x = 1,2 

Bài Toán Dễ Ẹc

\(\text{|x-2,2|=|0,2-x|}\)

\(\Leftrightarrow x-2,2=0,2-x\)

\(\Leftrightarrow x+x=0,2+2,2\)

\(\Leftrightarrow2x=2,4\)

\(\Leftrightarrow x=1,2\)

trường hợp 1 : x - 2,2 = -( 0,2 - x )

\(\Leftrightarrow\)x - 2,2 = x - 0,2

\(\Leftrightarrow\)2,2 = 0,2 vô lí !

trường hợp 2 :  x - 2,2 = 0,2 - x

\(\Leftrightarrow\)2x = 2,4

\(\Leftrightarrow\)x = 2,4 : 2 = 1,2

Vậy x = 1,2

a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)

    \(\dfrac{3}{7}\)\(x\)         = - \(\dfrac{17}{35}\) + 0,4

     \(\dfrac{3}{7}\)\(x\)       = - \(\dfrac{3}{35}\)

        \(x\)       = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)

        \(x\)       = - \(\dfrac{1}{5}\)

b, 0,2.(\(x\) - 3) +2,4 = 10

    0,2.(\(x\) - 3)          = 10 - 2,4

    0,2.(\(x\) - 3)          = 7,6

           \(x\) - 3            = 7,6:0,2

           \(x\) - 3            = 38

            \(x\)                = 38 + 3

             \(x\)                = 41

\(\dfrac{3}{7}x-0,4=\dfrac{-17}{35}\)

\(\dfrac{3}{7}x=\dfrac{-17}{35}+0,4=\dfrac{-34}{70}+\dfrac{28}{70}\)

\(\dfrac{3}{7}x=\dfrac{-6}{70}=\dfrac{-3}{35}\)

\(x=\dfrac{-3}{35}:\dfrac{3}{7}=\dfrac{-3}{35}\cdot\dfrac{7}{3}\)

\(x=\dfrac{3\cdot\left(-1\right)\cdot7}{5\cdot7\cdot3}=-\dfrac{1}{5}\)

b) \(0,2\left(x-3\right)+2,4=10\)

\(0,2\left(x-3\right)=10-2,4=7,6\)

\(x-3=7,6:0,2=38\)

\(x=38+3=41\)

c) \(\left(x:2,2\right)\cdot\dfrac{1}{6}=\dfrac{-3}{8}\cdot\left(0,5-1\dfrac{3}{5}\right)\)

\(\left(x:2,2\right)\cdot\dfrac{1}{6}=\dfrac{-3}{8}\cdot\dfrac{-11}{10}\)

\(x:2,2=\dfrac{-3}{8}\cdot\dfrac{-11}{10}\cdot\dfrac{6}{1}\)

\(x:2,2=\dfrac{-3\cdot\left(-11\right)\cdot2\cdot3}{2\cdot4\cdot10\cdot1}=\dfrac{99}{40}\)

\(x=\dfrac{99}{40}\cdot2,2=\dfrac{99}{40}\cdot\dfrac{22}{10}\)

\(x=\dfrac{99\cdot2\cdot11}{40\cdot2\cdot5}=\dfrac{1089}{200}\)

\(Ix-2,2I=I0,2-xI\)

\(=>\orbr{\begin{cases}x-2,2=0,2-x\\x-2,2=x-0,2\end{cases}}\)

\(=>\orbr{\begin{cases}x+x=2,2+0,2\\x-x=2,2-0,2\end{cases}}\)

\(=>\orbr{\begin{cases}2x=2,4\\0x=2\end{cases}}\)(0x=2 k thỏa mãn)

\(=>x=\frac{2,4}{2}\)

\(=>x=1,2\)

Vậy x= 1,2

Ta có: |x-2,2| = |0,2-x|

\(\Rightarrow\orbr{\begin{cases}x-2,2=0,2-x\\x-2,2=x-0,2\end{cases}\Rightarrow\orbr{\begin{cases}2x=2,4\\0x=2\left(voli\right)\end{cases}}}\)\(\Rightarrow x=1,2\)

Vậy x=1,2.

x = 1,2 nhe

\(|x-2,2|=|0,2+x|\)

\(\Leftrightarrow\orbr{\begin{cases}x-2,2=0\\0,2+x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0+2,2\\x=0-0,2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2,2\\x=-0,2\end{cases}}\)

chúc bạn học tốt:^-^

x-2.2= -(0,2+x) hoặc x-2.2= 0.2+x

x-2.2= -0,2-x hoặc x-2.2= 0.2+x

2x=2 or x-2.4=x(loại)

x=1

\(\left|x-2,2\right|=\left|0,2-x\right|\)

\(\Rightarrow\orbr{\begin{cases}x-2,2=0,2-x\\x-2,2+0,2-x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+x=0,2+2,2\\2=0\left(\text{vô lí}\right)\end{cases}}\)

\(\Rightarrow2x=2,4\)

\(\Rightarrow x=1,2\)

\(\left|x-2,2\right|=\left|0,2-x\right|\)

\(\Leftrightarrow x-2,2=0,2-x\)

\(\Leftrightarrow\)\(x+x=2,2+0,2\)

\(\Leftrightarrow\)\(2x=2,4\)

\(\Rightarrow\)\(x=1,2\)

Khó quá ạ.

 GIÚP MK VỚI !!!!

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