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\(\left(\dfrac{9}{x^2-9}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\) ( sửa đề \(x^3-9\) thành \(x^2-9\) )
\(=\left(\dfrac{9}{\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\left(\dfrac{9}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-3}{\left(x-3\right)\left(x+3\right)}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\dfrac{9+x-3}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\left(\dfrac{3\left(x-3\right)}{3x\left(x+3\right)}-\dfrac{x.x}{3x\left(x+3\right)}\right)\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{\left(x+6\right)3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(3x-9-x^2\right)}\)
\(=\dfrac{3x\left(x+6\right)}{\left(x-3\right)\left(3x-9-x^2\right)}\)
d. ĐKXĐ: x khác 1, x khác 3
\(\dfrac{x+5}{x-1}=\dfrac{x+1}{\left(x-3\right)}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+5\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\) \(\Leftrightarrow x^2+2x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+1+8=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\) (loại)
Vậy pt vô nghiệm
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
1)
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)
\(\Leftrightarrow x=105\)
b)
\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)
\(\Leftrightarrow50-x=0\)
\(\Leftrightarrow x=50\)
2)
\(\left(5x+1\right)^2=\left(3x-2\right)^2\)
\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)
b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)
\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)
\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)
\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)
\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
\(x^2-9=\left(x-3\right)\left(x+3\right)\)
\(\left(\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{x^2-9}\right):\left(1-\dfrac{1}{x+3}\right)=\left(\dfrac{3\left(x+2\right)}{x^2-9}\right):\left(\dfrac{x+2}{x+3}\right)=\left(x-3\right)\)