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1)\(-\frac{3}{2}\left(\frac{4}{5}-\frac{2}{3}\right)+x=4\left(x-\frac{2}{3}\right)\)
\(\Leftrightarrow-\frac{3}{2}\cdot\frac{2}{15}+x=4x-\frac{8}{3}\)
\(\Leftrightarrow-\frac{1}{5}+x=4x-\frac{8}{3}\)
\(\Leftrightarrow x-4x=-\frac{8}{3}+\frac{1}{5}\)
\(\Leftrightarrow-3x=-\frac{37}{15}\)
\(\Leftrightarrow x=\frac{37}{45}\)
2)\(2\left(\frac{3}{2}-x\right)-\frac{1}{3}=7x-\frac{1}{4}\)
\(\Leftrightarrow3-2x-\frac{1}{3}=7x-\frac{1}{4}\)
\(\Leftrightarrow-2x-7x=-\frac{1}{4}-3+\frac{1}{3}\)
\(\Leftrightarrow-9x=-\frac{35}{12}\)
\(\Leftrightarrow x=\frac{35}{108}\)
3)\(\frac{1}{5}\left(-\frac{3}{5}10\right)+5x=x-\frac{2}{3}\)
\(\Leftrightarrow-\frac{6}{5}+5x=x-\frac{2}{3}\)
\(\Leftrightarrow5x-x=-\frac{2}{3}+\frac{6}{5}\)
\(\Leftrightarrow4x=\frac{8}{15}\)
\(\Leftrightarrow x=\frac{2}{15}\)
4)\(-\frac{3}{2}\left(5-\frac{1}{6}\right)+4\left(x-\frac{1}{2}\right)=1\)
\(\Leftrightarrow-\frac{15}{2}+\frac{1}{4}+4x-2=1\)
\(\Leftrightarrow4x=1+\frac{15}{2}-\frac{1}{4}+2\)
\(\Leftrightarrow4x=\frac{41}{4}\)
\(\Leftrightarrow x=\frac{41}{16}\)
1
2(\(\frac{3}{4}\)-5x)=\(\frac{4}{5}\)-3x
=> \(\frac{6}{4}-10x=\frac{4}{5}-3x\)
=>\(-10x+3x=\frac{4}{5}-\frac{6}{4}\)
=> \(x=\frac{1}{10}\)
2 .
\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
=>\(\frac{3}{2}-1+4x=\frac{2}{3}-7x\)
=>\(11x=\frac{1}{6}\)
=>x=\(\frac{1}{66}\)
3.
\(3\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
=>\(\frac{3}{2}-3x+\frac{1}{3}=\frac{7}{6}-x\)
=>\(-2x=\frac{-2}{3}\)
=>\(\frac{1}{3}\)
4. câu 4 ko hiểu bạn ơi
khỏi chép lại đề ha
- 2 - 4x - 5x + \(\frac{3}{2}\)= \(\frac{7}{4}\)
\(\frac{7}{2}\)- 9x = \(\frac{7}{4}\)
-9x = \(\frac{7}{2}-\frac{7}{4}\)
-9x = \(\frac{7}{4}\)
x = \(\frac{7}{4}:\left(-9\right)\)
x = \(\frac{-7}{36}\)
- 3 - 2x - \(\frac{1}{3}=7x-\frac{1}{4}\)
-2x - 7x = \(\frac{-1}{4}-3+\frac{1}{3}\)
-9x = \(\frac{-35}{12}\)
x = \(\frac{-35}{12}:\left(-9\right)\)
x = \(\frac{35}{108}\)
- \(\frac{-15}{2}\)+ \(\frac{1}{4}\)+ 4x -2 = 1
4x = 1 + \(\frac{15}{2}-\frac{1}{4}+2\)
4x = \(\frac{41}{4}\)
x = \(\frac{41}{4}:4\)
x = \(\frac{41}{16}\)
1,\(2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
\(\frac{6}{4}-10x=\frac{4}{5}-3x\)
\(\frac{6}{4}+\frac{4}{5}=7x\)
\(\frac{23}{10}=7x\)
\(\frac{23}{70}=x\)
2,\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)
\(\frac{3}{2}-1-4x=\frac{2}{3}-7x\)
\(\frac{3}{2}-1-\frac{2}{3}=-3x\)
\(\frac{-1}{6}=-3x\)
\(\frac{1}{18}=x\)
3,\(3\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(\frac{3}{2}-3x+\frac{1}{3}=\frac{7}{6}-x\)
\(\frac{3}{2}-\frac{7}{6}+\frac{1}{3}=2x\)
\(\frac{2}{3}=2x\)
\(\frac{1}{3}=x\)
4,mình không hiểu a ở đây là gì
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 \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
a) |7x + 1| - |5x + 6| = 0
Vì |7x - 1| \(\ge\)0\(\forall\)x
|5x + 6|\(\ge\)0 \(\forall\)x
Do đó : |7x - 1| + |5x + 6| \(\ge\)0\(\forall\)x
Và |7x - 1| + |5x + 6| = 0
<=> 7x - 1 = 0 <=> x = 1/7
và 5x + 6 = 0 và x = -6/5 (vô lí)
=> x \(\in\varnothing\)
b) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
<=> \(\frac{3}{2}x+\frac{1}{2}=4x-1\)hoặc\(\frac{3}{2}x+\frac{1}{2}=-4x+1\)
<=>\(\frac{3}{2}x-4x=\frac{-1}{2}-1\)hoặc \(\frac{3}{2}x+4x=\frac{-1}{2}+1\)
<=> \(\frac{-5}{2}x=\frac{-3}{2}\)hoặc \(\frac{11}{2}x=\frac{1}{2}\)
<=>\(x=\frac{-3}{2}:\frac{-5}{2}\)hoặc \(x=\frac{1}{2}:\frac{11}{2}\)
<=> \(x=\frac{3}{5}\)hoặc \(x=\frac{1}{11}\)
a) Ta có: \(5x\left(\frac{1}{5}x-2\right)+3\left(6-\frac{1}{3}x^2\right)=12\)
\(\Leftrightarrow x^2-10x+18-x^2=12\)
\(\Leftrightarrow-10x+18=12\)
\(\Leftrightarrow-10x=-6\)
hay \(x=\frac{3}{5}\)
Vậy: \(x=\frac{3}{5}\)
b) Ta có: \(7x\left(x-2\right)-5\left(x-1\right)=7x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5-7x^2-3=0\)
\(\Leftrightarrow-19x+2=0\)
\(\Leftrightarrow-19x=-2\)
hay \(x=\frac{2}{19}\)
Vậy: \(x=\frac{2}{19}\)