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1.
b) \(B=\left|x+8\right|+\left|x+18\right|+\left|x+50\right|\)
Ta có:
\(B=\left|x+8\right|+\left|x+18\right|+\left|x+50\right|\ge\left(\left|x+8\right|+\left|-50-x\right|\right)+\left|x+18\right|\)
\(\Rightarrow B=\left(\left|x+8-50-x\right|\right)+\left|x+18\right|\)
\(\Rightarrow B=\left|-42\right|+\left|x+18\right|\)
\(\Rightarrow B=42+\left|x+18\right|\ge42\)
\(\Rightarrow MIN_B=42\) khi và chỉ khi:
\(\left\{{}\begin{matrix}x+8\ge0\\x+18=0\\x+50\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge-8\\x=-18\\x\ge-50\end{matrix}\right.\Rightarrow x=-18.\)
Vậy \(MIN_B=42\) khi \(x=-18.\)
3.
b) \(\left|x-3\right|-\left|2x+1\right|=0\)
\(\Rightarrow\left|x-3\right|=\left|2x+1\right|\)
\(\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2x=1+3\\x+2x=\left(-1\right)+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-1x=4\\3x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4:\left(-1\right)\\x=2:3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{-4;\frac{2}{3}\right\}.\)
Chúc bạn học tốt!
a) \(-\dfrac{3}{5}-x=-0,75\)
\(\Rightarrow-\dfrac{3}{5}-x=-\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{3}{5}\)
\(\Rightarrow x=\dfrac{15}{20}-\dfrac{12}{20}=\dfrac{8}{20}=\dfrac{2}{5}\)
b) \(1\dfrac{4}{5}=-0,15-x\)
\(\Rightarrow\dfrac{9}{5}=-\dfrac{3}{20}-x\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{9}{5}\)
\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{36}{20}=-\dfrac{39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{2}{5}+\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}+\dfrac{1}{3}-\dfrac{1}{3}=\dfrac{2}{5}\)
a) \(-\dfrac{3}{5}-x=-0,75\)
\(x=-\dfrac{3}{5}+0,75=\dfrac{3}{5}+\dfrac{3}{4}\)
\(x=\dfrac{27}{20}\)
________
b) \(1\dfrac{4}{5}=-0,15-x\)
\(=>-0,15-x=\dfrac{9}{5}\)
\(x=\dfrac{-3}{20}-\dfrac{9}{5}=\dfrac{-3}{20}-\dfrac{36}{20}\)
\(x=\dfrac{-39}{20}\)
c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)=\dfrac{6}{15}+\dfrac{5}{15}\)
\(x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(x=\dfrac{11}{15}-\dfrac{1}{3}=\dfrac{11}{15}-\dfrac{5}{15}\)
\(x=\dfrac{6}{15}=\dfrac{2}{5}\)
( 0,25x ) : 3 = 5/6 : 0,125
( 1/4x ) : 3 = 5/6 : 1/8
( 1/4x ) : 3 = 5/6 x 8
( 1/4x ) : 3 = 20/3
1/4x = 20/3 x 3
1/4x = 20
x = 20 : 1/4
x = 80
0,01 : 2,5 = ( 0,75x ) : 0,75
1/100 : 25/10 = (3/4x) : 3/4
1/100 x 10/25 = (3/4x) : 3/4
=>(3/4x) : 3/4 = 1/250
3/4x = 1/250 x 3/4
3/4x = 3/1000
x = 3/1000 : 3/4
x = 1/250
a) Ta có: \(x-\frac{2}{3}=0,75\)
\(\Leftrightarrow x-\frac{2}{3}=\frac{3}{4}\)
hay \(x=\frac{3}{4}+\frac{2}{3}=\frac{17}{12}\)
Vậy: \(x=\frac{17}{12}\)
b) Ta có: \(\frac{1}{3}+\frac{2}{3}x=-1\)
\(\Leftrightarrow\frac{2}{3}x=-1-\frac{1}{3}=-\frac{4}{3}\)
hay \(x=\frac{-4}{3}:\frac{2}{3}=\frac{-4}{3}\cdot\frac{3}{2}=-2\)
Vậy: x=-2
c) Ta có: \(\left|2x-3\right|-\frac{3}{4}=4,25\)
\(\Leftrightarrow\left|2x-3\right|=\frac{17}{4}+\frac{3}{4}=5\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=5\\2x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vậy: x∈{-1;4}
a, \(x-\frac{2}{3}=0.75\)
\(x=0.75-\frac{2}{3}\)
\(x=\frac{1}{12}\)
Vậy...
b, \(\frac{1}{3}+\frac{2}{3}\cdot x=-1\)
\(\frac{2}{3}\cdot x=-1-\frac{1}{3}=-\frac{4}{3}\)
\(x=-\frac{4}{3}:\frac{2}{3}=-2\)
Vậy x = -2
c, \(|2x-3|-\frac{3}{4}=4.25\)
\(|2x-3|=4.25-\frac{3}{4}=\frac{7}{2}\)
=> \(2x-3=\frac{7}{2}hay2x-3=-\frac{7}{2}\)
2x = \(\frac{7}{2}+3\) 2x = \(-\frac{7}{2}+3\)
2x = \(\frac{13}{2}\) 2x = \(-\frac{1}{2}\)
x = \(\frac{12}{2}:2=\frac{13}{4}\) x = \(-\frac{1}{2}:2\) = \(-\frac{1}{4}\)
Vậy...