Tìm x biết (x+2)^2+(x-3)^2-2.(x-1).(x+9)

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Asked: Tháng Mười 25, 20202020-10-25T02:41:28+00:00 2020-10-25T02:41:28+00:00In: Môn Toán

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Maris · Answer 1 · 2020-10-25T02:42:33+00:00

Đáp án:

Giải thích các bước giải :

(x + 2)^2 + (x – 3)^2 – 2(x – 1)(x + 9) = 9

=> x^2 + 4x + 4 + x^2 – 6x + 9 – 2[x(x + 9) – 1(x + 9)] = 9

=> x^2 + 4x + 4 + x^2 – 6x + 9 – 2(x^2 + 9x – x – 9) = 9

=> x^2 + 4x + 4 + x^2 – 6x + 9 – 2x^2 – 18x + 2x + 18 = 9

=> (x^2 + x^2 – 2x^2) + (4x – 6x -18x + 2x) + (4 + 9 + 18) = 9

=> -18x + 31 = 9

=> -18x = -22

=> 18x = 22 => x = 11/9

Sapo · Answer 2 · 2020-10-25T02:43:06+00:00

$(x+2)^2+(x-3)^2-2(x-1)(x+9)=9$
$⇒(x^2+4x+4)+(x^2-6x+9)-2(x^2+8x-9)=9$

$⇒x^2+4x+4+x^2-6x+9-2x^2-16x+18=9$

$⇒-18x+31=9$
$⇒-18x=-22$

`⇒x=11/9`