Tìm x, biết: \(|x - 3,2| + \left|2x - \frac{1}{5}\right| = x + 3\)
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\(\frac{x+5}{3}=\frac{x-1}{4}\)
\(\Rightarrow\left(x+5\right).4=\left(x-1\right).3\)
\(\Rightarrow4x+20=3x-3\)
\(\Rightarrow4x-3x=-3-20\Rightarrow x=-23\)
\(\frac{x+5}{3}=\frac{x-1}{4}\)
\(\Rightarrow\left(x+5\right)\cdot4=\left(x-1\right)\cdot3\)
\(4x+20=3x-3\)
\(4x-3x=-3-20\)
\(x=-23\)
Vậy \(x=-23\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
Để 1 phân số được xác định thì mẫu số của chúng phải khác 0
BÀI LÀM
ĐKXĐ: \(\left(x-1\right)\left(-2x+8\right)\ne0\)
\(\Leftrightarrow\)\(-2\left(x-1\right)\left(x-4\right)\ne0\)
\(\Leftrightarrow\) \(\orbr{\begin{cases}x-1\ne0\\x-4\ne0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
Vậy....
\(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x-1\right)\left(x+3\right)}\left(x\ne-3;x\ne1\right)\)
\(\Leftrightarrow\frac{x+2}{x+3}-\frac{x+1}{x-1}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2}{\left(x+3\right)\left(x-1\right)}-\frac{x^2+4x+3}{\left(x-1\right)\left(x+3\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-2-x^2-4x-3-4}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3x-9}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-3}{x-1}=0\)
=> PT vô nghiệm
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2-\left(\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}+\frac{1}{4}\right)\left(\frac{1}{x}-\frac{2}{3}-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{5}{12}\right)\left(\frac{1}{x}-\frac{11}{12}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{5}{12}=0\\\frac{1}{x}-\frac{11}{12}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{5}{12}\\\frac{1}{x}=\frac{11}{12}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}\)
Vậy....
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}\)
\(\left|x-3,2\right|+\left|2x-\frac{1}{5}\right|=x+3.\)
ĐK : \(x+3\ge0\Leftrightarrow x\ge-3\)
Th1 : \(x-3,2+2x-\frac{1}{5}=x+3\)
\(x-3,2+2x=x+\frac{16}{5}\)
\(x+2x=x+\frac{32}{5}\)
\(2x=\frac{32}{5}\)
\(\Leftrightarrow x=3,2\)(tm)
\(x-3,2+2x-\frac{1}{5}=3-x\)
\(x-3,2+2x=3-x+\frac{1}{5}\)
\(x-3,2+2x=\frac{16}{5}-x\)
\(x+2x=\frac{16}{5}-x+3,2\)
\(x+2x=\frac{32}{5}-x\)
\(2x=\frac{32}{5}-x-x\)
\(2x=\frac{32}{5}-2x\)
\(4x=\frac{32}{5}\)
\(x=1,6\)(tm)
Vậy \(x=1,6\)hoặc \(x=3,2\)