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Question

(X+2)5-(x-2)5=64.giúp mk vs ạ

Nguyễn Đức Trí · Accepted Answer

$\left(x+2\right)^5-\left(x-2\right)^5=64$

$\Rightarrow x^5+10x^4+40x^3+80x^2+80x+32-\left(x^5-10x^4+40x^3-80x^2+80x-32\right)=64$

$\Rightarrow20x^4+160x^2+64=64$

$\Rightarrow20x^4+160x^2=0$

$\Rightarrow20x^2\left(x^2+8\right)=0$

mà $x^2+8>0$

$\Rightarrow x^2=0\Rightarrow x=0$Câu trả lời: $\left(x+2\right)^5-\left(x-2\right)^5=64$

$\Rightarrow x^5+10x^4+40x^3+80x^2+80x+32-\left(x^5-10x^4+40x^3-80x^2+80x-32\right)=64$

$\Rightarrow20x^4+160x^2+64=64$

$\Rightarrow20x^4+160x^2=0$

$\Rightarrow20x^2\left(x^2+8\right)=0$

mà $x^2+8>0$

$\Rightarrow x^2=0\Rightarrow x=0$

✎﹏Kim Kim✧❤‿✶🌈(*•.¸♡ţę... · Answer

$\Leftrightarrow x^5+10x^4+40x^3+80x^2+80x+32-x^5+10x^4-40x^3+80x^2-80x+32=64$

$\Rightarrow20x^4+160x^2+54-64=0$

$\Rightarrow20x^4+160x^2=0$

$\Leftrightarrow20x^2\left(x^2+8\right)=0$

$\Leftrightarrow x=0$

Do $x^2+8=\ge0$(luôn đúng)

Vây: $x^2\ge-8$Câu trả lời: $\Leftrightarrow x^5+10x^4+40x^3+80x^2+80x+32-x^5+10x^4-40x^3+80x^2-80x+32=64$

$\Rightarrow20x^4+160x^2+54-64=0$

$\Rightarrow20x^4+160x^2=0$

$\Leftrightarrow20x^2\left(x^2+8\right)=0$

$\Leftrightarrow x=0$

Do $x^2+8=\ge0$(luôn đúng)

Vây: $x^2\ge-8$