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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: `(5x)^2+(8x-3)^2=(4x+5)^2` Tìm x

Tháng Tư 25, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: (5x)^2+(8x-3)^2=(4x+5)^2
Tìm x

Comments ( 1 )

  1. hiennhi98
    hiennhi98
    25 Tháng Tư, 2022 at 5:06 sáng
    Reply
    (5x)^2 +(8x-3)^2 =(4x+5)^2
    <=> 25x^2 +( 64x^2-48x + 9) =16x^2+40x+25
    <=> 25x^2 +64x^2 -48x + 9 – 16x^2 -40x -25 =0
    <=> (25x^2+64x^2 -16x^2) + (-48x -40x) +(9 -25) =0
    <=>73x^2 -88x -16 =0
    <=> 73(x^2 -88/73x – 16/73) =0
    <=> 73(x^2 – 2 . x . 44/73 +  1936/5329 -3104/5329) =0
    <=> 73(x – 44/73)^2 – 3104/73 =0
    <=> 73(x-44/73)^2 = 3104/73
    <=> (x-44/73)^2 =3104/5329  
    <=>  \(\left[ \begin{array}{l}x-\dfrac{44}{73} = \sqrt{\dfrac{3104}{5329}}\\x-\dfrac{44}{73} = -\sqrt{\dfrac{3104}{5329}}\end{array} \right.\)  
    <=>  \(\left[ \begin{array}{l}x =\sqrt{\dfrac{3104}{5329}} + \dfrac{44}{73} = \dfrac{\sqrt{3104}}{73} + \dfrac{44}{73}\\x=\dfrac{44}{73} – \sqrt{\dfrac{3104}{5329}} =\dfrac{44}{73} -\dfrac{\sqrt{3104}}{73} \end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}x=\dfrac{\sqrt{3104} +44}{73}\\x= \dfrac{44-\sqrt{3104} }{73}\end{array} \right.\)   
    Vậy S = {\frac{\sqrt{3104} +44}{73} ; \frac{44-\sqrt{3104} }{73}}

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222-9+11+12:2*14+14 = ? ( )

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