a: \(2^{300}=8^{100}\)

\(3^{200}=9^{100}\)

mà 8<9

nên \(2^{300}< 3^{200}\)

b: \(3^{500}=243^{100}\)

\(7^{300}=343^{100}\)

mà 243<243

nên \(3^{500}< 7^{300}\)

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

\(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)

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

\(7^{300}=\left(7^3\right)^{100}=343^{100}>243^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

Giải chi tiết giúp mình ạ~

\(\left(d\right):202^{303}=\left(202^3\right)^{101}=8242408^{101}>303^{202}=\left(303^2\right)^{101}=91809^{101}\)

\(\left(e\right):107^{50}=\left(107^2\right)^{25}=11449^{25}< 73^{75}=\left(73^3\right)^{25}=389017^{25}\)

Bài 8:

a) \(2^{225}=\left(2^3\right)^{75}=8^{75}\)

\(3^{150}=\left(3^2\right)^{75}=9^{75}\)

Vì \(8^{75}< 9^{75}\Rightarrow2^{225}< 3^{150}\)

b) \(2^{91}=\left(2^{13}\right)^7=8192^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\)

Vì \(8192^7>3125^7\Rightarrow2^{91}>5^{35}\)

c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)

Bài 6: 

a: \(15=\sqrt{225}>\sqrt{200}\)

b: \(27=9\sqrt{9}>9\sqrt{5}\)

c: \(-24=-\sqrt{576}< -\sqrt{540}=-6\sqrt{15}\)

a) \(2^x=16=2^4\Rightarrow x=4\)

b) \(x^3=27=3^3\Rightarrow x=3\)

c) \(x^{50}=x\Rightarrow x\left(x^{49}-1\right)=0\Rightarrow x=0\) hay \(x=1\)

d) \(\left(x-2\right)^2=16=4^2\Rightarrow x-2=4\) hay \(x-2=-4\)

\(\Rightarrow x=6\) hay \(x=-2\)

a) \(2^{300}=2^{3.100}=8^{100}\)

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

vì \(8^{100}< 9^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) \(3^{500}=3^{5.100}=243^{100}\)

\(7^{300}=7^{3.100}=343^{100}\)

vì \(243^{100}< 343^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

`@` `\text {Ans}`

`\downarrow`

`1,`

`a,`

`2^x = 16`

`=> 2^x = 2^4`

`=> x = 4`

Vậy, `x = 4`

`b,`

`x^3 = 27`

`=> x^3 = 3^3`

`=> x = 3`

Vậy, `x = 3`

`c,`

\(x^{50}=x\)

`=>`\(x^{50}-x=0\)

`=>`\(x\left(x^{49}-1\right)=0\)

`=>`\(\left[{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x^{49}=1\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy, `x \in {0; 1}`

`d,`

`(x-2^2)=16`

`=> x - 2^2 = 16`

`=> x = 16 + 2^2`

`=> x = 20`

Vậy, `x = 20`

`2,`

`a,`

Ta có:

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

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

Vì `8 < 9 =>`\(8^{100}< 9^{100}\)

`=>`\(2^{300}< 3^{200}\)

Vậy, \(2^{300}< 3^{200}\)

`b,`

Ta có:

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

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

Vì `243 < 343 =>`\(243^{100}< 343^{100}\)

`=>`\(3^{500}< 7^{300}\)

Vậy, \(3^{500}< 7^{300}.\)

a: \(2^{333}=8^{111}< 9^{111}=3^{222}\)

giúp minh

A

A?

A

a)

\(\begin{array}{l}{( - 3)^2}.{( - 3)^4} = 9.81 = 729\\ {( - 3)^6} = ( - 3).( - 3).( - 3).( - 3).( - 3).( - 3)\\ = 9.9.9 = 729\end{array}\)

Vậy \({( - 3)^2}.{( - 3)^4}\) = \({( - 3)^{6}}\)

b)

\(\begin{array}{l}0,6{}^3:0,{6^2} = 0,216:0,36 = 0,6\end{array}\)

Vậy \(0,6{}^3:0,{6^2}\) = \(0,{6}\)