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2)
a) \(\dfrac{1}{x}.\dfrac{6x}{y}\)
\(=\dfrac{6x}{xy}\)
\(=\dfrac{6}{y}\)
b) \(\dfrac{2x^2}{y}.3xy^2\)
\(=\dfrac{2x^2.3xy^2}{y}\)
\(=\dfrac{6x^3y^2}{y}\)
\(=6x^3y\)
c) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}\)
\(=\dfrac{15x.2y^2}{7y^3.x^2}\)
\(=\dfrac{30xy^2}{7x^2y^3}\)
\(=\dfrac{30}{7xy}\)
d) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
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: \(=\dfrac{1-2x+3+2y+2y-4}{6x^3y}=\dfrac{-2x+4y}{6x^3y}=\dfrac{-2\left(x-2y\right)}{6x^3y}=\dfrac{-x+2y}{3x^3y}\)
b: \(=\dfrac{x^2-2+2-x}{x\left(x-1\right)^2}=\dfrac{x\left(x-1\right)}{x\left(x-1\right)^2}=\dfrac{1}{x-1}\)
c: \(=\dfrac{3x+1+x^6-3x}{x^2-3x+1}\)
\(=\dfrac{x^6+1}{x^2-3x+1}\)
d: \(=\dfrac{x^2+38x+4+3x^2-4x-2}{2x^2+17x+1}\)
\(=\dfrac{4x^2+34x+2}{2x^2+17x+1}=2\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}=\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9-6x}{x\left(x-3\right)}=\dfrac{x^2-6x+9}{x\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}=\dfrac{x-3}{x}\)
a) \(x-1-\dfrac{x^2-4}{x+1}=\dfrac{\left(x-1\right)\left(x+1\right)-\left(x^2-4\right)}{x+1}=\dfrac{x^2-1-x^2+4}{x+1}=\dfrac{3}{x+1}\)
b) \(\dfrac{3x-1}{2\left(3x+1\right)}+\dfrac{3x+1}{2\left(3x-1\right)}-\dfrac{6x}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{\left(3x-1\right)^2+\left(3x+1\right)^2-12x}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{9x^2-6x+1+9x^2+6x+1-12x}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{18x^2-12x+2}{2\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{9x^2-6x+1}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{3x-1}{3x+1}\)
c: \(=\dfrac{1}{x-2}-\dfrac{x+4}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x\left(x+2\right)}\)
\(=\dfrac{x\left(x+2\right)-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x-x^2-6x+4}{x\left(x-2\right)\left(x+2\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
d: Sửa đề: \(\dfrac{2x^2+1}{x^3+1}-\dfrac{x-1}{x^2-x+1}-\dfrac{1}{x+1}\)
\(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
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: \(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{6x^2y^2}\cdot\dfrac{3xy}{x+y}\)
\(=\dfrac{x-y}{2xy}\)