Question

1)  Tim x

3^x +3^x+1 +3^x+2=351

2) So sanh

a)25^15 va 8^10×3^30

b(0,1)^10 va (0,3)^20

Answer 1

\(1)\) \(3^x+3^{x+1}+3^{x+2}=351\)

\(\Leftrightarrow\)\(3^x.1+3^x.3+3^x.3^2=351\)

\(\Leftrightarrow\)\(3^x\left(1+3+3^2\right)=351\)

\(\Leftrightarrow\)\(3^x.13=351\)

\(\Leftrightarrow\)\(3^x=\frac{351}{13}\)

\(\Leftrightarrow\)\(3^x=27\)

\(\Leftrightarrow\)\(3^x=3^3\)

\(\Leftrightarrow\)\(x=3\)

Vậy \(x=3\)

Chúc bạn học tốt ~ 

Answer 2

\(2)\) 

\(a)\) Ta có : 

\(25^{15}=\left(5^2\right)^{15}=5^{2.15}=5^{30}\)

\(8^{10}.3^{30}=\left(2^3\right)^{10}.3^{30}=2^{30}.3^{30}=\left(2.3\right)^{30}=6^{30}\)

Vì \(5^{30}< 6^{30}\) nên \(25^{15}< 8^{10}.3^{30}\)

Vậy \(25^{15}< 8^{10}.3^{30}\)

\(b)\) Ta có : 

\(\left(0,3\right)^{20}=\left[\left(0,3\right)^2\right]^{10}=\left(0,09\right)^{10}\)

Vì \(\left(0,1\right)^{10}>\left(0,09\right)^{10}\) nên \(\left(0,1\right)^{10}>\left(0,3\right)^{20}\)

Vậy \(\left(0,1\right)^{10}>\left(0,3\right)^{20}\)

Chúc bạn học tốt ~