\(28+\left(2x+3\right)^3=92\Leftrightarrow\left(2x+3\right)^3=92-28=64=4^3\)
\(\Rightarrow2x+3=4\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\( 28 + {\left( {2x + 3} \right)^3} = 92\\ \Leftrightarrow 28 + 4{x^2} + 12x + 9 + 92\\ \Leftrightarrow 37 + 4{x^2} + 12x - 92 = 0\\ \Leftrightarrow - 55 + 4{x^2} + 12x = 0\\ \Leftrightarrow 4{x^2} + 12x - 55 = 0\\ \Leftrightarrow 4{x^2} + 22x - 10x - 55 = 0\\ \Leftrightarrow 2x\left( {2x + 11} \right) - 5\left( {2x + 11} \right) = 0\\ \Leftrightarrow \left( {2x + 11} \right)\left( {2x - 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} 2x + 11 = 0\\ 2x - 5 = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = - \dfrac{{11}}{2}\\ x = \dfrac{5}{2} \end{array} \right. \)
\(28+\left(2x+3\right)^3=92\)
=> \(\left(2x+3\right)^3=92-28\)
=> \(\left(2x+3\right)^3=64\)
=> \(\left(2x+3\right)^3=4^3\)
=> \(2x+3=4\)
=> \(2x=4-3\)
=> \(2x=1\)
=> \(x=1:2\)
=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}.\)
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28 + (2x + 3 )\(^3\) = 92
=> ( 2x + 3 )\(^3\) = 92 - 28 = 64
=> ( 2x + 3 )\(^3\) = 4\(^3\)
=> 2x + 3 = 4
=> 2x = 4 - 3 = 1
=> x = \(\frac{1}{2}\)
28 + ( 2x + 3 )3 = 92
<=> ( 2x + 3 )3 = 92 - 28
<=> ( 2x + 3 )3 = 64
<=> ( 2x + 3 )3 = 43
<=> 2x + 3 = 4
<=> 2x = 4 - 3 = 1
<=> x = 1/2
Vậy....