Question

[Kiều Tuấn Nhật Minh]

help me 

1 tìm x

2 chứng minh rằng

và bài này 

ai nhanh mik tick cho

Answer 1

[Yuri Sweet[𝕿𝖊𝖆𝖒 𝕹𝖊𝖕𝖆𝖑]]

1.Điều kiện : \(x\ge0\)

\(\Rightarrow\hept{\begin{cases}x+3,4>0\\x+2,4>0\\x+7,2>0\end{cases}}\Rightarrow\hept{\begin{cases}\left|x+3,4\right|=x+3,4\\\left|x+2,4\right|=x+2,4\\\left|x+7,2\right|=x+7,2\end{cases}}\)

\(\Rightarrow\left|x+3,4\right|+\left|x+2,4\right|+\left|x+7,2\right|=x+3,4+x+2,4+x+7,2\)

                                                                                \(=3x+13=4x\)

\(\Rightarrow4x-3x=13\)

\(\Rightarrow x=13\)

Vậy \(x=13\)

2.\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)

\(=3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)\)

\(=3^n\left(27+3\right)+2^n\left(8+4\right)\)

\(=3^n.30+2^n.12\)

\(=6\left(3^n.5+2^n.2\right)⋮6\)

Answer 2

[Yuri Sweet[𝕿𝖊𝖆𝖒 𝕹𝖊𝖕𝖆𝖑]]

4.a)

• \(3^{34}=3^{30+4}=3^{30}.3^4=3^{3.10}.3^4=\left(3^3\right)^{10}.3^4=27^{10}.3^4\)

\(5^{20}=5^{2.10}=\left(5^2\right)^{10}=25^{10}\)

Vì \(27^{10}>25^{10}\Rightarrow27^{10}.3^4>25^{10}\)

hay \(3^{34}>5^{20}\)

• \(17^{20}=17^{4.5}=\left(17^4\right)^5=83521^5>71^5\)

b)\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)

Vì \(8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)