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