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kisihoangtoc

$(x+3)^4+(x-3)^2=162$
\Leftrightarrow$(x^4+12x^3+54x^2+108x+81)+(x^4-12x^3+54x^2-108x+81)=162$
\Leftrightarrow$x^4+54x^2=0$
\Leftrightarrow$x^2(x^2+54)=0$
\Rightarrow $x=0$
 
H

hacongdinh2001

dễ

$(x+3)^4+(x-3)^4=162$
\Rightarrow $x^4+81+x^4-81 =162$
\Rightarrow $2.x^4=162$
$ x^4=162:2$
$ x^4=81$
vì $3^4 = 81$
\Rightarrow x =3
Chú ý latex
 
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H

huynhbachkhoa23

kisihoangtoc said:
$(x+3)^4+(x-3)^4=162$
\Leftrightarrow$(x^4+12x^3+54x^2+108x+81)+(x^4-12x^3+54x^2-108x+81)=162$
\Leftrightarrow$x^4+54x^2=0$
\Leftrightarrow$x^2(x^2+54)=0$
\Rightarrow $x=0$
Bấm để xem đầy đủ nội dung ...

Nếu muốn làm nhanh hơn thì theo cách này.

Áp dụng BDT Cauchy-Schwarz:

$(x+3)^4+(x-3)^4 \ge \dfrac{\left [(x+3)^2+(3-x)^2\right ]^2}{2} \ge \dfrac{\left [ \dfrac{(x+3+3-x)^2}{2} \right ]^2}{2}=162$

$\to x+3=3-x \leftrightarrow x=0$
 
K

khanhlinh2018

Sử dụng tam giác Pascal khai triển:
[TEX](a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 [/TEX]
[TEX] (x + 3)^4 + (x - 3)^4 =162 [/TEX]
[tex]\Leftrightarrow( x^4 + 12x^3 + 54x^2 + 108x^3 + 81) + ( x^4 - 12x^3 + 54x^2 - 108x^3 + 81) =162 [/TEX]
[tex]\Leftrightarrow 2x^4 + 108x^2 + 162 = 162[/TEX]
[tex]\Leftrightarrow 2x^4 + 108x^2 = 0[/TEX]
[tex]\Leftrightarrow 2x^2( x^2 + 90) =0[/TEX]
[tex]\Leftrightarrow 2x^2 = 0[/TEX]
x^2 + 90 = 0
[tex]\Leftrightarrow x = 0[/TEX] (chọn)
[tex] x^2 = -90 [/TEX] (loại vì x^2[tex] \ge[/TEX] 0)
Vậy x= 0
 
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