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

Toán Lớp 8: tìm x 5 – $(2x – 3)^{2}$ – $(x + 1)^{2}$ – 15(x + 4)(x – 4) = – 10

Home/ Toán Lớp 8: tìm x 5 – $(2x – 3)^{2}$ – $(x + 1)^{2}$ – 15(x + 4)(x – 4) = – 10

Toán Lớp 8: tìm x 5 – $(2x – 3)^{2}$ – $(x + 1)^{2}$ – 15(x + 4)(x – 4) = – 10

Tháng Chín 16, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: tìm x
5 – $(2x – 3)^{2}$ – $(x + 1)^{2}$ – 15(x + 4)(x – 4) = – 10

Comments ( 1 )

  1. truongthanhhung
    truongthanhhung
    16 Tháng Chín, 2022 at 12:08 sáng
    Reply
    5 – (2x-3)^2 – (x+1)^2 – 15(x+4)(x-4) = -10
    => 5 – (4x^2 – 12x + 9) – (x^2 + 2x + 1) – 15(x^2 – 16) = -10
    => 5 – 4x^2 + 12x – 9 – x^2 – 2x – 1 – 15x^2 + 240 = -10
    => – 4x^2 + 12x – x^2 – 2x – 15x^2 = -10 – 5 + 9 + 1 – 240
    => – 20x^2 + 10x = -245
    => 20x^2 – 10x – 245 = 0
    => 20x^2 – 10x + 5/4 = 985/4
    => (sqrt{20}x-sqrt{5}/2)^2= 985/4
    => \(\left[ \begin{array}{l} \sqrt{20}x – \frac{\sqrt{5}}{2} = -\frac{\sqrt{985}}{2}\\\sqrt{20}x – \frac{\sqrt{5}}{2} = \frac{\sqrt{985}}{2}\end{array} \right.\) 
    => \(\left[ \begin{array}{l} \sqrt{20}x = -\frac{\sqrt{985}-\sqrt{5}}{2}\\\sqrt{20}x = \frac{\sqrt{985}+\sqrt{5}}{2}\end{array} \right.\) 
    => \(\left[ \begin{array}{l} x = -\frac{\sqrt{985}-\sqrt{5}}{2\sqrt{20}}\\x = \frac{\sqrt{985}+\sqrt{5}}{2\sqrt{20}}\end{array} \right.\) 
    => \(\left[ \begin{array}{l} x = \frac{1-\sqrt{197}}{4}\\x = \frac{1+\sqrt{197}}{4}\end{array} \right.\) 

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

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