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a)\(\dfrac{12x^3y^2}{18xy^5}\)=\(\dfrac{2x^2}{3y^3}\)
b)\(\dfrac{15x.\left(x+5\right)^2}{20x^2.\left(x+5\right)}\)=\(\dfrac{3.5x\left(x+5\right)}{4x.5x.\left(x+5\right)}\)=\(\dfrac{3\left(x+5\right)}{4x}\)
1)\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\frac{2y}{5\left(x+y\right)^2}\)
2) \(\frac{15x\left(x+y\right)^2}{20x^2\left(x+5\right)}=\frac{3\left(x^2+2xy+y^2\right)}{4x\left(x+5\right)}=\frac{3\left(x+y\right)^2}{4x^2+20x}\)
3) \(\frac{15x\left(x-y\right)}{3\left(y-x\right)}=\frac{5x\left(x-y\right)}{-3\left(x-y\right)}=-\frac{5x}{3}\)
4)\(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{\left(y-x\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x+y\right)}{\left(x-y\right)^2}\)
\(a,=\dfrac{2y^4}{3x\left(2x-3y\right)}\\ b,=-\dfrac{2y\left(3x-1\right)^2}{3x^2}\\ c,=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)}{2x+3}\\ d,=\dfrac{5x\left(x-2y\right)}{-2\left(x-2y\right)^3}=-\dfrac{5x}{2\left(x-2y\right)^2}\)
\(P=\dfrac{15x^5y^3-10x^3y^2+20x^4y^4}{5x^2y^2}\)
\(=\dfrac{15x^5y^3}{5x^2y^2}-\dfrac{10x^3y^2}{5x^2y^2}+\dfrac{20x^4y^4}{5x^2y^2}\)
\(=3x^3y-2x+4x^2y^2\)
Khi x=-1 và y=2 thì \(P=3\cdot\left(-1\right)^3\cdot2-2\cdot\left(-1\right)+4\cdot\left(-1\right)^2\cdot2^2\)
\(=-6+2+16=4+16=20\)
1) \(\dfrac{15-5x}{5x^2-15x}=\dfrac{5\left(3-x\right)}{5x\left(x-3\right)}=-\dfrac{5\left(x-3\right)}{5x\left(x-3\right)}=-\dfrac{1}{x}\)
Chọn A
2) \(\dfrac{x\left(x-5\right)}{x^2+25}=\dfrac{x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{x}{x+5}\)
\(A=0\Leftrightarrow\dfrac{x}{x+5}=0\Leftrightarrow x=0\)
Chọn B
3) \(\dfrac{2x-5}{5-2x}=-\dfrac{5-2x}{5-2x}=-1\)
Chọn D
b: \(\dfrac{\left(3x+2\right)^2-\left(x+2\right)^2}{x^3-x^2}\)
\(=\dfrac{\left(3x+2+x+2\right)\left(3x+2-x-2\right)}{x^2\left(x-1\right)}\)
\(=\dfrac{\left(4x+4\right)\cdot2x}{x^2\left(x-1\right)}\)
\(=\dfrac{8x\left(x+1\right)}{x^2\left(x-1\right)}=\dfrac{8}{x}\)
a: Ta có: \(\left(x-2\right)^2-\left(2x-1\right)^2+\left(3x-1\right)\left(x-5\right)\)
\(=x^2-4x+4-4x^2+4x-1+3x^2-15x-x+5\)
\(=-16x+8\)
b: Ta có: \(\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
=27x-55
\(a,\dfrac{10x^3y^2}{20xy^5}=\dfrac{x^2}{2y^3}\\ b,\dfrac{15\left(x+5\right)^2}{20x\left(x+5\right)}=\dfrac{3\left(x+5\right)}{4x}\)
a) \(\dfrac{x^2}{2y^3}\)
b) \(\dfrac{3\left(x+5\right)}{4}\)=\(\dfrac{3x+15}{4}\)