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a) TH1: \(x< 2,5\) , ta có:
\(2,5-x=1,3\)
\(x=2,5-1,3=1,2\)
TH2: \(x\ge2,5\), ta có:
\(x-2,5=1,3\)
\(x=1,3+2,5=3,8\)
Vậy \(\orbr{\begin{cases}x=1,2\\x=3,8\end{cases}}\)
b) \(1,6-\left|x-0,2\right|=0\)
\(\Rightarrow\left|x-0,2\right|=1,6\)
TH1: \(x< 0,2\) ta có:
\(0,2-x=1,6\Rightarrow x=0,2-1,6=-1,4\)
TH2: \(x\ge0,2\) ta có:
\(x-0,2=1,6\Rightarrow x=1,6+0,2=1,8\)
Vậy \(\orbr{\begin{cases}x=-1,4\\x=1,8\end{cases}}\)
a) \(12,5.4-x=4,3+2,6.\)
\(\Rightarrow12,5.4-x=6,9\)
\(\Rightarrow50-x=6,9\)
\(\Rightarrow x=50-6,9=43,1\)
b) \(\left(x+0,9\right).\left(1-0,4\right)=2412\)
\(\Rightarrow\left(x+0,9\right).0,6=2412\)
\(\Rightarrow x+0,9=2412:0,6=4020\)
\(\Rightarrow x=4020-0,9=4019,1\)
Ủng hộ mik nha!
\(a,\left[0,75x+\frac{5}{2}\right]-\frac{4}{7}-\left[-\frac{1}{3}\right]=-\frac{5}{6}\)
\(\Leftrightarrow\left[0,75x+\frac{5}{2}\right]-\frac{4}{7}+\frac{1}{3}=-\frac{5}{6}\)
\(\Leftrightarrow\left[0,75x+\frac{5}{2}\right]=-\frac{5}{6}-\frac{1}{3}+\frac{4}{7}\)
\(\Leftrightarrow0,75x+\frac{5}{2}=-\frac{25}{42}\)
\(\Leftrightarrow0,75x=-\frac{65}{21}\Leftrightarrow x=-\frac{260}{63}\)
\(b,\left[4x-9\right]\left[2,5+-\frac{7}{3}x\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x-9=0\\2,5+-\frac{7}{3}x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)
Vậy \(x\in\left\{\frac{9}{4};\frac{15}{14}\right\}\)
Tham khảo nhé phamthiminhanh
a) \(x:0,1+x:0,2=5,1\)
\(\Leftrightarrow x:\left(0,1+0,2\right)=5,1\)
\(\Leftrightarrow x:0,3=5,1\)
\(\Leftrightarrow x=5,1.0,3\)
\(\Leftrightarrow x=1,53\)
b) \(5:x-2,6:x=2,5\)
\(\Leftrightarrow\left(5-2,6\right):x=2,5\)
\(\Leftrightarrow2,4:x=2,5\)
\(\Leftrightarrow x=2,4:2,5\)
\(\Leftrightarrow x=0,96\)
\(a)\) \(x:0,1+x:0,2=5,1\)
\(x:\left(0,1+0,2\right)=5,1\)
\(x:0,3=5,1\)
\(x=5,1.0,3\)
\(x=1,53\)
\(b)5:x-2,6:x=2,5\)
\(x:\left(5-2,6\right)=2,5\)
\(x:2,4=2,5\)
\(x=2,5.2,4\)
\(x=6\)