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Tìm x
\(a,2x-25\%=\frac{1}{2}\)
\(b,\left(\frac{3x}{7}+1\right).\left(-0,25\right)=\frac{-1}{28}\)
\(\)
a) \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4}{5}\)
\(\Leftrightarrow2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4}{5}\)
\(\Leftrightarrow2\times\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{4}{5}\)
\(\Leftrightarrow2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4}{5}\)
\(\Leftrightarrow2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{4}{5}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{4}{5}:2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{2}{5}-\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{4}{10}-\frac{5}{10}=\frac{-1}{10}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{4}{10}-\frac{5}{10}=\frac{1}{-10}\)
\(\Leftrightarrow x+1=-10\)
\(\Leftrightarrow x=-10-1\)
\(\Leftrightarrow x=-11\)
Hông chắc !!! <3
b) Đề khó hiểu vậy, nếu đề là : \(x+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)thì làm như sau nha
\(x+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)
\(\Leftrightarrow x+\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right)=1\)
\(\Leftrightarrow x+1=1\)
\(\Leftrightarrow x=1-1\)
\(\Leftrightarrow x=0\)
Rất vui vì giúp đc bạn <3
Cô mk giao thế, bó tay.com. Ko bỏ trị tuyệt đối đi vô lý như thế chứ
\(\left|x-3\right|=2x+4\)
\(\left|x-3\right|=2x+2\cdot2\)
\(\left|x-3\right|=2\left(x+2\right)\)
\(\Rightarrow\orbr{\begin{cases}x-3=-\left[2\cdot\left(x+2\right)\right]\\x-3=2\left(x+2\right)\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x-3=-\left[2x+4\right]\\x-3=2x+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=-2x-4\\x=2x+2+3\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-2x-4+3\\x=2x+5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2x-1\\x=2x+5\end{cases}}\) \(.....................\)
1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)
\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)
\(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)
\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)
\(x=\frac{45}{4}+\frac{9}{4}\)
\(x=\frac{27}{2}\)
GIẢI
( x + 1/5 ) ^2 - 9/25 = 0
( x + 1/5 ) ^2 = 0 + 9/25
( x + 1/5 ) ^2 = 9/25
( x + 1/5 ) ^2 = ( 3/5 )^2
x + 1/5 = 3/5
x = 3/5 - 1/5
x = 2/5
HỌC TỐT ^_^
\(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
Bài làm
a) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\) b) \(\left(x-\frac{1}{2}\right)^2=\frac{4}{25}\)
=> \(\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\) => \(\left(x-\frac{1}{2}\right)^2=\left(\frac{2}{5}\right)^2\)
=> \(x-\frac{1}{2}=\frac{1}{3}\) => \(x-\frac{1}{2}=\frac{2}{5}\)
\(x=\frac{1}{3}+\frac{1}{2}\) \(x=\frac{2}{5}+\frac{1}{2}\)
\(x=\frac{2}{6}+\frac{3}{6}\) \(x=\frac{4}{10}+\frac{5}{10}\)
\(x=\frac{5}{6}\) \(x=\frac{9}{10}\)
Vậy \(x=\frac{5}{6}\) Vậy \(x=\frac{9}{10}\)
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