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a) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\) MTC: \(2x\left(x+3\right)\)
\(=\dfrac{3x}{2x\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-\left(x-6\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
b) \(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{5x+6}{4-x^2}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{x^2-4}\)
\(=\dfrac{4}{x+2}+\dfrac{2}{x-2}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\) MTC: \(\left(x-2\right)\left(x+2\right)\)
\(=\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4\left(x-2\right)+2\left(x+2\right)-\left(5x+6\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{4x-8+2x+4-5x-6}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x-10}{\left(x-2\right)\left(x+2\right)}\)
c) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{2x\left(2x-1\right)}\) MTC: \(2x\left(2x-1\right)\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}-\dfrac{3x-2}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{-2x+1}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{-1}{2x}\)
d) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(=\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2}{x^2+x+1}-\dfrac{1}{x-1}\) MTC: \(\left(x-1\right)\left(x^2+x+1\right)\)
\(=\dfrac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{2\left(x-2\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x^2+2\right)+2\left(x-2\right)-\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2+2+2x-4-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{-3+x}{\left(x-1\right)\left(x^2+x+1\right)}\)
a)\(\frac{3+2x}{2+x}-1=\frac{2-x}{2+x}\) (x khác -2)
\(\Leftrightarrow\frac{3+2x}{2+x}-\frac{2-x}{2+x}=1\)
\(\Leftrightarrow\frac{1+3x}{2+x}=1\)
\(\Leftrightarrow1+3x=2+x\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
b) \(\frac{5-2x}{3}+\frac{x^2-1}{3}x-1=\frac{\left(x-2\right)\left(1-3x\right)}{9x-3}\) (x khác 1/3)
\(\Leftrightarrow\frac{x^3-3x+5}{3}+\frac{\left(x-2\right)\left(3x-1\right)}{3\left(3x-1\right)}=1\)
\(\Leftrightarrow\frac{x^2-2x+3}{3}=1\)
\(\Leftrightarrow x\left(x-2\right)=0\Leftrightarrow\left[\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\frac{1}{\left(3-2x\right)^2}-\frac{4}{\left(3+2x\right)^2}=\frac{3}{9-4x^2}\) (x khác +- 3/2)
\(\Leftrightarrow\frac{\left(3+2x\right)^2}{\left(3+2x\right)^2\left(3-2x\right)^2}-\frac{4\left(3-2x\right)^2}{\left(3+2x\right)^2\left(3-2x\right)^2}=\frac{9}{\left(3+2x\right)^2\left(3-2x\right)^2}\)
\(\Leftrightarrow9+12x+4x^2-4\left(9-12x+4x^2\right)-9=0\)
\(\Leftrightarrow-12x^2+60x-36=0\)
\(\Leftrightarrow-12\left(x^2-5x+3\right)=0\Leftrightarrow x^2-5x+3=0\)
\(\Rightarrow\Delta=b^2-4ac=25-12=13>0\)
\(x_1=\frac{-b+\sqrt{\Delta}}{2ac}=\frac{5+\sqrt{13}}{6}\)
\(x_2=\frac{5-\sqrt{13}}{6}\)
d) \(\frac{1}{x^2+2x+1}=\frac{4}{x+2x^2+x^3}=\frac{5}{2x+2x^2}\)
\(\Leftrightarrow\frac{x^2+2x+1}{1}=\frac{x+2x^2+x^3}{4}=\frac{2x+2x^2}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\frac{x^2+2x+1}{1}=\frac{x+2x^2+x^3}{4}=\frac{2x+2x^2}{5}=\frac{x^2+2x+1-\left(x+2x^2+x^3\right)+2x+2x^2}{1-4+5}\)
(dấu bằng thứ nhất của câu d là dấu cộng à???)
b: \(\Leftrightarrow\dfrac{2}{\left(x+7\right)\left(x-3\right)}=\dfrac{3x+21}{\left(x-3\right)\left(x+7\right)}\)
=>3x+21=2
=>x=-19/3
d: \(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=8\)
\(\Leftrightarrow4x^2+4x+1-4x^2+4x-1=8\)
=>8x=8
hay x=1
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
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: \(\dfrac{2x^2+1}{x^3+1}-\dfrac{x-1}{x^2-x+1}-\dfrac{1}{x+1}\)
\(=\dfrac{2x^2+1-x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\)
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) \(\dfrac{2x^2}{x-y}.\dfrac{y}{5x^3}\)
\(=\dfrac{2x^2.y}{\left(x-y\right).5x^3}\)
\(=\dfrac{2y}{5x\left(x-y\right)}\)
a) \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)
\(=\dfrac{x^3+2x}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)
\(=\dfrac{x^3+2x}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+2x+2x\left(x+1\right)+\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+2x+2x^2+2x+x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{x^3+3x^2+3x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)^3}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2}{x^2-x+1}\)
b) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{4x^2-2x}\)
\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{2x\left(2x-1\right)}\) MTC: \(2x\left(2x-1\right)\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}-\dfrac{3x-2}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(1-3x\right)\left(2x-1\right)+2x\left(3x-2\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{\left(2x-1-6x^2+3x\right)+\left(6x^2-4x\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)
\(=\dfrac{-2x+1}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(2x-1\right)}{2x\left(2x-1\right)}\)
\(=\dfrac{-1}{2x}\)