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a) \(-\frac{3}{4}x+\frac{1}{6}x=1-2\frac{5}{9}\)
\(\left(-\frac{3}{4}+\frac{1}{6}\right).x=1-\frac{23}{9}\)
\(-\frac{7}{12}.x=-\frac{14}{9}\)
\(x=-\frac{14}{9}:\left(-\frac{7}{12}\right)\)
\(x=\frac{8}{3}\)
Vậy x = ...
b) \(\left|2x-\frac{3}{8}\right|+2\frac{3}{4}=3\frac{1}{16}\)
\(\left|2x-\frac{3}{8}\right|+\frac{11}{4}=\frac{49}{16}\)
\(\left|2x-\frac{3}{8}\right|=\frac{49}{16}-\frac{11}{4}\)
\(\left|2x-\frac{3}{8}\right|=\frac{5}{16}\)
\(\Rightarrow\left|2x-\frac{3}{8}\right|\in\text{{}\frac{5}{16};-\frac{5}{16}\)}
Nếu, \(2x-\frac{3}{8}=\frac{5}{16}\)
\(2x=\frac{11}{16}\)
\(x=\frac{11}{32}\)
Nếu, \(2x-\frac{3}{8}=-\frac{5}{16}\)
\(2x=\frac{1}{16}\)
\(x=\frac{1}{32}\)
Vậy \(x\in\text{{}\frac{1}{32};\frac{11}{32}\)}
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a,\(\frac{1}{3}x-1=\frac{2}{3}\)\(< =>\frac{1}{3}x=\frac{2}{3}+1\)\(< =>x=\frac{5}{3}:\frac{1}{3}\)\(< =>x=5\)
câu b,d tương tự
câu c thì mk sợ lm theo cách của mk thì bn chưa học
bn muốn mk giải tiếp để tham khảo thì mk sẽ giải
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c) \(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{2};3\right\}\)
e) \(2\left|\frac{1}{2}x-\frac{1}{3}\right|-\frac{3}{2}=\frac{1}{4}\)
\(\Leftrightarrow2\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{4}+\frac{3}{2}=\frac{7}{4}\)
\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}:2=\frac{7}{4}.\frac{1}{2}=\frac{7}{8}\)
\(\Rightarrow\left[{}\begin{matrix}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\left(-\frac{7}{8}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{29}{12}\\x=\frac{-13}{12}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{29}{12};\frac{-13}{12}\right\}\)
Mấy bài này ko quá khó, tải MathPhoto trong đt về nó tự lm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\frac{3}{4}.x-\frac{9}{16}\right).\left(\frac{1}{3}+\frac{-3}{5}:x\right)=0\)
<=> \(\hept{\begin{cases}\frac{3}{4}.x-\frac{9}{16}=0\\\frac{1}{3}-\frac{3}{5}.\frac{1}{x}=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\\frac{3}{5x}=\frac{1}{3}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{9}{5}\end{cases}}\)
\(\left(x-\frac{1}{3}\right)\left(\frac{2}{5}+x\right)>0\)
<=> \(\hept{\begin{cases}x-\frac{1}{3}>0\\x+\frac{2}{5}>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-\frac{1}{3}< 0\\x+\frac{2}{5}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x>\frac{1}{3}\\x>\frac{-2}{5}\end{cases}}\)hoặc \(\hept{\begin{cases}x< \frac{1}{3}\\x< \frac{-2}{5}\end{cases}}\)
<=>\(x>\frac{1}{3}\)hoặc \(x< \frac{-2}{5}\)
câu c tương tự nha
học tốt
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a) \(\left|2x-1\right|=5\)
\(2x-1=5\)
\(2x=6\)
\(x=3\)
b) \(x:4\frac{1}{3}=-2,5\)
\(x:\frac{4}{3}=\frac{-5}{2}\)
\(x=\frac{-5}{2}.\frac{4}{3}=\frac{-10}{3}\)
c) \(x^3-\frac{9}{10}.x=0\)
\(xx^2-\frac{9}{10}.x=0\)
\(x\left(x^2-\frac{9}{10}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x^2-\frac{9}{10}=0\Rightarrow x^2=\frac{9}{10}\Rightarrow x=\sqrt{\frac{9}{10}}\end{cases}}\)
Câu c có 2 kết quả là \(\hept{\begin{cases}x=0\\x=\sqrt{\frac{9}{10}}\end{cases}}\)
Câu b nhầm
\(x:4\frac{1}{3}=-2,5\)
\(x:\frac{13}{3}=\frac{-5}{2}\)
\(x=\frac{-5}{2}.\frac{13}{3}=\frac{-65}{6}\)
Lời giải:
$3\frac{1}{3}x+16\frac{3}{4}=-13,25$
$\frac{10}{3}x+16,75=-13,25$
$\frac{10}{3}x=-13,25-16,75=-30$
$x=-30:\frac{10}{3}=-9$