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a)\(\dfrac{3}{10}\)-x=\(\dfrac{25}{30}\)-\(\dfrac{4}{30}\)
\(\dfrac{3}{10}-x=\dfrac{7}{10}\)
x = \(\dfrac{3}{10}-\dfrac{7}{10}\)
x=\(\dfrac{-4}{10}\)
b)\(\dfrac{-5}{8}+x=\dfrac{4}{9}-\dfrac{63}{9}\)
\(\dfrac{-5}{9}+x=\dfrac{-59}{9}\)
\(x=\dfrac{-59}{9}-\dfrac{-5}{9}\)
\(x=\dfrac{-64}{9}\)
c)=>2.18=(x-3).(x-3)
=>36=(x-3)\(^2\)
=>6\(^2\)=(x-3)\(^2\)
6= x-3
x=6+3=9
a.\(\dfrac{-4}{5}-\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{2}{7}\)
\(\left(\dfrac{2}{3}x+1\dfrac{1}{4}\right)=\dfrac{-4}{5}-\dfrac{2}{7}=\dfrac{-38}{35}\)
\(\dfrac{2}{3}x=\dfrac{-38}{35}-1\dfrac{1}{4}\)
\(\dfrac{2}{3}x=\dfrac{-327}{140}\Rightarrow x=\dfrac{-327}{140}:\dfrac{2}{3}=\dfrac{-981}{280}\)
Vậy \(x=\dfrac{-981}{280}\)
b. \(\dfrac{5}{6}+\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}\)
\(\left(\dfrac{3}{4}-\dfrac{1}{2}:x\right)=\dfrac{-2}{3}-\dfrac{5}{6}=\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{3}{4}-\dfrac{-3}{2}\)
\(\dfrac{1}{2}:x=\dfrac{9}{4}\Rightarrow x=\dfrac{1}{2}:\dfrac{9}{4}=\dfrac{2}{9}\)
Vậy \(x=\dfrac{2}{9}\)
c. \(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right):\dfrac{3}{4}=0,7\)
\(\left(\dfrac{4}{5}x-1\dfrac{1}{3}\right)=0,7.\dfrac{3}{4}=\dfrac{21}{40}\)
\(\dfrac{4}{5}x=\dfrac{21}{40}+1\dfrac{1}{3}=\dfrac{223}{120}\)
\(\Rightarrow x=\dfrac{223}{120}:\dfrac{4}{5}=\dfrac{223}{96}\)
Vậy \(x=\dfrac{223}{96}\)
d. \(\dfrac{5}{6}-\dfrac{3}{4}x=1\dfrac{1}{3}+0,5x\)
\(0,5x+\dfrac{3}{4}x=\dfrac{5}{6}-1\dfrac{1}{3}\)
\(\dfrac{5}{4}x=\dfrac{-1}{2}\Rightarrow x=\dfrac{-1}{2}:\dfrac{5}{4}=\dfrac{-2}{5}\)
Vậy \(x=\dfrac{-2}{5}\)
a/ \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{5}=\dfrac{5}{6}-\dfrac{-5}{12}\)
\(\Leftrightarrow\dfrac{1}{2}x-\dfrac{1}{5}=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{29}{20}\)
\(\Leftrightarrow x=\dfrac{29}{10}\)
Vậy ...
b/ \(\left(4x-3\right)\left(\dfrac{5}{4}x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{8}{5}\end{matrix}\right.\)
Vậy .....
c/ \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)
\(\Leftrightarrow\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=-\dfrac{9}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=-\dfrac{19}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-\dfrac{38}{21}\end{matrix}\right.\)
Vậy ......
d/ \(\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\dfrac{8}{125}\)
\(\Leftrightarrow\left(\dfrac{3}{5}x-\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\)
\(\Leftrightarrow\dfrac{3}{5}x-\dfrac{1}{2}=\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{3}{5}x=\dfrac{9}{10}\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy ...
a. \(\dfrac{5}{6}-\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{-5}{12}\)
\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{6}-\dfrac{-5}{12}\)
\(\left(\dfrac{3}{6}x-\dfrac{1}{5}\right)=\dfrac{5}{4}\)
\(\dfrac{3}{6}x=\dfrac{5}{4}+\dfrac{1}{5}\)
\(\dfrac{3}{6}x=\dfrac{29}{20}\)
\(x=\dfrac{29}{20}:\dfrac{3}{6}\)
\(x=\dfrac{29}{10}\)
Vậy...
b. \(\left(4x-3\right).\left(\dfrac{5}{4}x+2\right)=0\)
\(\left[{}\begin{matrix}4x-3=0\\\dfrac{5}{4}x+2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}4x=3\\\dfrac{5}{4}x=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-8}{5}\end{matrix}\right.\)
Vậy ...
c. \(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|-\dfrac{3}{4}=1,5\)
\(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=1,5+\dfrac{3}{4}\)
\(\left|\dfrac{7}{8}x-\dfrac{2}{3}\right|=\dfrac{9}{4}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{9}{4}\\\dfrac{7}{8}x-\dfrac{2}{3}=\dfrac{-9}{4}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{8}x=\dfrac{35}{12}\\\dfrac{7}{8}x=\dfrac{-19}{12}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=\dfrac{-38}{21}\end{matrix}\right.\)
Vậy...
Cho \(A=\dfrac{\dfrac{-5}{8}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{3}{4}+\dfrac{1}{7}.\dfrac{1}{2}+\dfrac{15}{8}}{a+\dfrac{5}{6}-\left(\dfrac{-1}{3}\right)}\)
a) Rút gọn A?
b) Tính A khi a=75%
c) Tìm a để A=50%
d) Tìm a thuộc Z để A là số nguyên.
e) Với a = bao nhiêu để A có giá trị bằng với giá trị của biểu thức:
\(B=\dfrac{\dfrac{2}{3}.\dfrac{15}{6}+\left(-0,5\right)^3}{\dfrac{1}{9}.6^2-5\dfrac{1}{3}}\)
Giải
a, Ta có:
\(A=\dfrac{\dfrac{-5}{8}.\dfrac{3}{7}+\dfrac{3}{7}.\dfrac{3}{4}+\dfrac{3}{7}.\dfrac{1}{6}+\dfrac{1}{8}.15}{a+\dfrac{5}{6}+\dfrac{1}{3}}\)
\(A=\dfrac{\dfrac{3}{7}.\left(\dfrac{-5}{8}+\dfrac{3}{4}+\dfrac{1}{6}\right)+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{3}{7}.\dfrac{7}{24}+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{1}{8}+\dfrac{1}{8}.15}{a+\dfrac{7}{6}}\)
\(A=\dfrac{\dfrac{1}{8}.\left(15+1\right)}{a+\dfrac{7}{6}}\)
\(A=\dfrac{2}{a+\dfrac{7}{6}}\)
b, Thay \(a=75\%\) vào \(A\), ta được:
\(A=\dfrac{2}{75\%+\dfrac{7}{6}}\)
\(A=\dfrac{2}{\dfrac{3}{4}+\dfrac{7}{6}}\)
\(\Rightarrow A=\dfrac{23}{12}\)
c, Ta có: \(\dfrac{2}{a+\dfrac{7}{6}}=50\%\)
\(\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{1}{2}\)
\(\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{2}{4}\)
\(\Rightarrow a+\dfrac{7}{6}=4\)
\(\Rightarrow a=\dfrac{17}{6}\)
d, Để \(A\in Z\Rightarrow2⋮a+\dfrac{7}{6}\)
\(\Rightarrow a+\dfrac{7}{6}\in\left\{\pm1;\pm2\right\}\)
\(\circledast,a+\dfrac{7}{6}=1\Rightarrow a=\dfrac{-1}{6}\)
\(\circledast,a+\dfrac{7}{6}=-1\Rightarrow a=\dfrac{-13}{6}\)
\(\circledast,a+\dfrac{7}{6}=2+\Rightarrow a=\dfrac{5}{6}\)
\(\circledast,a+\dfrac{7}{6}=-2\Rightarrow a=\dfrac{-19}{6}\)
\(a\in\varnothing\) khi \(A\in Z\)
e, Ta có:
\(B=\dfrac{5}{3}+\dfrac{-1}{8}\Rightarrow B=\dfrac{37}{24}\)
\(\Rightarrow\dfrac{2}{a+\dfrac{7}{6}}=\dfrac{37}{24}\)
\(a+\dfrac{7}{6}=\dfrac{37}{24}.2\)
\(a+\dfrac{7}{6}=\dfrac{37}{12}\)
\(\Rightarrow a=\dfrac{23}{12}\)
Chúc bạn học thiệt giỏi nha!!! 
)
K chép lại đề, lm luôn nhé:
*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)
\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{3}{4}\)
* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)
=> K có gt x nào t/m đề
* Đề sai
* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)
\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)
\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)
\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)
* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)
* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)
\(\Rightarrow-7x=-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)
a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)
b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)
\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)
\(\Rightarrow x\in\varnothing\)
c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.
d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)
e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)
\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)
\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)
g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)
\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)
\(th1:x=0\)
\(th2:x=\dfrac{-3}{5}\)
h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)
\(-5x+\dfrac{-1}{2}=2x\)
\(\dfrac{-1}{2}=2x+5x\)
\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)
a) \(\left(2x-3\right)\left(6-2x\right)=0\)
\(\circledast\)TH1: \(2x-3=0\\ 2x=0+3\\ 2x=3\\ x=\dfrac{3}{2}\)
\(\circledast\)TH2: \(6-2x=0\\ 2x=6-0\\ 2x=6\\ x=\dfrac{6}{2}=3\)
Vậy \(x\in\left\{\dfrac{3}{2};3\right\}\).
b) \(\dfrac{1}{3}x+\dfrac{2}{5}\left(x-1\right)=0\)
\(\dfrac{1}{3}x=0-\dfrac{2}{5}\left(x-1\right)\)
\(\dfrac{1}{3}x=-\dfrac{2}{5}\left(x-1\right)\)
\(-\dfrac{2}{5}-\dfrac{1}{3}=-x\left(x-1\right)\)
\(-\dfrac{11}{15}=-x\left(x-1\right)\)
\(\Rightarrow x=1.491631652\)
Vậy \(x=1.491631652\)
c) \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\circledast\)TH1: \(3x-1=0\\ 3x=0+1\\ 3x=1\\ x=\dfrac{1}{3}\)
\(\circledast\)TH2: \(-\dfrac{1}{2}x+5=0\\ -\dfrac{1}{2}x=0-5\\ -\dfrac{1}{2}x=-5\\ x=-5:-\dfrac{1}{2}\\ x=10\)
Vậy \(x\in\left\{\dfrac{1}{3};10\right\}\).
d) \(\dfrac{x}{5}=\dfrac{2}{3}\\ x=\dfrac{5\cdot2}{3}\\ x=\dfrac{10}{3}\)
Vậy \(x=\dfrac{10}{3}\).
e) \(\dfrac{x}{3}-\dfrac{1}{2}=\dfrac{1}{5}\\ \)
\(\dfrac{x}{3}=\dfrac{1}{5}+\dfrac{1}{2}\)
\(\dfrac{x}{3}=\dfrac{7}{10}\)
\(x=\dfrac{3\cdot7}{10}\)
\(x=\dfrac{21}{10}\)
Vậy \(x=\dfrac{21}{10}\).
f) \(\dfrac{x}{5}-\dfrac{1}{2}=\dfrac{6}{10}\)
\(\dfrac{x}{5}=\dfrac{6}{10}+\dfrac{1}{2}\)
\(\dfrac{x}{5}=\dfrac{11}{10}\)
\(x=\dfrac{5\cdot11}{10}\)
\(x=\dfrac{55}{10}=\dfrac{11}{2}\)
Vậy \(x=\dfrac{11}{2}\).
g) \(\dfrac{x+3}{15}=\dfrac{1}{3}\\ x+3=\dfrac{15}{3}=5\\ x=5-3\\ x=2\)
Vậy \(x=2\).
h) \(\dfrac{x-12}{4}=\dfrac{1}{2}\\ x-12=\dfrac{4}{2}=2\\ x=2+12\\ x=14\)
Vậy \(x=14\).
a)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
a) \(\dfrac{1}{3}x-\dfrac{1}{2}=\dfrac{3}{4}x+\dfrac{1}{15}\)
\(\Rightarrow\dfrac{1}{3}x-\dfrac{3}{4}x=\dfrac{1}{2}+\dfrac{1}{15}\)
\(\Rightarrow\dfrac{4}{12}x-\dfrac{9}{12}x=\dfrac{15}{30}+\dfrac{2}{30}\)
\(\Rightarrow\dfrac{-5}{12}x=\dfrac{17}{30}\)
\(\Rightarrow x=\dfrac{-102}{75}\)
\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)
\(\Rightarrow\left(x-\dfrac{2}{9}\right)^3=\dfrac{64}{729}\)
\(\Rightarrow x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(\Rightarrow x=\dfrac{2}{3}\)