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21 tháng 8 2020

\(\left(x-\frac{3}{5}\right)=\frac{2}{5}×-\frac{1}{3}\)

\(\left(x-\frac{3}{5}\right)=-\frac{2}{165}\)

\(x=-\frac{2}{165}+\frac{3}{5}\)

\(x=\frac{97}{165}\)

vậy \(x=\frac{97}{165}\)

\(x×\left(\frac{3}{7}+\frac{2}{3}\right)=\frac{10}{21}\)

\(x×\frac{23}{21}=\frac{10}{21}\)

\(x=\frac{10}{21}:\frac{23}{21}\)

\(x=\frac{10}{23}\)

vậy \(x=\frac{10}{23}\)

21 tháng 8 2020

\(\left(x-\frac{3}{5}\right):\frac{-1}{3}=\frac{2}{5}\)

=> \(x-\frac{3}{5}=\frac{2}{5}\cdot\left(-\frac{1}{3}\right)=-\frac{2}{15}\)

=> \(x=-\frac{2}{15}+\frac{3}{5}=-\frac{2}{15}+\frac{9}{15}=\frac{7}{15}\)

\(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\)

=> \(\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\)

=> \(-\frac{5}{21}x=\frac{10}{21}\)

=> \(x=\frac{10}{21}:\frac{-5}{21}=\frac{10}{21}\cdot\frac{-21}{5}=-2\)

Hai bài của ☆luffy cute☆ đều sai hết , xem xét lại đi nhé

25 tháng 9 2019

2x + 13/6 =8/27

2x            = 8/27 - 13/6

2x            = - 101/54

x            = - 101/54 : 2

x              = - 101/108

20 tháng 9 2020

a) \(\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\Rightarrow\frac{28}{27}=x:\frac{3}{7}\Rightarrow x=\frac{4}{9}\)

b) \(\left(x-\frac{4}{7}\right)^3=343\Rightarrow\left(x-\frac{4}{7}\right)^3=7^3\Rightarrow x-\frac{4}{7}=7\Rightarrow x=\frac{53}{7}\)

c) \(x^5=x^3\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)

e) \(\left(x-1\right)^4=16\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^4=2^4\\\left(x-1\right)^4=\left(-2\right)^4\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x-1=2\\x-1=\left(-2\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

13 tháng 11 2019

b) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

\(\Rightarrow\frac{3}{5}x=\left(-\frac{1}{7}\right)+\frac{1}{2}\)

\(\Rightarrow\frac{3}{5}x=\frac{5}{14}\)

\(\Rightarrow x=\frac{5}{14}:\frac{3}{5}\)

\(\Rightarrow x=\frac{25}{42}\)

Vậy \(x=\frac{25}{42}.\)

c) \(5-\left|3x-1\right|=3\)

\(\Rightarrow\left|3x-1\right|=5-3\)

\(\Rightarrow\left|3x-1\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=3\\3x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3:3\\x=\left(-1\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{1;-\frac{1}{3}\right\}.\)

d) \(\left(1-2x\right)^2=9\)

\(\Rightarrow\left(1-2x\right)^2=\left(\pm3\right)^2\)

\(\Rightarrow1-2x=\pm3.\)

\(\Rightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-2\right):2\\x=4:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{-1;2\right\}.\)

Chúc bạn học tốt!

13 tháng 11 2019

Chương I : Số hữu tỉ. Số thực

\(2^x:1+2^x:2+...+2^x:49=2^{49}-1\)

\(2^x.1+2^x.\frac{1}{2}+...+2^x.\frac{1}{49}=2^{49}-1\)

\(2^x.\left(1+\frac{1}{2}+...+\frac{1}{49}\right)=2^{49}-1\)

3 tháng 3 2020

Đặt: \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\)

=> \(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}\)

=> \(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^{49}}\right)\)

=> \(A=1-\frac{1}{2^{49}}=\frac{2^{49}-1}{2^{49}}\)

\(2^{x-1}+2^{x-2}+2^{x-3}+...+2^{x-49}=2^{49}-1\)

<=> \(\frac{2^x}{2}+\frac{2^x}{2^2}+\frac{2^x}{2^3}+...+\frac{2^x}{2^{49}}=2^{49}-1\)

<=> \(2^x\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}\right)=2^{49}-1\)

<=> \(2^x.\frac{2^{49}-1}{2^{49}}=2^{49}-1\)

<=> \(2^x=2^{49}\)

<=> x = 49.

23 tháng 9 2021

\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)