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a, \(\left(\frac{5}{8}.0,6-5:3\frac{1}{3}\right).\left(\frac{1}{5}-1,4\right).\left(-5\right)^2.x=120\)
\(\left(\frac{5}{8}.\frac{3}{5}-5:\frac{10}{3}\right)\left(\frac{1}{5}-\frac{7}{5}\right).25x=120\)
\(\left(\frac{3}{8}-\frac{3}{2}\right)\left(-\frac{6}{5}\right).25x=120\)
\(\left(-\frac{9}{8}\right).\left(-\frac{6}{5}\right).25x=120\)
\(\frac{27}{20}.25x=120\)\(25x=120:\frac{27}{20}=\frac{800}{9}\)
\(x=\frac{800}{9}:25=\frac{32}{9}\)
đợi tí đi nấu cơm đã
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a) \(14:\frac{0,4x+0,6}{x}=7\)
\(\frac{0,4x+0,6}{x}=2\)
0,4x + 0,6 = 2.x
2x - 0,4x = 0,6
1,6x = 0,6
x = 0,375
b) \(\left(160\%+\frac{2}{3}x-x\right).12=660\)
\(\left(160\%+\frac{2}{3}x-x\right)=55\)
\(x\left(\frac{2}{3}-1\right)=53,4\)
\(-\frac{1}{3}x=\frac{267}{5}\)
\(x=\frac{267}{5}.\frac{3}{-1}\)
\(x=-160,2\)
c) \(1:\frac{1.2.3.4.....31}{2.2.2.3.2.4.....2.32}=2^x\)
\(1:\frac{1.2.3.4.....31}{2^{31}.2.3.4.....31.2^5}=2^x\)
\(1:\frac{1}{2^{36}}=2^x\)
\(2^{36}=2^x\)
\(x=36\)
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\(a,\left[\frac{4}{5}+\frac{2}{3}\right]:\frac{1}{5}-1,4\cdot\left[\frac{-5}{7}\right]^2\)
\(=\left[\frac{4\cdot3}{15}+\frac{2\cdot5}{15}\right]:\frac{1}{5}-1,4\cdot\frac{-5}{7}\cdot\frac{-5}{7}\)
\(=\left[\frac{12}{15}+\frac{10}{15}\right]:\frac{1}{5}-\frac{14}{10}\cdot\frac{25}{49}\)
\(=\frac{22}{15}:\frac{1}{5}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22}{15}\cdot\frac{5}{1}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22\cdot5}{15\cdot1}-\frac{7\cdot25}{5\cdot49}=\frac{22\cdot1}{3\cdot1}-\frac{1\cdot5}{1\cdot7}=\frac{22}{3}-\frac{5}{7}\)
= ...
Tự tính
Bài 2 : \(a,3-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}\)
\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}+3\)
\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{2}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{2}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{23}{3}\\x=\frac{-19}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{23}{3};\frac{-19}{3}\right\}\)
b, \(0,6-160\%< x\le3\frac{2}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{160}{100}< x\le\frac{11}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{8}{5}< x\le\frac{11}{3}\cdot\frac{18}{22}\)
\(\Rightarrow0,6-1,6< x\le3\)
\(\Rightarrow-1< x\le3\)
\(\Rightarrow x\in\left\{0;1;2;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)
\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)
\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(A=\frac{11}{3}\)(1)
+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)
\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)
\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)
Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)
=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)
=> \(\frac{11}{3}-x=1\)
=> \(x=\frac{11}{3}-1=\frac{8}{3}\)
Vậy x = 8/3
\(\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
\(\Rightarrow\left(-0,6x-\frac{1}{2}\right).\frac{3}{4}=\frac{1}{3}+\left(-1\right)=-\frac{2}{3}\)
\(\Rightarrow-0,6x-\frac{1}{2}=-\frac{2}{3}:\frac{3}{4}=-\frac{8}{9}\)
\(\Rightarrow-0,6x=-\frac{8}{9}+\frac{1}{2}=-\frac{7}{18}\)
\(\Rightarrow x=-\frac{7}{18}:\left(-0,6\right)=\frac{35}{54}\)
\(\left(-0,6.x-\frac{1}{2}\right)\times\frac{3}{4}-\left(-1\right)=\) \(\frac{1}{3}\)
\(\left(0,6x-\frac{1}{2}\right)\times\frac{3}{4}+1=\frac{1}{3}\)
\(\left(-0,6.x-\frac{1}{2}\right)\times\frac{3}{4}=\frac{1}{3}-1\)
\(\left(-0,6.x-\frac{1}{2}\right)\times\frac{3}{4}=\frac{-2}{3}\)
\(-0,6.x-\frac{1}{2}=\frac{-2}{3}:\frac{3}{4}\)
\(-0,6.x-\frac{1}{2}=\frac{-8}{9}\)
\(-0,6.x=\frac{-8}{9}+\frac{1}{2}\)
\(-0,6.x=\frac{-7}{18}\)
\(x=\frac{-7}{18}:\left(-0,6\right)=\frac{35}{54}\)