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

0 bình luận về “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”

  1. Đáp án:

    c. \(x =  \pm \sqrt 2 \)

    Giải thích các bước giải:

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

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