Question

Answer 1

\(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(=>\left[{}\begin{matrix}\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\\\dfrac{1}{5}-\dfrac{3}{2}x=-\dfrac{3}{2}\end{matrix}\right.\\ =>\left[{}\begin{matrix}\dfrac{3}{2}x=-\dfrac{13}{10}\\\dfrac{3}{2}x=\dfrac{17}{10}\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=-\dfrac{13}{15}\\x=\dfrac{17}{15}\end{matrix}\right.\)

Answer 2

=1

 

Answer 3

\(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

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

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

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

\(\Leftrightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Leftrightarrow x=-\dfrac{13}{15}\)

Answer 4

\(\left(\dfrac{1}{5}-\dfrac{3}{2}.x\right)^2=\dfrac{9}{4}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}.x=\dfrac{1}{5}-\dfrac{3}{2}\\\dfrac{3}{2}.x=\dfrac{1}{5}-\left(-\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}.x=-\dfrac{13}{10}\\\dfrac{3}{2}.x=\dfrac{17}{10}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{10}:\dfrac{3}{2}\\x=\dfrac{17}{10}:\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{15}\\x=\dfrac{17}{15}\end{matrix}\right.\)

 Vậy ................

Answer 5

    =(1/5-3/2x)=3/2

=> 1/5-3/2x=+-3/2

=> th1 :  1/5-3/2x=3/2

                    3/2x=1/5-3/2

                    3/2x=-13/10

                         x=-13/10:3/2

                         x=-13/15

     th2  :  1/5-3/2x=-3/2

                     3/2x=1/5-(-3/2)

                     3/2x=17/10

                          x=17/10:3/2

                          x= 17/15

Vậy x \(\in\) (-13/15 hoặc 17/15 )

hôm trước vừa làm bài kiểm tra cô chữa bảo là vậy

Answer 6

\(c,\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-13}{15}\\x=\dfrac{17}{15}\end{matrix}\right.\)