[Toán 9] Giải phương trình

T

thanhlan9

Đk x [TEX]\geq[/TEX]-1
Pt [TEX]\Leftrightarrow (x+1)^2-(x+1)+12\sqrt[2]{x+1}-36=0[/TEX]
Đặt t=[TEX]\sqrt[2]{x+1} (t \geq 0)[/TEX]
Pt trở thành [TEX]t^4-t^2+12t-36=0 \Leftrightarrow t=2 x=3[/TEX]
 
L

lp_qt

$x^2+x+12\sqrt{x+1}=36$

\Leftrightarrow $x^2+2x+1=x+1-12\sqrt{x+1}+36$

\Leftrightarrow $(x+1)^2=(\sqrt{x+1}-6)^2$

\Leftrightarrow $\begin{bmatrix}x+1=\sqrt{x+1}-6 & \\ x+1=-\sqrt{x+1}+6
& \end{bmatrix}$

\Leftrightarrow ...
 
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