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, \(3^{500}< 7^{300}.\)

Giải:

a) A=17¹⁸+1/17¹⁹+1

17A=17¹⁹+17/17¹⁹+1

17A=17¹⁹+1+16/17¹⁹+1

17A=1+16/17¹⁹+1

Tương tự:

B=17¹⁷+1/17¹⁸+1

17B=17¹⁸+17/17¹⁸+1

17B=17¹⁸+1+16/17¹⁸+1

17B=1+16/17¹⁸+1

Vì 16/17¹⁹+1<16/17¹⁸+1 nên 17A<17B

⇒A<B

b) A=10⁸-2/10⁸+2

    A=10⁸+2-4/10⁸+2

    A=1+-4/10⁸+2

Tương tự:

B=10⁸/10⁸+4

B=10⁸+4-4/10⁸+1

B=1+-4/10⁸+1

Vì -4/10⁸+2>-4/10⁸+1 nên A>B

c)A=20¹⁰+1/20¹⁰-1

   A=20¹⁰-1+2/20¹⁰-1

   A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-3>2/20¹⁰-1 nên B>A

⇒A<B

Chúc bạn học tốt!

a: \(98^{10}\cdot A=\dfrac{98^{98}+98^{10}}{98^{98}+1}=1+\dfrac{98^{10}-1}{98^{98}+1}\)

\(98^{10}\cdot B=\dfrac{98^{99}+98^{10}}{98^{99}+1}=1+\dfrac{98^{10}-1}{98^{99}+1}\)

98^88+1>98^99+1

=>A<B

b: \(\dfrac{1}{2022^2}\cdot C=\dfrac{2022^{2023}+1}{2022^{2023}+2022^2}=1+\dfrac{1-2022^2}{2022^{2023}+2022^2}\)

\(\dfrac{1}{2022^2}\cdot D=\dfrac{2022^{2021}+1}{2022^{2021}+2022^2}=1+\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

2022^2023>2022^2021

=>2022^2023+2022^2>2022^2021+2022^2

=>\(\dfrac{2022^2-1}{2022^{2023}+2022^2}< \dfrac{2022^2-1}{2022^{2021}+2022^2}\)

=>\(\dfrac{1-2022^2}{2022^{2023}+2022^2}>\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

=>C>D

a) Ta có: 1,(81) = 1,8181…

Vì 1,8181… > 1,812 nên -1,8181… < -1,812 hay -1,(81) < -1,812

b) Ta có: \(2\frac{1}{7}\) = 2,142857….

Vì 2,142857….> 2,142 nên \(2\frac{1}{7}\) > 2,142

c) Vì 48,075… < 48,275… nên - 48,075…. > – 48,275…

d) Vì 5 < 8 nên \(\sqrt 5 \) < \(\sqrt 8 \)

a: -1,(81)>-1,812

b: 2+1/7>2,142

c: -48,075...>-48,275...

d: \(\sqrt{5}< \sqrt{8}\)

a) <
b) <
c) >
d) =

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,Ta có:\(2=\sqrt{4}\)

Vì \(\sqrt{4}>\sqrt{3}\)

\(\Rightarrow2>\sqrt{3}\)

b,Ta có:\(6=\sqrt{36}\)

Vì \(\sqrt{36}< \sqrt{41}\)

\(\Rightarrow6< \sqrt{41}\)

c,Ta có:\(7=\sqrt{49}\)

Vì \(\sqrt{49}>\sqrt{47}\)

\(\Rightarrow7>\sqrt{47}\)

a) 2 =√4 > √3 ;    

b) 6=√36 < √41 ;    

c) 7=√49 > √47

a) Ta có: A=45.47

          => A=(44+1).47

          => A=44.47+47

Lại có: B=44.48

=> B=44.(47+1)

=> B=44.47+44

Vì 47>44 => A>B

b) Ta có: C=27.58-31

              => C= (26+1).58-31

              => C=26.58+58-31

              => C=26.58+26

Vì 26.58+26=58.26+27 => C=D

c) Ta có: E= 2005.2005

            => E= (1995+5).2005

           => E=1995.2005+5.2005

Lại có: F=1995.2010               ( Mình nghĩ đề đúng phải là 2010)

        => F=1995.(2005+5)

        => F=1995.2005+1995.5

Vì 1995.5<2005.5 => F<E

Thắc mắc j cứ nhắn tin vào mik nhá pn!

a) A=(44+1).47= 44.47 + 47

B=44.(47+1)=44.47 + 44 

Vì 44.47 = 44.47 ; 44<47

=> A>B

- Ice Wings đúng rồi.