\(\dfrac{2y+x}{2y^{2^{ }}-xy}\)+\(\dfrac{8x}{x^{2^{ }}-4y^2}\)+\(\dfrac{2y-x}{2y^{2^{ }}+xy}\)
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a: =-4xyz^2
b: =-9x^2y
c: =16x^2y^2
d: =1/6x^2y^3
e: =13/6x^3y^2
f: =7/12x^4y
a) -xyz² - 3xz.yz
= -xyz² - 3xyz²
= -4xyz²
b) -8x²y - x.(xy)
= -8x²y - x²y
= -9x²y
c) 4xy².x - (-12x²y²)
= 4x²y² + 12x²y²
= 16x²y²
d) 1/2 x²y³ - 1/3 x²y.y²
= 1/2 x²y³ - 1/3 x²y³
= 1/6 x²y³
e) 3xy(x²y) - 5/6 x³y²
= 3x³y² - 5/6 x³y²
= 13/6 x³y²
f) 3/4 x⁴y - 1/6 xy.x³
= 3/4 x⁴y - 1/6 x⁴y
= 7/12 x⁴y
a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2
b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y
=>A-B=12xy^2-14x^2y
c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2
=>A-B=-5x^2y^3-x^3y^2
d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2
bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)
\(=\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x-x^2-2x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4x}{x^2-4}\)
Bài 2:
1: \(x^2y^2-8-1\)
\(=x^2y^2-9\)
\(=\left(xy-3\right)\left(xy+3\right)\)
2: \(x^3y-2x^2y+xy-xy^3\)
\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)
\(=xy\left(x^2-2x+1-y^2\right)\)
\(=xy\left[\left(x-1\right)^2-y^2\right]\)
\(=xy\left(x-1-y\right)\left(x-1+y\right)\)
3: \(x^3-2x^2y+xy^2\)
\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)
\(=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\)
4: \(x^2+2x-y^2+1\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
5: \(x^2+2x-4y^2+1\)
\(=\left(x^2+2x+1\right)-4y^2\)
\(=\left(x+1\right)^2-4y^2\)
\(=\left(x+1-2y\right)\left(x+1+2y\right)\)
6: \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
Đề bài sai, đề đúng thì phân thức đằng sau dấu chia phải là:
\(\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
Thu gọn đa thức:
\(C=-\dfrac{1}{2}x^2y-2xy+\dfrac{1}{2}x^2y-xy+xy-\dfrac{1}{3}x+\dfrac{1}{2}+x-0,25\)
\(=x^2y\left(-\dfrac{1}{2}+\dfrac{1}{2}\right)+xy\left(-2-1+1\right)+x\left(-\dfrac{1}{3}+1\right)+\dfrac{1}{2}-\dfrac{1}{4}\)
\(=-2xy+\dfrac{2}{3}x+\dfrac{1}{4}\)
a: \(\dfrac{xy^2}{xy-y}=\dfrac{y\cdot xy}{y\cdot\left(x-1\right)}=\dfrac{xy}{x-1}\)
=>Hai phân thức này bằng nhau
b: \(\dfrac{xy+y}{x}=\dfrac{y\left(x+1\right)}{x}\)
\(\dfrac{xy+x}{y}=\dfrac{x\left(y+1\right)}{y}\)
Vì \(\dfrac{y\left(x+1\right)}{x}\ne\dfrac{x\left(y+1\right)}{y}\)
nên hai phân thức này không bằng nhau
c: \(\dfrac{-6}{4y}=\dfrac{-6:2}{4y:2}=\dfrac{-3}{2y}\)
\(\dfrac{3y}{-2y^2}=\dfrac{-3y}{2y^2}=\dfrac{-3y}{y\cdot2y}=\dfrac{-3}{2y}\)
Do đó: \(\dfrac{-6}{4y}=\dfrac{3y}{-2y^2}\)
=>Hai phân thức này bằng nhau
\(A=x^2y^3\left(\dfrac{1}{5}+\dfrac{2}{3}-\dfrac{3}{4}+1\right)=\dfrac{67}{60}x^2y^3\)
\(B=x^6y^3\cdot\dfrac{1}{4}x^2y^4z^2=\dfrac{1}{4}x^8y^7z^2\)
\(A+B=\dfrac{67}{60}x^2y^3+\dfrac{1}{4}x^8y^7z^2\)
\(A-B=\dfrac{67}{60}x^2y^3-\dfrac{1}{4}x^8y^7z^2\)
A=x2y3(15+23−34+1)=6760x2y3A=x2y3(15+23−34+1)=6760x2y3
B=x6y3⋅14x2y4z2=14x8y7z2B=x6y3⋅14x2y4z2=14x8y7z2
A+B=6760x2y3+14x8y7z2A+B=6760x2y3+14x8y7z2
A−B=6760x2y3−14x8y7z2
\(=\dfrac{2y+x}{y\left(2y-x\right)}+\dfrac{8x}{\left(x-2y\right)\left(x+2y\right)}+\dfrac{2y-x}{y\left(2y+x\right)}\)
\(=\dfrac{2y+x}{y\left(2y-x\right)}-\dfrac{8x}{\left(2y-x\right)\left(x+2y\right)}+\dfrac{2y-x}{y\left(2y+x\right)}\)
\(=\dfrac{\left(2y+x\right)^2}{y\left(2y-x\right)\left(2y+x\right)}-\dfrac{8xy}{y\left(2y-x\right)\left(x+2y\right)}+\dfrac{\left(2y-x\right)^2}{y\left(2y+x\right)\left(2y-x\right)}\)
\(=\dfrac{\left(2y+x\right)^2-8xy+\left(2y-x\right)^2}{y\left(2y-x\right)\left(2y+x\right)}\)
\(=\dfrac{8y^2-8xy+2x^2}{\left(y\right)\left(2y-x\right)\left(2y+x\right)}\)
Phân tích trên tử ta có:
\(=2\left(\left(2y\right)^2+4xy+x^2\right)=2\left(2y+x\right)^2\)
\(=\dfrac{2\left(2y+x\right)^2}{y\left(2y+x\right)\left(2y-x\right)}=\dfrac{2\left(2y+x\right)}{y\left(2y-x\right)}\)
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