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\(\left|2x+1\right|+\left|x+8\right|=4x\) (*)
+)Xét \(x\ge-8\Rightarrow\)\(\begin{cases}2x+1\ge0\Rightarrow\left|2x+1\right|=2x+1\\x+8\ge0\Rightarrow\left|x+8\right|=x+8\end{cases}\) thì (*) thành:
\(2x+1+x+8=4x\)
\(\Rightarrow3x+9=4x\)
\(\Rightarrow x=9\) (thỏa mãn)
+)Xét \(-\frac{1}{2}\le x< -8\)\(\Rightarrow\begin{cases}x\ge-\frac{1}{2}\Rightarrow2x+1\ge0\Rightarrow\left|2x+1\right|=2x+1\\x< -8\Rightarrow x+8< 0\Rightarrow\left|x+8\right|=-\left(x+8\right)=-x-8\end{cases}\) thì (*)
thành: \(2x+1+\left(-x-8\right)=4x\)
\(\Leftrightarrow x-7=4x\)
\(\Leftrightarrow-3x=7\)
\(\Leftrightarrow x=-\frac{7}{3}\)( không thỏa mãn)
+)Xét \(x< -\frac{1}{2}\Rightarrow\)\(\begin{cases}2x+1< 0\Rightarrow\left|2x+1\right|=-\left(2x+1\right)=-2x-1\\x+8< 0\Rightarrow\left|x+8\right|=-\left(x+8\right)=-x-8\end{cases}\) thì (*) thành:
\(\left(-2x-1\right)+\left(-x-8\right)=4x\)
\(\Leftrightarrow-3x-9=4x\)
\(\Leftrightarrow-7x=9\)
\(\Leftrightarrow x=-\frac{9}{7}\) (không thỏa mãn)
Ta có :
\(3x=2y\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{2x}{4}\)
ADTCDTSBN , ta có :
\(\frac{x}{2}=\frac{y}{3}=\frac{2x}{4}=\frac{y-2x}{3-4}=\frac{5}{-1}=-5\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=-5\\\frac{y}{3}=-5\end{cases}\Rightarrow\hept{\begin{cases}x=-5.2=-10\\y=-5.3=-15\end{cases}}}\)
Vậy \(x=-10;y=-15\)
\(\left(2x-1\right)^7-\left(2x-1\right)^5=0\Leftrightarrow\left(2x-1\right)^5\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(2x-1\right)^5\left(2x-1-1\right)\left(2x-1+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^5\left(2x-2\right)2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^5=0\\2x-2=0\\2x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=0\end{matrix}\right.\)
\(\left(2x-1\right)^7-\left(2x-1\right)^5=0\)
\(\Leftrightarrow\left(2x-1\right)^5\left(\left(2x-1\right)^2-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^5\left(4x^2-4x+1-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^5\left(4x^2-4x\right)=0\)
\(\Leftrightarrow\left(2x-1\right)^54x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x-1=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=1\\x=0\end{matrix}\right.\)