a ) <

b ) <

-2021/2020<1/2

a)2022/2021<2021/2020

1.

b) \(3^x+3^{x+2}=2430\)

\(\Rightarrow3^x.1+3^x.3^2=2430\)

\(\Rightarrow3^x.\left(1+3^2\right)=2430\)

\(\Rightarrow3^x.10=2430\)

\(\Rightarrow3^x=2430:10\)

\(\Rightarrow3^x=243\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5.\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)

Chúc bạn học tốt!

a, A = 3⁵⁰⁰ = (3⁵)¹⁰⁰ = 243¹⁰⁰
B = 7³⁰⁰ = (7³)¹⁰⁰ = 343¹⁰⁰
Mà 243¹⁰⁰ < 343¹⁰⁰
=> A < B
@nguyễn thi trà giang

a) \(A=3^{500}=\left(3^5\right)^{100}=243^{100}\)

\(B=7^{300}=\left(7^3\right)^{100}=343^{100}\)

Vì \(243^{100}< 343^{100}\Rightarrow3^{500}< 7^{300}\)

\(\Rightarrow A< B\)

b) \(A=303^{202}=\left(303^2\right)^{101}=91809^{101}\)

\(B=202^{303}=\left(202^3\right)^{101}=8242408^{101}\)

Vì \(91809^{101}< 8242408^{101}\Rightarrow303^{202}< 202^{303}\)

\(\Rightarrow A< B\)

c) \(A=3^{21}=3\cdot3^{20}=3\cdot\left(3^2\right)^{10}=3\cdot9^{10}\)

\(B=2^{31}=2\cdot2^{30}=2\cdot\left(2^3\right)^{10}=2\cdot8^{10}\)

Ta có: \(3>2;9^{10}>8^{10}\Rightarrow3\cdot9^{10}>2\cdot8^{10}\Rightarrow3^{21}>2^{31}\)

\(\Rightarrow A< B\)

Ừ lớp 5 hỏi thế này có giải cũng chẳng hiểu=> thôi