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3,47<3,x9<3,82
x=4,5,6,7
25,41<x<32,1
x=26,27,28,29,30,31,32
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a) Ta có: x 2 = 2 2 nên x = 2.
b) Ta có: x 2 = 5 2 nên x = 5.
c) Ta có: 3 . x 5 = 3 nên x 5 = 1 . Do đó x = 1.
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\(176:x^2+79=10\cdot9\)
\(\Leftrightarrow176:x^2=11\)
\(\Leftrightarrow x=4\)
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Bài 1:
a: Ta có: \(48751-\left(10425+y\right)=3828:12\)
\(\Leftrightarrow y+10425=48751-319=48432\)
hay y=38007
b: Ta có: \(\left(2367-y\right)-\left(2^{10}-7\right)=15^2-20\)
\(\Leftrightarrow2367-y=1222\)
hay y=1145
Bài 2:
Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
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\(1440:\left[120-\left(3.9x\right)\right]=120\) \(120+\left[\left(999-9x\right):60\right].24=480\)
\(120-\left(3.9x\right)=1440:120\) \(\left[\left(999-9x\right):60\right].24=360\)
\(120-\left(3.9x\right)=12\) \(\left(999-9x\right):60=15\)
\(3.9x=120-12\) \(999-9x=900\)
\(3.9x=108\) \(9x=999-900\)
\(9x=108:3\) \(9x=99\)
\(9x=36\) \(x=99:9\)
\(x=4\) \(x=11\)