bài 5:

1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)

2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)

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

3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)

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

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

4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)

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

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

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

5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)

\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)

\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)

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

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

Bài 3:

1: \(9x^3-xy^2\)

\(=x\cdot9x^2-x\cdot y^2\)

\(=x\left(9x^2-y^2\right)\)

\(=x\left(3x-y\right)\left(3x+y\right)\)

2: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

3: \(x^2-3xy-6x+18y\)

\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)

\(=x\left(x-3y\right)-6\left(x-3y\right)\)

\(=\left(x-3y\right)\left(x-6\right)\)

4: \(6xy-x^2+36-9y^2\)

\(=36-\left(x^2-6xy+9y^2\right)\)

\(=36-\left(x-3y\right)^2\)

\(=\left(6-x+3y\right)\left(6+x-3y\right)\)

5: \(x^4-6x^2+5\)

\(=x^4-x^2-5x^2+5\)

\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)

\(=\left(x^2-5\right)\left(x^2-1\right)\)

\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)

6: \(9x^2-6x-y^2+2y\)

\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)

\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)

\(=\left(3x-y\right)\left(3x+y-2\right)\)

a) ĐKXĐ: \(x\ne0\)

 \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)

\(=\dfrac{2\left(4x+1\right)+2x-3}{6x}\)

\(=\dfrac{10x-1}{6x}\)

b) ĐKXĐ: \(x,y\ne0\)

 \(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)

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

\(=\dfrac{x-y}{2xy}\)

a) Ta có: \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)

\(=\dfrac{2\left(4x+1\right)}{6x}+\dfrac{2x-3}{6x}\)

\(=\dfrac{8x+2+2x-3}{6x}\)

\(=\dfrac{10x-1}{6x}\)

b) Ta có: \(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)

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

\(=\dfrac{x-y}{2xy}\)

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

a: Khi x=1/3 thì \(A=3\cdot\dfrac{1}{9}-6\cdot\dfrac{1}{3}+5=\dfrac{10}{3}\)

Khi x=-2 thì \(A=3\cdot4-6\cdot\left(-2\right)+5=12+12+5=29\)

b: Trường hợp 1: x=2; y=-3

\(C=2\cdot2^2-3\cdot2\cdot\left(-3\right)+4\cdot\left(-3\right)^2=62\)

Trường hợp 2: x=-2; y=-3

\(C=2\cdot\left(-2\right)^2-3\cdot\left(-2\right)\cdot\left(-3\right)+4\cdot\left(-3\right)^2=26\)

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