9.7}+\dfrac{4}{5}\)
\(=-\dfrac{13}{5}+\dfrac{8}{9}+\dfrac{4}{5}\)
\(=\left(-\dfrac{13}{5}+\dfrac{4}{5}\right)+\dfrac{8}{9}\)
\(=-\dfrac{9}{5}+\dfrac{8}{9}\)
\(=\dfrac{-81+40}{45}=-\dfrac{41}{45}\)
3)\(\left(7\dfrac{5}{9}+2\dfrac{2}{3}\right)-5\dfrac{2}{9}\)
\(=\left(7\dfrac{5}{9}-5\dfrac{2}{9}\right)+2\dfrac{2}{3}\)
\(=2\dfrac{1}{3}+2\dfrac{2}{3}\)
\(=5\)
\(\dfrac{2}{67}-\left(\dfrac{3}{7}+\dfrac{2}{67}\right)\\ =\dfrac{2}{67}-\dfrac{215}{469}\\ =\dfrac{-3}{7}\)
-472 + (235 - 28) - (35 - 350)
= -472 + 235 - 28 - 35 + 350
= (-472 - 28) + (235 - 35) + 350
= -500 + 200 + 350
= -500 + 550
= 50