Chọn đáp án A.

a) 0,(26)<0,261

b) 0,15>0,14(9)

a: 0,(26)<0,261

b: 0,15>0,14(9)

a) Ta có :\(20< 25\Rightarrow\sqrt{20}< \sqrt{25}\Leftrightarrow2\sqrt{5}< 5\)

b) Ta có : \(\dfrac{16}{9}< 12\Rightarrow\sqrt{\dfrac{16}{9}}< \sqrt{12}\Leftrightarrow\dfrac{1}{3}\cdot\sqrt{16}< \sqrt{12}\)

a: \(2\sqrt{5}=\sqrt{20}\)

\(5=\sqrt{25}\)

mà 20<25

nên \(2\sqrt{5}< 5\)

b: \(\dfrac{1}{3}\cdot\sqrt{16}=\sqrt{\dfrac{1}{9}\cdot16}=\sqrt{\dfrac{16}{9}}\)

\(\sqrt{12}=\sqrt{\dfrac{108}{9}}\)

mà 16<9

nên \(\dfrac{1}{3}\sqrt{16}< \sqrt{12}\)

a: \(\dfrac{\sqrt{81}}{\sqrt{16}}=\dfrac{9}{4}=\dfrac{36}{16}< \dfrac{81}{16}\)

b: \(\sqrt{16+25}=\sqrt{41}< 9=\sqrt{16}+\sqrt{25}\)

A=1+2+2^2+2^3+....+2^9

2A=2+2^2+2^3+....+2^10

2A-A=2^10-1

A=2^10-1/2

B=5.2^8=(2^2+1).2^8=2^10+2^8

=>B>A

2A = 2(1 + 2 + 2² + .... + 2⁹ )

= 2 + 2² + 2³ + ..... + 2¹⁰

2A - A = (2 + 2² + 2³ + ..... + 2¹⁰) - (1 + 2 + 2² + .... + 2⁹ )

A = 2¹⁰ - 1  

B = 5.2⁸ = (2² + 1).2⁸ = 2¹⁰ + 2⁸

2¹⁰ - 1 < 2¹⁰ + 2⁸

=> A < B

\(1,\\ a,2< 3\Rightarrow2^{30}< 3^{30}\Rightarrow-2^{30}>-3^{30}\\ b,6^{10}=6^{2\cdot5}=\left(6^2\right)^5=36^5>35^5\left(36>35\right)\)

\(2,\\ a,\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\dfrac{3^{10}\cdot5^5\cdot3^5}{5^6\cdot3^{14}}=\dfrac{3}{5}\\ b,\left(8x-1\right)^{2x+1}=5^{2x+1}\\ \Leftrightarrow8x-1=5\\ \Leftrightarrow x=\dfrac{3}{4}\)

Bài 2: 

a: Ta có: \(\dfrac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}\)

\(=\dfrac{-3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^{14}}\)

\(=-\dfrac{3}{5}\)

b: Ta có: \(\left(8x-1\right)^{2x+1}=5^{2x+1}\)

\(\Leftrightarrow8x-1=5\)

\(\Leftrightarrow8x=6\)

hay \(x=\dfrac{3}{4}\)

Bài 1: 

a: \(-2^{30}=-8^{10}\)

\(-3^{30}=-27^{10}\)

mà 8<27

nên \(-2^{30}>-3^{30}\)

b: \(35^5=35^5\)

\(6^{10}=36^5\)

mà 35<36

nên \(35^5< 6^{10}\)

a) 107⁵⁰ và 73⁷⁵

107⁵⁰ = ( 107² )²⁵ 

73⁷⁵ = ( 75³ )²⁵

Mà 107² < 75³ 

Vậy 107⁵⁰ < 73⁷⁵

3 phần sau tương tự

câu c là (2\(^2\))\(^3\)và 2 mũ 2 mũ 3 nha

giúp mk nhanh 1 chút mk cần rất rất gấp

\(a,2^{24}=\left(2^3\right)^8=8^8< 9^8=\left(3^2\right)^8=3^{16}\\ b,16^{11}=\left(2^4\right)^{11}=2^{44}< 2^{45}=\left(2^5\right)^9=32^9\\ \Rightarrow\left(-16\right)^{11}>\left(-32\right)^9\\ c,\left(2^2\right)^3=2^6< 2^8=2^{2^3}\)

Ta có: A = \(\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)
A = \(\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)

A = \(-1+1+\frac{1}{2}\)

A = \(\frac{1}{2}\)

B = \(\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)

B = \(\frac{9}{16}+\frac{8}{27}+1+\frac{7}{16}-\frac{19}{27}\)

B = \(\left(\frac{9}{16}+\frac{7}{16}\right)+1+\left(\frac{8}{27}-\frac{19}{27}\right)\)

B = \(1+1-\frac{11}{27}\)

B = \(\frac{43}{27}\)

Mà 1/2 < 43/27 (Vì 1/2 < 1; 43/27 > 1)

=> A < B

                       Giải

\(A=\frac{-2}{11}+\frac{6}{7}+\frac{1}{2}+\frac{-9}{11}+\frac{1}{7}\)

\(\Leftrightarrow A=\left(\frac{-2}{11}+\frac{-9}{11}\right)+\left(\frac{6}{7}+\frac{1}{7}\right)+\frac{1}{2}\)

\(\Leftrightarrow A=\frac{-11}{11}+\frac{7}{7}+\frac{1}{2}\)

\(\Leftrightarrow A=-1+1+\frac{1}{2}\)

\(\Leftrightarrow A=\frac{1}{2}< 1\left(1\right)\)

\(B=\left(\frac{9}{16}+\frac{8}{27}\right)+\left(1+\frac{7}{16}+\frac{-19}{27}\right)\)

\(\Leftrightarrow B=\left(\frac{9}{16}+\frac{7}{16}\right)+\left(\frac{8}{27}+\frac{-19}{27}\right)+1\)

\(\Leftrightarrow B=\frac{16}{16}+\frac{-11}{27}+1\)

\(\Leftrightarrow B=1+\frac{-11}{27}+1\)

\(\Leftrightarrow B=2+\frac{-11}{27}\)

\(\Leftrightarrow B=\frac{43}{27}\)\(>1\left(2\right)\)

Từ (1) và (2) suy ra A < B