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a, 0,4 : x = x : 0,9
<=> x2 = 0,4 . 0,9
<=> x2 = 0,36
<=> x = 0,6 hoặc -0,6
b, \(13\frac{1}{3}\div1\frac{1}{3}=26\div\left(2x-1\right)\)
\(\Leftrightarrow\frac{40}{3}\div\frac{4}{3}=26\div\left(2x-1\right)\)
\(\Leftrightarrow10=26\div\left(2x-1\right)\)
\(\Leftrightarrow2x-1=\frac{13}{5}\)
\(\Leftrightarrow2x=\frac{18}{5}\)
\(\Leftrightarrow x=\frac{9}{5}\)
c, \(0,2\div1\frac{1}{5}=\frac{2}{3}\div\left(6x+7\right)\)
\(\Leftrightarrow\frac{1}{5}\div\frac{6}{5}=\frac{2}{3}\div\left(6x+7\right)\)
\(\Leftrightarrow\frac{1}{6}=\frac{2}{3}\div\left(6x+7\right)\)
\(\Leftrightarrow6x+7=4\)
\(\Leftrightarrow6x=-3\)
\(\Leftrightarrow x=\frac{-1}{2}\)
d, \(\frac{37-x}{x+13}=\frac{3}{7}\)
\(\Leftrightarrow7\left(37-x\right)=3\left(x+13\right)\)
\(\Leftrightarrow259-7x=3x+39\)
\(\Leftrightarrow-10x=-220\)
\(\Leftrightarrow x=22\)
a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)
a, \(\frac{3}{5}\left(2x-\frac{1}{3}\right)+\frac{4}{15}=\frac{12}{30}\)
\(\Leftrightarrow\frac{3}{5}\left(2x-\frac{1}{3}\right)=\frac{2}{15}\)
\(\Leftrightarrow2x-\frac{1}{3}=\frac{2}{9}\)
\(\Leftrightarrow2x=\frac{5}{9}\)
\(\Leftrightarrow x=\frac{5}{18}\)
b,\(\left(-0,2\right)^x=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{-1}{5}\right)^x=\left(\frac{-1}{5}\right)^2\)
\(\Leftrightarrow x=2\)
c,\(\left|x-1\right|-\frac{3}{12}=\left(-\frac{1}{2}\right)^2\)
\(\Leftrightarrow\left|x-1\right|-\frac{3}{12}=\frac{1}{4}\)
\(\Leftrightarrow\left|x-1\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{1}{2}\\x-1=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)
\(a,\frac{3}{5}\left(2x-\frac{1}{3}\right)=\frac{12}{30}-\frac{4}{15}\)
\(\frac{3}{5}\left(2x-\frac{1}{3}\right)=\frac{2}{15}\)
\(2x-\frac{1}{3}=\frac{2}{9}\)
\(x=\frac{5}{18}\)
\(b,\left(-0,2\right)^x=\frac{1}{25}\)
\(\left(-0,2\right)^x=\left(-\frac{1}{5}\right)^2\)
\(\left(-0,2\right)^x=\left(-0,2\right)^2\)
\(x=2\)
c,/x-1/=1/2
Nếu
\(x-1\ge0\)
\(x\ge1\)
suy ra x-1=1/2
x=3/2(thỏa mãn điều kiện )
nếu \(x-1\le0\)
\(x\le1\)
suy ra x-1=-1/2
x=1/2 (thỏa mãn điều kiện )
Vậy ...
nha !!!