(2x-5)(2x+1)=(2x-5)(x+4)
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52 - 2x = -11
25 - 2x = - 11
2x = 25 - ( - 11 )
2x = 36
x = 36 : 2
x = 18
Vẫy x = 18
`5^2 - 2x = -11`
`=> 25 - 2x = -11`
`=> 2x = 25 - (-11) `
`=> 2x = 25 + 11`
`=> 2x = 36`
`=> x = 36 : 2`
`=> x = 18`
Vậy ...

\(\dfrac{16}{2^x}\) = 2
\(\dfrac{2^4}{2^x}\) = 2
2\(4-x\) = 21
4 - \(x=1\)
\(x=4-1\)
\(x=2\)
Vậy \(x=3\)
Cách hai: \(\dfrac{16}{2^x}\) = 2
2\(^x\) = 16 : 2
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)

`270 - (3^2 xx 2 + 7^15 : 7^13)`
`= 270 - (9 xx 2 + 7^(15-13))`
`= 270 - (18 + 7^2)`
`= 270 - (18 + 49)`
`= 270 - 67`
`= 203`

( - 1127 - 785 ) - ( 85 - 127 + 215 )
= - 1127 - 785 - 85 + 127 - 215
= ( - 1127 + 127 ) - ( 785 + 215 ) - 85
= - 1000 - 1000 - 85
= - 2000 - 85
= - 2085
`(-1127-785)-(85-127+215)`
`= -1127-785-85+127-215`
`= -(1127-127)-(785+215)-85`
`= -1000-1000-85`
`= -2085`

a; (\(\sqrt{45}\) - \(\sqrt{125}\) + \(\sqrt{20}\)) : \(\sqrt{5}\)
= (\(\sqrt{9.5}\) - \(\sqrt{25.5}\) + \(\sqrt{4.5}\)):\(\sqrt{5}\)
= (3\(\sqrt{5}\) - 5\(\sqrt{5}\) + 2\(\sqrt{5}\)): \(\sqrt{5}\)
= (- 2\(\sqrt{5}\) + 2\(\sqrt{5}\)) : \(\sqrt{5}\)
= 0 : \(\sqrt{5}\)
= 0


`(2x - 5)(2x + 1) = (2x - 5)(x + 4)`
`(2x - 5)(2x + 1) - (2x - 5)(x +4) = 0`
`(2x - 5)[(2x + 1) - (x + 4)]=0`
`(2x - 5)(2x + 1 - x - 4) = 0`
`(2x - 5)(x - 3) = 0`
\(\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\\ \left[{}\begin{matrix}2x=5\\x=3\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
(2\(x-5\)).(2\(x+1\)) = (2\(x-5\)).(\(x+4\))
(2\(x-5\))(2\(x+1\)) - (\(2x-5\)).(\(x+4\)) = 0
(2\(x-5\))[2\(x+1\) - \(x-4\)] = 0
(2\(x-5\)).[(2\(x-x\)) - (4 - 1)] = 0
(2\(x\) - 5).[\(x\) - 3] = 0
\(\left[{}\begin{matrix}2x-5=0\\x-3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\) {\(\dfrac{5}{2}\); 3}