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, \(3^{500}< 7^{300}.\)

Bài 1:

   D     =      5  + 5² + 5³+...+ 5¹⁰⁰

5.D     =             5² + 5³+...+5 ¹⁰⁰ + 5¹⁰¹

5D - D = 5¹⁰¹ - 5

4D       = 5¹⁰¹ - 5

  D      = \(\dfrac{5^{101}-5}{4}\)

Bài 2:

So sánh 

a, 54⁴ = (2.3³)⁴ = 2⁴.3¹²  

    21¹² = (3.7)¹² = 3¹².7¹²

Vì 2⁴ < 7¹² nên 54⁴ < 21¹²

b, 3³⁹ và 11²¹

    3³⁹   =   (3¹³)³

   11²¹ = (11⁷)³

     3¹³ = (3²)⁶.3 = 9⁶.3 < 9⁷ < 11⁷ 

Vậy 3³⁹  < 11²¹

1) \(D=5+5^2+5^3+...+5^{100}\)

\(\Rightarrow D+1=1+5+5^2+5^3+...+5^{100}\)

\(\Rightarrow D+1=\dfrac{5^{100+1}-1}{5-1}\)

\(\Rightarrow D+1=\dfrac{5^{101}-1}{4}\)

\(\Rightarrow D=\dfrac{5^{101}-1}{4}-1=\dfrac{5^{101}-5}{4}=\dfrac{5\left(5^{100}-1\right)}{4}\)

2)

a) \(21^{12}=\left(21^3\right)^4=9261^4>54^4\Rightarrow54^4< 21^{12}\)

b) \(3^{39}< 3^{40}=\left(3^2\right)^{20}=9^{20}< 11^{20}< 11^{21}\)

\(\Rightarrow3^{39}< 11^{21}\)

c) \(201^{60}=\left(201^4\right)^{15}=\text{1632240801}^{15}\)

\(398^{45}=\left(398^3\right)^{15}=\text{63044792}^{15}< \text{1632240801}^{15}\)

\(201^{60}>398^{45}\)

a)

Ta có : A = 27⁵ = (3³)⁵ = 3¹⁵

            B = 243³ = (3⁵)³ = 3¹⁵

Vì 3¹⁵ = 3¹⁵ => A = B

b )

Ta có : A = 2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰

            B = 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰

Vì 8¹⁰⁰ < 9¹⁰⁰ => A<B

a) Ta có:

\(\dfrac{16}{9}\)=\(\dfrac{48}{27}\)                 \(\dfrac{24}{13}=\dfrac{48}{26}\)

Vì 27>26

➝\(\dfrac{48}{27}>\dfrac{48}{26}hay\dfrac{16}{9}>\dfrac{24}{13}\)

So sánh:

a) 16/9 và 24/13 

 Ta có \(\dfrac{16}{9}=\dfrac{208}{117}\) và \(\dfrac{24}{13}=\) \(\dfrac{216}{117}\)

\(\Rightarrow\dfrac{216}{117}>\dfrac{208}{117}\Rightarrow\dfrac{24}{13}>\dfrac{16}{9}\)

b) 27/82 và 26/75

Ta có \(\dfrac{27}{82}\approx0,33\) và \(\dfrac{26}{75}\approx0,35\)

\(\Rightarrow9,35>0,33\Rightarrow\dfrac{26}{75}>\dfrac{27}{82}\)

a) Ta có: \(\dfrac{16}{9}=\dfrac{16\cdot13}{9\cdot13}=\dfrac{208}{117}\)

\(\dfrac{24}{13}=\dfrac{24\cdot9}{13\cdot9}=\dfrac{216}{117}\)

mà \(\dfrac{208}{117}< \dfrac{216}{117}\)

nên \(\dfrac{16}{9}< \dfrac{24}{13}\)

a: 3>2

=>5/3<5/2

b: 23/21>1

21/23<1

=>23/21>21/23

c: -15/-17>0

16/-19<0

=>-15/-17>16/-19

a: Ta có: \(3^{2020}=3^{2018}\cdot3^2=3^{2018}\cdot9\)

mà 9<10

nên \(3^{2020}< 10\cdot3^{2018}\)

a: \(4\sqrt{7}=\sqrt{4^2\cdot7}=\sqrt{112}\)

\(3\sqrt{13}=\sqrt{3^2\cdot13}=\sqrt{117}\)

mà 112<117

nên \(4\sqrt{7}< 3\sqrt{13}\)

b: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(\dfrac{1}{4}\sqrt{84}=\sqrt{\dfrac{1}{16}\cdot84}=\sqrt{\dfrac{21}{4}}\)

\(6\sqrt{\dfrac{1}{7}}=\sqrt{36\cdot\dfrac{1}{7}}=\sqrt{\dfrac{36}{7}}\)

mà \(\dfrac{21}{4}>\dfrac{36}{7}\)

nên \(\dfrac{1}{4}\sqrt{84}>6\sqrt{\dfrac{1}{7}}\)

d: \(3\sqrt{12}=\sqrt{3^2\cdot12}=\sqrt{108}\)

\(2\sqrt{16}=\sqrt{16\cdot2^2}=\sqrt{64}\)

mà 108>64

nên \(3\sqrt{12}>2\sqrt{16}\)

c: \(3=\sqrt{16}-1< \sqrt{17}-1\)