[Toán 8] Giải phương trình

V

vanmanh2001

$(x+1)^4 = 1$
\Rightarrow $(x+1)^4 - 1 = 0$
\Rightarrow $[(x+1)^2 - 1] [(x+1)^2 + 1 ] = 0$
Vì $(x+1)^2 + 1 > 0$ \Rightarrow (x+1 - 1)(x+1 + 1) = 0 \Rightarrow $x(x+2) = 0$
\Rightarrow $x = 0$ hoặc $x = -2$
=)
 
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H

hocsinhchankinh

$(x+1)^4=1$
\Leftrightarrow$(x+1)^4-1=0$
\Leftrightarrow$[(x+1)^2+1][(x+1)^2-1]=0$
vì (x+1)^2+2\geq1 nên
(x+1+1)(x+1-1)=0
\Leftrightarrowx=0
x=-2

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