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a) \(\left|x\left(x-7\right)\right|=x\)
\(\Rightarrow\orbr{\begin{cases}x\left(x-7\right)=x\\x\left(x-7\right)=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=6\end{cases}}}\)
b) \(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)
\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow4x+2,8=5x\)
\(\Leftrightarrow x=2,8\)
\(a.\)\(\left|x.\left(x-7\right)\right|=x\)( Đk: \(x\ge0\))
\(\Leftrightarrow\orbr{\begin{cases}x.\left(x-7\right)=x\\x.\left(x-7\right)=-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=x:x\\x-7=-x:x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1+7\\x=-1+7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=8\\x=6\end{cases}}\)
\(b.\)\(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)( Đk: \(5x\ge0\Leftrightarrow x\ge0\))
\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)
\(\Leftrightarrow\left(x+x+x+x\right)+\left(-1,1+1,2+1,3+1,4\right)=5x\)
\(\Leftrightarrow4x+2,8=5x\)
\(\Leftrightarrow2,8=5x-4x\)
\(\Leftrightarrow x=2,8\)
\(c.\)\(7^{x+2}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.7^{x+3}+2.7^{x-1}=345\)
\(\Leftrightarrow7^{x-1}.\left(7^{x+3}+2\right)=345\)
\(......................\)
Đến đây mk ko bt làm nữa, tự lm nhé !
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
Ta có : 2xy - x - y = 2
<=> 2xy - x = 2 + y
<=> x(2y - 1) = y + 2
=> x = \(\frac{y+2}{2y-1}\)
Vì x nguyên nên \(\frac{y+2}{2y-1}\) nguyên
Ta có ; \(\frac{y+2}{2y-1}=\frac{2y+4}{2y-1}=\frac{\left(2y-1\right)+5}{2y-1}=\frac{2y-1}{2y-1}+\frac{5}{2y-1}=1+\frac{5}{2y-1}\)
Để \(\frac{y+2}{2y-1}\) nguyên thì \(\frac{5}{2y-1}\) nguyên
Suy ra : 2y - 1 \(\in\) Ư(5) = {-5;-1;1;5}
Ta có bảng :
| 2y - 1 | -5 | -1 | 1 | 5 |
| 2y | -4 | 0 | 2 | 6 |
| y | -2 | 0 | 1 | 3 |
| x | 0 | -2 | 3 | 1 |
Cảm ơn bạn nhá 
Nhưng mà \(\dfrac{y+2}{2y-1}\) làm sao mà bằng \(\dfrac{2y+4}{2y-1}\)
Phải \(2x\) mới bằng \(\dfrac{2y+4}{2y-1}\) được chứ 
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
