\(a,\left(2x-1\right)^3=-8\)

\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)

\(\Rightarrow2x-1=-2\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-\dfrac{1}{2}\)

\(b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)

\(\Rightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

\(\Rightarrow x=-\dfrac{1}{4}\)

\(c,\left(2x+3\right)^2=\dfrac{9}{121}\)

\(\Rightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\)

\(\Rightarrow2x+3=\dfrac{3}{11}\)

\(\Rightarrow2x=-\dfrac{30}{11}\)

\(\Rightarrow x=-\dfrac{15}{11}\)

\(d,\left(2x-1\right)^3=-\dfrac{8}{27}\)

\(\Rightarrow\left(2x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)

\(\Rightarrow2x-1=-\dfrac{2}{3}\)

\(\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)

\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\)

\(\left(2x+3\right)^2=\dfrac{9}{121}\Leftrightarrow\left(2x+3\right)^2=\left(\dfrac{3}{11}\right)^2\Leftrightarrow2x+3=\dfrac{3}{11}\Leftrightarrow x=\dfrac{-15}{11}\)

\(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\Leftrightarrow2x-1=-2\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\)

\(\text{Câu 1 :}\) \(\text{Tìm }x\)

\(\text{a) }\left(2x-1\right)^3=-8\\ \Leftrightarrow\left(2x-1\right)^3=-2^3\\ \Leftrightarrow2x-1=-2\\ \Leftrightarrow2x=-1\\ \Leftrightarrow x=-\dfrac{1}{2}\\ \text{Vậy }x=-\dfrac{1}{2}\)

\(\text{b) }\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\\ \Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\\ \Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{2}\\ \Leftrightarrow x=-\dfrac{1}{4}\\ \text{Vậy }x=-\dfrac{1}{4}\)

\(\text{Câu 2 : So Sánh}\)

\(\text{a) }99^{20}\text{ và }9999^{10}\\ \text{Ta có : }99^{20}=\left(99^2\right)^{10}=9801^{10}\\ \text{Mà }9801^{10}< 9999^{10}\\ \Rightarrow99^{20}< 9999^{10}\\ \\ \text{Vậy }99^{20}< 9999^{10}\)

\(\text{b) }3^{4000}\text{ và }9^{2000}\\ \text{Ta có : }3^{4000}=\left(3^2\right)^{2000}=9^{2000}\\ \text{ Mà }9^{2000}=9^{2000}\\ \Rightarrow3^{4000}=9^{2000}\\ \text{Vậy }3^{4000}=9^{2000}\)

\(\text{c) }2^{332}\text{ và }3^{223}\\ \text{Ta có : }2^{332}< 2^{333}\\ 3^{223}< 3^{222}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ (1) }\\ \text{Ta lại có : }2^{333}=\left(2^3\right)^{111}=8^{111}\\ 3^{222}=\left(3^2\right)^{111}=9^{111}\\ \\ \text{Mà }8^{111}< 9^{111}\\ \Rightarrow2^{333}< 3^{222}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\left(2\right)\\ \text{Từ }\left(1\right)\text{ và }\left(2\right)\text{ suy ra : }2^{332}< 2^{333}< 3^{222}< 3^{223}\\ \Rightarrow2^{332}< 3^{223}\\ \text{Vậy }2^{332}< 3^{223}\)

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a) ta có: 2²²⁵ = (2³)⁷⁵ = 8⁷⁵

3¹⁵⁰ = (3²)⁷⁵ = 9⁷⁵ > 8⁷⁵

=> ...

b) ta có: 2⁹¹ > 2⁷⁵ = (2³)²⁵ = 8²⁵ > 3²⁵

=> ...

c) ta có: 27⁸ = (3³)⁸ = 3²⁴

81⁴ = (3⁴)⁴ = 3¹⁶ < 3²⁴

=>...

d)ta có:  2³³² < 2³³³ = (2³)¹¹¹ = 8¹¹¹

3²²³ > 3²²² = (3²)¹¹¹ = 9¹¹¹ > 8¹¹¹

=>...

e)C1: ta có: 9²⁰⁰⁰ = (3²)²⁰⁰⁰ = 3⁴⁰⁰⁰

C2: ta có: 3⁴⁰⁰⁰ = (3²)²⁰⁰⁰ = 9²⁰⁰⁰

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

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

\(8< 9=>....\)

a: \(2^{225}=8^{75}\)

\(3^{150}=9^{75}\)

mà 8<9

nên \(2^{225}< 3^{150}\)

b: \(2^{91}=8192^7\)

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

mà 8192>3125

nên \(2^{91}>5^{35}\)

\(2^{91}=\left(2^{13}\right)^7=73728^7\)

\(5^{35}=\left(5^5\right)^7=3125^7\) nhỏ hơn \(73728^7\)

\(\Rightarrow2^{91}\) lớn hơn \(5^{35}\)

\(b,3^{400}=\left(3^4\right)^{100}=81^{100}\\ 4^{300}=\left(4^3\right)^{100}=64^{100}\\ Vì:81^{100}>64^{100}\left(Do:81>64\right)\\ \Rightarrow3^{400}>4^{300}\)

\(d,10^{20}=\left(10^2\right)^{10}=100^{10}\\ Vì:100^{10}>20^{10}\left(Do:100>20\right)\\ \Rightarrow10^{20}>20^{10}\)

a) \(2^{91}\)và \(5^{35}\)

Ta có :

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

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

Vì \(8192^7>3125^7\)nên \(2^{91}>5^{35}\)

b) \(3^{4000}\)và \(9^{2000}\)

Ta có :

\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)

\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)

Vì \(81^{1000}=81^{1000}\)nên \(3^{4000}=9^{2000}\)

\(2^{91}\)và  \(5^{35}\)

Ta có : 

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

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

Vì \(8192>3125\)nên \(2^{91}>5^{35}\)

\(3^{4000}\)và  \(9^{2000}\)

Ta có : 

\(3^{4000}=\left(3^4\right)^{1000}=81^{1000}\)

\(9^{2000}=\left(9^2\right)^{1000}=81^{1000}\)

Vì \(81=81\)nên \(3^{4000}=9^{2000}\)

a/

\(2^{91}=\left(2^7\right)^{13}=128^{13}\)

\(5^{35}< 5^{39}=\left(5^3\right)^{13}=125^{13}\)

\(\Rightarrow2^{91}>5^{35}\)

b/ \(3^{4000}=\left(3^2\right)^{2000}=9^{2000}\)

c/

\(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

\(\Rightarrow3^{223}>2^{332}\)

a) Ta có :

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

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

     Mà 8^75 < 9^75 => 2^225<3^150

b) Ta có 

        2^91=(2^13)^7=8192^7

        3^35=(3^5)^7=243^7

mà 8192^7<243^7=> 2^91<3^35

c) 3^4000=(3^2)^2000=9^2000

d) 2^332 < 2^333=2^3^111=8^111

3^223>3^222=9^111

=>2^332<3^223

2|}}dasKJLFDJHLSKAfhsdklfjdlsa;fjdsafjdsa;fjdsl;fjlsa;fjadskljfdlfjdskfjl;+)2349890432483085439-

a)10³⁰và2¹⁰⁰ 

\(\Leftrightarrow\left(2^5\right)^{30}\)và \(2^{100}\)

\(=2^{150}\)và \(2^{100}\)

vậy \(10^{30}>2^{100}\)

b)333⁴⁴⁴và444³³³   

tự làm 

a)10³⁰và2¹⁰⁰ 

⇔(25)30và 2100

=2150và 2100

vậy 1030>2100

n