Question

Tìm x :

\(\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2-\dfrac{1}{25}=0\)

P/s : Rảnh rỗi sinh nông nổi

Answer 1

\(\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2-\dfrac{1}{25}=0\\ \Leftrightarrow\left(\dfrac{x}{10}-\dfrac{3}{2}+\dfrac{1}{5}\right)\left(\dfrac{x}{10}-\dfrac{3}{2}-\dfrac{1}{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{10}-\dfrac{13}{10}=0\\\dfrac{x}{10}-\dfrac{17}{10}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=13\\x=17\end{matrix}\right.\)

Answer 2

\(\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2-\dfrac{1}{25}=0\)

\(\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2=\dfrac{1}{25}\)

\(\dfrac{x}{10}-\dfrac{3}{2}=\pm\dfrac{1}{5}\)

\(\left[{}\begin{matrix}\dfrac{x}{10}-\dfrac{3}{2}=\dfrac{1}{5}\\\dfrac{x}{10}-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x-15=2\\x-15=-2\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=2+15\\x=-2+15\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=17\\x=13\end{matrix}\right.\)

Vậy \(x_1=17;x_2=13\)

Answer 3

Ta có: \(\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2-\dfrac{1}{25}=0\\ =>\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2=0+\dfrac{1}{25}=\dfrac{1}{25}\\ Mà:\left\{{}\begin{matrix}\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\left(-\dfrac{1}{5}\right)^2=\dfrac{1}{25}\end{matrix}\right.< =>\left\{{}\begin{matrix}\dfrac{x}{10}-\dfrac{3}{2}=\dfrac{1}{5}\\\dfrac{x}{10}-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\\ =>\left\{{}\begin{matrix}\dfrac{x}{10}=\dfrac{1}{5}+\dfrac{3}{2}=\dfrac{17}{10}\\\dfrac{x}{10}=\dfrac{13}{10}\end{matrix}\right.\\ =>\left\{{}\begin{matrix}x=\dfrac{17.10}{10}=17\\x=\dfrac{13.10}{10}=13\end{matrix}\right.=>\left\{{}\begin{matrix}x=17\\x=13\end{matrix}\right.\)

Vậy: \(x\in\left\{13;17\right\}\)

Answer 4

\(\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2=\dfrac{1}{25}=\left(-\dfrac{1}{5}\right)^2=\left(\dfrac{1}{5}\right)^2\)

TH1 : \(\dfrac{x}{10}-\dfrac{3}{2}=\dfrac{1}{5}\)

=> \(\dfrac{x}{10}=\dfrac{1}{5}+\dfrac{3}{2}=\dfrac{17}{10}\)

=> x = 17

TH2 : \(\dfrac{x}{10}-\dfrac{3}{2}=-\dfrac{1}{5}\)

=> \(\dfrac{x}{10}=-\dfrac{1}{5}+\dfrac{3}{2}=\dfrac{13}{10}\)

Vậy x = 17 hoặc 13

Answer 5

\(\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2-\dfrac{1}{25}=0\)

\(\Rightarrow\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2=0+\dfrac{1}{25}=\dfrac{1}{25}\)

\(\Rightarrow\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2=\left(\dfrac{1}{5}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\left(-\dfrac{1}{5}\right)^2=\dfrac{1}{25}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{x}{10}-\dfrac{3}{2}=\dfrac{1}{5}\\\dfrac{x}{10}-\dfrac{3}{2}=-\dfrac{1}{5}\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}\dfrac{x}{10}=\dfrac{1}{5}+\dfrac{3}{2}=\dfrac{17}{10}\\\dfrac{x}{10}=-\dfrac{1}{5}+\dfrac{3}{2}=\dfrac{13}{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{17\cdot10}{10}=17\\x=\dfrac{13\cdot10}{10}=13\end{matrix}\right.\)

Vậy \(x\in\left\{13;17\right\}\).

Answer 6

\(\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2-\dfrac{1}{25}=0\)

<=> \(\left(\dfrac{x}{10}-\dfrac{3}{2}\right)^2-\left(\dfrac{1}{5}\right)^2=0\)

<=> \(\left(\dfrac{x}{10}-\dfrac{3}{2}+\dfrac{1}{5}\right)\left(\dfrac{x}{10}-\dfrac{3}{2}-\dfrac{1}{5}\right)=0\)

<=> \(\left(\dfrac{x}{10}-\dfrac{13}{10}\right)\left(\dfrac{x}{10}-\dfrac{17}{10}\right)=0\)

<=> \(\left(\dfrac{x-13}{10}\right)\left(\dfrac{x-17}{10}\right)=0\)

Vậy \(\dfrac{x-13}{10}=0\) hoặc \(\dfrac{x-17}{10}=0\)

1) \(\dfrac{x-13}{10}=0\)

<=> x - 13 =0

<=> x = 13

2) \(\dfrac{x-17}{10}=0\)

<=> x - 17 =0

<=> x = 17

Vậy x= 13 hoặc x =17

Answer 7

ủa , bộ ... đâu mà k tick ... zậy \(\Sigma\)(=\(\Delta\)=)