[Toán 9] Phương trình

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vansang02121998

$(2x+1)^2(x+1)x=105$

$\Leftrightarrow (2x+1)^2(2x+2).2x=420$

Đặt $2x+1=a$

$\Leftrightarrow a^2(a+1)(a-1)=420$

$\Leftrightarrow a^4-a^2-420=0$

$\Leftrightarrow a^2-21=0$

$\Leftrightarrow 2x+1=\pm \sqrt{21}$

Cách khác

$(2x+1)^2(x+1)x=105$

$\Leftrightarrow (x^2+x-5)(4x^2+4x+21)=0$
 
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