Đáp án:
c. \(x = \pm \sqrt 2 \)
Giải thích các bước giải:
\(\begin{array}{l}
a.(2x + 5)(2x - 7) - {(4x - 3)^2} = 16\\
\to 4{x^2} - 14x + 10x - 35 - \left( {16{x^2} - 24x + 9} \right) = 16\\
\to - 12{x^2} + 20x - 60 = 0\\
\to - 4\left( {3{x^2} - 5x + 15} \right) = 0\\
\to 3{x^2} - 5x + 15 = 0\\
\to 3{x^2} - 2.x\sqrt 3 .\dfrac{5}{{2\sqrt 3 }} + {\left( {\dfrac{5}{{2\sqrt 3 }}} \right)^2} + \dfrac{{155}}{{12}} = 0\\
\to {\left( {x\sqrt 3 - \dfrac{5}{{2\sqrt 3 }}} \right)^2} + \dfrac{{155}}{{12}} = 0\left( {vô lý} \right)\\
Do:{\left( {x\sqrt 3 - \dfrac{5}{{2\sqrt 3 }}} \right)^2} + \dfrac{{155}}{{12}} > 0\forall x \in R\\
b.{(3x + 5)^2} - (2x - 3)(2x + 3) = 5{(2 - x)^2}\\
\to 9{x^2} + 30x + 25 - \left( {4{x^2} - 9} \right) = 5\left( {4 - 4x + {x^2}} \right)\\
\to 50x = - 14\\
\to x = \dfrac{{ - 7}}{{25}}\\
c.(8{x^2} + 3)(8{x^2} - 3) - {(8{x^2} - 1)^2} = 22\\
\to 64{x^4} - 9 - \left( {64{x^4} - 16{x^2} + 1} \right) = 22\\
\to 16{x^2} = 32\\
\to {x^2} = 2\\
\to x = \pm \sqrt 2 \\
d.{(5x - 7)^2} - (4x - 3)(4x + 3) = {(2 - 3x)^2}\\
\to 25{x^2} - 70x + 49 - 16{x^2} + 9 = 4 - 12x + 9{x^2}\\
\to 58x = 54\\
\to x = \dfrac{{27}}{{29}}
\end{array}\)
