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\(\left(3x-1\right)\left(\frac{2}{3}x+\frac{1}{5}\right)>0\)
<=> \(\orbr{\begin{cases}3x-1>0\\\frac{2}{3}x+\frac{1}{5}>0\end{cases}}\)và \(\orbr{\begin{cases}3x-1< 0\\\frac{2}{3}x+\frac{1}{5}< 0\end{cases}}\)
<=> \(\orbr{\begin{cases}x>\frac{1}{3}\\x>\frac{-3}{10}\end{cases}}\)và\(\orbr{\begin{cases}x< \frac{1}{3}\\x< \frac{-3}{10}\end{cases}}\)
<=> \(x>\frac{1}{3}\)và \(x< \frac{-3}{10}\)
<=> \(x\)thuộc rỗng
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: =>-3/2+x-7=5-1/3x+4/15
=>4/3x=413/30
hay x=413/40
b: \(\Leftrightarrow5-\dfrac{3}{2}x=-\dfrac{22}{3}\cdot\dfrac{-11}{8}=\dfrac{121}{12}\)
=>3/2x=-61/12
hay x=-61/18
c: (3x+2)2+|3x+2y|=0
=>3x+2=0 và 3x=-2y
=>x=-2/3 và -2y=-2
=>(x,y)=(-2/3;1)
( 3x - 34 ). 63 = 65
( 3x - 34 ) = 65 : 63
(3x - 3^4 ) = 6^3 = 216
3x = 216 - 81 = 135
x = 135 : 3 = 45
60%x+\(\dfrac{2}{3}\)x=\(\dfrac{1}{3}\).6\(\dfrac{1}{3}\)
60%x+\(\dfrac{2}{3}\)x=\(\dfrac{1}{3}\).\(\dfrac{19}{3}\)
60%x+\(\dfrac{2}{3}\)x=\(\dfrac{19}{9}\)
\(\dfrac{3}{5}\) x+\(\dfrac{2}{3}\)x=\(\dfrac{19}{9}\)
(\(\dfrac{3}{5}+\dfrac{2}{3}\)) x=\(\dfrac{19}{9}\)
\(\dfrac{19}{15}\) x=\(\dfrac{19}{9}\)
x=\(\dfrac{19}{9}:\dfrac{19}{15}\)
x=\(\dfrac{5}{3}\)
\(\left(3x-4\right)\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(3x-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-4=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-1\end{cases}}\)
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)<=>\(\dfrac{\left(2x-3\right).2}{6}-\dfrac{3.3}{6}=\dfrac{5-2x}{6}-\dfrac{1.3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}=\dfrac{5-2x}{6}-\dfrac{3}{6}\)
<=>\(\dfrac{4x-6}{6}-\dfrac{9}{6}-\dfrac{5-2x}{6}+\dfrac{3}{6}=0\)
<=>\(\dfrac{4x-6-9-5+2x+3}{6}=\dfrac{4x-17}{6}=0\)
<=>\(4x-17=0\)
<=>\(4x=17\)<=>\(x=\dfrac{17}{4}\)
a) \(-1< \dfrac{5x}{13}< 0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5x}{13}>-1\\\dfrac{5x}{13}< 0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>-\dfrac{13}{5}\\x< 0\end{matrix}\right.\)
Vậy \(x\in\left\{-\dfrac{13}{5},0\right\}\)
b) \(3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(\Rightarrow3\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}\)
\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{27}\)
\(\Leftrightarrow3x-\dfrac{1}{2}=-\dfrac{1}{3}\)
\(\Leftrightarrow3x=-\dfrac{1}{3}+\dfrac{1}{2}\)
\(\Leftrightarrow3x=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{18}\)
Vậy \(x=\dfrac{1}{18}\)
Xử câu b trc =)
b) \(3\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\)
\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=-\dfrac{1}{9}:3=-\dfrac{1}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{1}{2}\right)^3=\left(-\dfrac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\dfrac{1}{2}=-\dfrac{1}{3}\)
\(\Leftrightarrow3x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\)
\(\Leftrightarrow x=\dfrac{1}{6}:3=\dfrac{1}{18}\)
`@` `\text {Ans}`
`\downarrow`
`1/3x^3 - 3x = 0`
`=> x(1/3x^2 - 3) = 0`
`=>`\(\left[{}\begin{matrix}x=0\\\dfrac{1}{3}x^2-3=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\\dfrac{1}{3}x^2=3\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x^2=3\div\dfrac{1}{3}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x^2=\left(\pm3\right)^2\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)
Vậy, `x \in {0; 3; -3}.`
\(\dfrac{1}{3}x^3-3x=0\Rightarrow\dfrac{1}{3}x\left(x^2-9\right)=0\)
\(\Rightarrow x=0\) hay \(x^2-9=0\)
\(\Rightarrow x=0\) hay \(x^2=9=3^2\)
\(\Rightarrow x=0\) hay \(x=\pm3\)