Tim:m
m+m X\(\frac{1}{3}\):\(\frac{4}{48}\)+m:\(\frac{4}{14}\)=330
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m) \(\frac{3}{4}-x=\frac{1}{5}\)
\(x=\frac{3}{4}-\frac{1}{5}\)
\(x=\frac{11}{20}\)
Vậy \(x=\frac{11}{20}\)
o) \(\frac{3}{5}x+\frac{1}{4}=\frac{1}{10}\)
\(\frac{3}{5}x=\frac{1}{10}-\frac{1}{4}\)
\(\frac{3}{5}x=-\frac{3}{20}\)
\(x=\left(-\frac{3}{20}\right):\frac{3}{5}\)
\(x=-\frac{1}{4}\)
Vậy \(x=-\frac{1}{4}\)
p) \(-\frac{4}{3}x+\frac{3}{2}=\frac{5}{6}\)
\(-\frac{4}{3}x=\frac{5}{6}-\frac{3}{2}\)
\(-\frac{4}{3}x=-\frac{2}{3}\)
\(x=\left(-\frac{2}{3}\right):\left(-\frac{4}{3}\right)\)
\(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
m) \(\frac{3}{4}\) - x = \(\frac{1}{5}\)
x= \(\frac{3}{4}\) - \(\frac{1}{5}\)
x= \(\frac{11}{20}\)
o) \(\frac{3}{5}\)x + \(\frac{1}{4}\) = \(\frac{1}{10}\)
\(\frac{3}{5}\)x = \(\frac{1}{10}\) - \(\frac{1}{4}\)
\(\frac{3}{5}\)x = -\(\frac{3}{20}\)
x = \(\frac{-3}{20}\):\(\frac{3}{5}\)
x = \(\frac{-1}{4}\)
p) \(\frac{-4}{3}\)x + \(\frac{3}{2}\) = \(\frac{5}{6}\)
\(\frac{-4}{3}\)x = \(\frac{5}{6}\) - \(\frac{3}{2}\)
\(\frac{-4}{3}\)x = \(\frac{-2}{3}\)
x = \(\frac{-2}{3}\) : \(\frac{-4}{3}\)
x = \(\frac{1}{2}\)
CHÚC BẠN HỌC TỐT
1) \(\frac{3}{4}-x=\frac{1}{5}\)<=> x=\(\frac{3}{4}-\frac{1}{5}=\frac{11}{20}\)
2) \(\frac{7}{2}+2x=-\frac{3}{4}\)<=> 2x=\(-\frac{3}{4}-\frac{7}{2}=-\frac{17}{4}\)<=> \(x=-\frac{17}{8}\)
3) \(\frac{1}{3}+\frac{2x}{3}=\frac{4}{3}\)<=> 1+2x=4<=> 2x=3<=> x=3/2
1) \(\frac{3}{4}-x=\frac{1}{5}\)
\(x=\frac{3}{4}-\frac{1}{5}\)
\(x=\frac{11}{20}\)
2)\(\frac{7}{2}+2x=\frac{-3}{4}\)
\(2x=\frac{-3}{4}-\frac{7}{2}\)
\(2x=-\frac{17}{4}\)
\(x=-\frac{17}{4}:2\)
\(x=-\frac{17}{8}\)
3)\(\frac{1}{3}+\frac{2}{3}x=\frac{4}{3}\)
\(\frac{2}{3}x=\frac{4}{3}-\frac{1}{3}\)
\(\frac{2}{3}x=1\)
\(x=1:\frac{2}{3}\)
\(x=\frac{3}{2}\)
^...^ ^_^
g) Đặt k = \(\frac{x-1}{2}\) = \(\frac{y-2}{3}\) = \(\frac{z-3}{4}\)
=> \(\begin{cases}x-1=2k\\y-2=3k\\z-3=4k\end{cases}\)
=> \(\begin{cases}x=2k+1\\y=3k+2\\z=4k+3\end{cases}\)
=> x - 2y + 3z = 2k+1 - 6k - 4 + 12k + 9 = 8k + 6
=> 8k + 6 = 14
=> k = 1
=> \(\begin{cases}x=2\\y=5\\z=7\end{cases}\)
mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn
Bài 1:
a) Ta có: \(\frac{-5}{8}+x=\frac{4}{9}\)
\(\Leftrightarrow x=\frac{4}{9}-\frac{-5}{8}=\frac{32}{72}-\frac{-45}{72}\)
hay \(x=\frac{77}{72}\)
Vậy: \(x=\frac{77}{72}\)
b) Ta có: \(1\frac{3}{4}\cdot x+1\frac{1}{2}=-\frac{4}{5}\)
\(\Leftrightarrow\frac{7}{4}\cdot x+\frac{3}{2}=-\frac{4}{5}\)
\(\Leftrightarrow\frac{7}{4}\cdot x=-\frac{4}{5}-\frac{3}{2}=-\frac{23}{10}\)
\(\Leftrightarrow x=\frac{-23}{10}:\frac{7}{4}=\frac{-23}{10}\cdot\frac{4}{7}\)
hay \(x=-\frac{46}{35}\)
Vậy: \(x=-\frac{46}{35}\)
c) Ta có: \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\)
\(\Leftrightarrow\frac{3}{4}x=\frac{2}{4}\)
\(\Leftrightarrow x=\frac{2}{4}:\frac{3}{4}=\frac{2}{4}\cdot\frac{4}{3}\)
hay \(x=\frac{2}{3}\)
Vậy: \(x=\frac{2}{3}\)
d) Ta có: \(x\cdot\left(\frac{1}{4}+\frac{1}{5}\right)-\left(\frac{1}{7}+\frac{1}{8}\right)=0\)
\(\Leftrightarrow x\cdot\frac{9}{20}-\frac{15}{56}=0\)
\(\Leftrightarrow x\cdot\frac{9}{20}=\frac{15}{56}\)
\(\Leftrightarrow x=\frac{15}{56}:\frac{9}{20}=\frac{15}{56}\cdot\frac{20}{9}\)
hay \(x=\frac{25}{42}\)
Vậy: \(x=\frac{25}{42}\)
e) Ta có: \(\frac{3}{35}-\left(\frac{3}{5}+x\right)=\frac{2}{7}\)
\(\Leftrightarrow\frac{3}{35}-\frac{3}{5}-x=\frac{2}{7}\)
\(\Leftrightarrow\frac{-18}{35}-x=\frac{2}{7}\)
\(\Leftrightarrow-x=\frac{2}{7}-\frac{-18}{35}=\frac{2}{7}+\frac{18}{35}=\frac{4}{5}\)
hay \(x=-\frac{4}{5}\)
Vậy: \(x=-\frac{4}{5}\)
f) Ta có: \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\)
\(\Leftrightarrow\frac{1}{7}\cdot\frac{1}{x}=\frac{3}{14}-\frac{3}{7}=\frac{-3}{14}\)
\(\Leftrightarrow\frac{1}{x}=\frac{-3}{14}:\frac{1}{7}=-\frac{3}{14}\cdot7=-\frac{3}{2}\)
\(\Leftrightarrow x=\frac{1\cdot2}{-3}=\frac{2}{-3}=-\frac{2}{3}\)
Vậy: \(x=-\frac{2}{3}\)
g) Ta có: \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{3}:2=\frac{1}{6}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{1}{6}\right\}\)
a) \(\dfrac{-1}{5}< \dfrac{1}{x}< \dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{1}{-5}< \dfrac{1}{x}< \dfrac{1}{7}\)
\(\Rightarrow\) \(x< -5\) và \(x< 7\left(x\in Z\right)\)
b) \(\dfrac{14}{5}< \dfrac{x}{5}< 4\)
\(\Leftrightarrow\dfrac{14}{5}< \dfrac{x}{5}< \dfrac{20}{5}\)
\(\Rightarrow14< x< 20\) mà \(x\in Z\)
\(\Rightarrow x\in\left\{15;16;17;18;19\right\}\)
c) \(\dfrac{1}{3}< \dfrac{9}{x}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{9}{27}< \dfrac{9}{x}< \dfrac{9}{18}\)
\(\Rightarrow18< x< 27\) mà \(x\in Z\)
\(\Rightarrow x\in\left\{19;20;...;25;26\right\}\)
m= 660/17