Bài 4: Quy đồng mẫu thức nhiều phân thức

Bài 1:

a: \(\dfrac{\left(x+y\right)^2-\left(x-y\right)^2}{\left(y+z\right)^2-\left(y-z\right)^2}\)

\(=\dfrac{\left(x+y-x+y\right)\left(x+y+x-y\right)}{\left(y+z+y-z\right)\left(y+z-y+z\right)}\)

\(=\dfrac{2y\cdot2x}{2y\cdot2z}=\dfrac{xy}{yz}=\dfrac{x}{z}\)

b: \(\dfrac{\left(x+y\right)^2-\left(xy+1\right)^2}{\left(x+1\right)\left(y-1\right)}\)

\(=\dfrac{\left(x+y-xy-1\right)\left(x+y+xy+1\right)}{\left(x+1\right)\left(y-1\right)}\)

\(=\dfrac{\left[x\left(1-y\right)-\left(1-y\right)\right]\left(x+1\right)\left(y+1\right)}{\left(x+1\right)\left(y-1\right)}\)

\(=\dfrac{-\left(x-1\right)\left(y-1\right)\left(y+1\right)}{y-1}=-\left(x-1\right)\left(y+1\right)\)

VD3:

\(\dfrac{x^6-3x^4y^2+3x^2y^4-y^6}{x^2-y^2}=\dfrac{\left(x^2-y^2\right)^3}{x^2-y^2}\)

\(=\left(x^2-y^2\right)^2\)

\(=x^4-2x^2y^2+y^4\)

\(VD1:\dfrac{4x^3-8x^2-x+2}{2x+1}=\dfrac{4x^2\left(x-2\right)-\left(x-2\right)}{\left(2x+1\right)}=\dfrac{\left(4x^2-1\right)\left(x-2\right)}{\left(2x+1\right)}=\dfrac{\left(2x-1\right)\left(2x+1\right)\left(x-2\right)}{2x+1}=\left(2x-1\right)\left(x-2\right)\\ VD2:\dfrac{x^5-2x^4+x^3}{x^4-2x^3+x^2}=\dfrac{x\left(x^4-2x^3+x^2\right)}{x^4-2x^3+x^2}=x=VP\)

\(\dfrac{x+1}{2x^2-x^4}=\dfrac{x+1}{x^2\left(2-x^2\right)}=\dfrac{-\left(x+1\right)\left(x^4+2x^2+4\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\)

\(\dfrac{x}{x^4+2x^2+4}=\dfrac{x}{x^4+2x^2+4}=\dfrac{x^3\left(x^2-2\right)}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\)

\(\dfrac{2x-1}{x^7-8x}=\dfrac{2x-1}{x\left(x^6-8\right)}=\dfrac{2x-1}{x\left(x^2-2\right)\left(x^4+2x^2+4\right)}=\dfrac{2x^2-x}{x^2\left(x^2-2\right)\left(x^4+2x^2+4\right)}\)

`MTC:2x(x+2)^3`

Có:

  `[2x]/[(x+2)^3]=[4x^2]/[(x+2)^3]`

  `[x-2]/[2x(x+2)^2]=[(x-2)(x+2)]/[2x(x+2)^3]=[x^2-4]/[2x(x+2)^3]`

\(\dfrac{2x}{\left(x+2\right)^3}=\dfrac{4x^2}{2x\left(x+2\right)^3}\)

\(\dfrac{x-2}{2x\left(x+2\right)^2}=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^3}\)

14:

a: \(\dfrac{x}{x^3+1}=\dfrac{x}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x^2}{x\left(x+1\right)\left(x^2-x+1\right)}\)

\(\dfrac{x+1}{x^2+x}=\dfrac{1}{x}=\dfrac{x^3+1}{x\left(x+1\right)\left(x^2-x+1\right)}\)

\(\dfrac{x+2}{x^2-x+1}=\dfrac{x\left(x+2\right)\left(x+1\right)}{x\left(x+1\right)\left(x^2-x+1\right)}\)

b: \(\dfrac{x-1}{x+1}=\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\)

\(\dfrac{x+1}{x-1}=\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{1}{x^2-1}=\dfrac{1}{\left(x+1\right)\left(x-1\right)}\)

c: \(\dfrac{x}{x^3-xy^2}=\dfrac{x}{x\left(x-y\right)\left(x+y\right)}=\dfrac{1}{\left(x-y\right)\left(x+y\right)}=\dfrac{x^2-y^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

\(\dfrac{1}{\left(x+y\right)^2}=\dfrac{\left(x-y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

\(\dfrac{1}{\left(x-y\right)^2}=\dfrac{\left(x+y\right)^2}{\left(x-y\right)^2\cdot\left(x+y\right)^2}\)

d: \(\dfrac{x^2+xy}{\left(x+y\right)^2}=\dfrac{x\left(x+y\right)}{\left(x+y\right)^2}=\dfrac{x}{x+y}=\dfrac{x^2-xy}{\left(x+y\right)\left(x-y\right)}\)

\(\dfrac{y^2-xy}{\left(x-y\right)^2}=\dfrac{-y\left(x-y\right)}{\left(x-y\right)^2}=\dfrac{-y}{x-y}=\dfrac{-y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)

\(\dfrac{2xy}{x^2-y^2}=\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\)

e: \(\dfrac{5x^2}{x^2+5x+6}=\dfrac{5x^2\left(x+5\right)}{\left(x+5\right)\left(x+2\right)\left(x+3\right)}\)

\(\dfrac{2x+3}{x^2+7x+10}=\dfrac{\left(2x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)}\)

\(-5=\dfrac{-5\left(x+2\right)\left(x+5\right)\left(x+3\right)}{\left(x+2\right)\left(x+5\right)\left(x+3\right)}\)

\(=\dfrac{x+1}{2x}-\dfrac{4x-6}{5x\left(x-1\right)}-\dfrac{2x}{5\left(x-1\right)}\)

\(=\dfrac{x+1}{2x}+\dfrac{-4x+6-2x^2}{5x\left(x-1\right)}\)

\(=\dfrac{5\left(x-1\right)\left(x+1\right)}{10x\left(x-1\right)}+\dfrac{-8x+12-4x^2}{10x\left(x-1\right)}\)

\(=\dfrac{5x^2-5-8x+12-4x^2}{10x\left(x-1\right)}\)

\(=\dfrac{x^2-8x+7}{10x\left(x-1\right)}=\dfrac{x-7}{10x}\)

a; MTC=35(x+1)

b: MTC=3(x+1)

c: MTC=12x(x-1)

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