[Toán 9] $x^2+4x+7=(x+4)\sqrt{x^2+7}$

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${x}^{2}+4x+7=(x+4)\sqrt{{x}^{2}+7}$
\Leftrightarrow $({x}^{2}+4x+7)^2 = [(x+4)\sqrt{{x}^{2}+7} ]^2$
\Leftrightarrow $x^4 + (4x)^2 + 7^2 + 2.x^2.4x + 2.4x.7 + 2.x^2.7 = (x^2 + 8x + 16)(x^2 + 7)$
\Leftrightarrow $x^4 + 16x^2 + 49 + 8x^3 + 56x + 14x^2 = x^4 + 8x^3 + 16x^2 + 7x^2 + 56x + 112$
\Leftrightarrow $x^4 + 8x^3 + 30x^2 + 56x + 49 = x^4 + 8x^3 + 23x^2 + 56x + 112$
\Leftrightarrow $7x^2 = 63$
\Leftrightarrow $x^2 = 9$
\Leftrightarrow $x = 3 hoặc x = -3$
 
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