$2,5 - 3x = 5,5 .2022^0$
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2022 - 2,2 x 5,5 - 2,2 x 3,5 - 2,2
2022 - 2,2 x ( 5,5 + 3,5 + 1)
= 2022 - 2,2 x 10
= 2022 - 22
= 2000
2 022-2,2∙5,5-2,2∙3,5-2,2
=2 022-2,2∙(5,5-3,5 -1)
=2 022-2,2∙( 2 -1)
=2 022-2,2∙ 1
=2 022- 2,2
= 2 019,8
b)\(5.5:x=\frac{13}{15}\)
\\(x=5.5:\frac{13}{15}=\frac{11}{2}x\frac{15}{13}=\frac{165}{26}\)
c)\(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)
\(\frac{3x}{7}+1=\frac{1}{7}\)
\(\frac{3x}{7}=-\frac{6}{7}\)
\(\Rightarrow x=6:3=2\)
a/ \(\left|2x-1,6\right|-2,3=1,4\)
\(\Leftrightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5,3\\2x=-2,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
Vậy ....
b/ \(5,4-\left|3x-1,2\right|=5,5\)
\(\Leftrightarrow\left|3x-1,2\right|=-0,1\)
Mà \(\left|3x-1,2\right|\ge0\)
\(\Leftrightarrow x\in\varnothing\)
c/ \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow3,7=4x-2x\)
\(\Leftrightarrow2x=3,7\)
\(\Leftrightarrow x=1,85\)
Vậy ....
d/ \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\) (loại)
Vậy ..
a, \(\left|2x-1,6\right|-2,3=1,4\)
\(\Rightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
b,\(5,4-\left|3x-1,2\right|=5,5\)
\(\Rightarrow\left|3x-1,2\right|=-0,1\) (vô lí)
Vì \(\left|x\right|\ge0\) mà \(\left|3x-1,2\right|< 0\)
Vậy, không có giá trị của x thỏa mãn.
c, \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow x+x+1,3+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow2x-4x=-3,7\)
\(\Leftrightarrow-2x=-3,7\)
\(\Leftrightarrow x=\dfrac{3,7}{2}\)
d, \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\)
c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0
(\(x\) - 2022).(\(x\) - 4) = 0
\(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)
\(=\left(7,5+2,5\right)+\left(6,5+3,5\right)+\left(5,5+4,5\right)\\ =10+10+10=30\)
\(=\left(7,5+2.5\right)+\left(6,5+3,5\right)+\left(5,5+4,5\right)=10+10+10=30\)
\(2,5-3x=5,5.2022^0\)
\(=>2,5-3x=5,5.1\)
\(=>2,5-3x=5,5\)
\(=>3x=2,5-5,5\)
\(=>3x=-3\)
\(=>x=\left(-3\right):3\)
\(=>x=\dfrac{-3}{3}=-1\)
Vậy...
\(#NqHahh\)
\(2,5-3x=5,5\cdot2022^0\)
\(2,5-3x=5,5\cdot1\)
\(2,5-3x=5,5\)
\(3x=2,5-5,5\)
\(3x=-3\)
\(x=-3:3\)
\(x=-1\)
Vậy \(x=-1\)