A Lớn hơn

A = \(\dfrac{10^{20}+3}{10^{21^{ }}+3}\)

B = \(\dfrac{10^{21}+4}{10^{22}+4}\) < 1

\(\Rightarrow\) B < \(\dfrac{10^{21}+4+6}{10^{22}+4+6}\) 

\(\Rightarrow\) B < \(\dfrac{10^{21}+10}{10^{22}+10}\)

\(\Rightarrow\) B < \(\dfrac{10\left(10^{20}+1\right)}{10\left(10^{21}+1\right)}\)

\(\Rightarrow\) B < \(\dfrac{10^{20}+1}{10^{21}+1}\) < \(\dfrac{10^{21}+1+2}{10^{22}+1+2}\)

\(\Rightarrow\) B < \(\dfrac{10^{21}+3}{10^{22}+3}\)

\(\Rightarrow\) B < A

\(A=\left(\frac{20}{5}+\frac{27}{9}\right)\times\frac{21}{10}=\left(4+3\right)\times\frac{21}{10}=7\times\frac{21}{10}=\frac{147}{10}\)

\(B=\left(\frac{13}{6}-\frac{3}{8}\right)\times\frac{11}{22}\)

\(B=\left(\frac{52}{24}-\frac{9}{24}\right)\times\frac{11}{22}\)

\(B=\frac{43}{24}\times\frac{1}{2}=\frac{43}{48}\)

Dễ thấy \(A=\frac{147}{10}>1\)

Mà \(B=\frac{43}{48}< 1\)

=> tự so sánh

Bài 2: a: <

        b: >

        c :=

Thông cảm nhé bài 1 mình ko biết

Bài 2 : So sánh: 

10⁷⁵⁰⁰ = 10^12.625 = ( 10¹²)⁶²⁵

11⁵⁰⁰⁰ = 11^8.625 = ( 11⁸)⁶²⁵ 

  Dễ thấy 10¹² > 11⁸ nên 10⁷⁵⁰⁰ > 11⁵⁰⁰⁰ 

Chúc bạn học tốt!!!

a: =(21/11+12/11)-(9/19+3/19+6/19)=3-18/19=39/19

b: =(2/3+4/3)+(3/4+7/4)+(4/5+16/5)

=2+4+10/4

=6+5/2

=17/2

a)

Có: \(2>1>0\)

\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)

b) Có: \(0< \sqrt{3}< 3\)

\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)

c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)

\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)

d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)

\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)

a: 2=1+1<1+căn 2

b: 4=1+3>1+căn 3

c: -2căn 11=-căn 44

-10=-căn 100

mà 44<100

nên -2 căn 11>-10

d: 12=3*4=3*căn 16>3*căn 11

a: \(\frac{3^{10}\cdot\left(-5\right)^{21}}{\left(-5\right)^{20}\cdot3^{12}}=\frac{-3^{10}\cdot5^{21}}{3^{12}\cdot5^{20}}=\frac{-5}{3^2}=-\frac59\)

b: \(\frac{-11^5\cdot13^7}{11^5\cdot13^8}=\frac{-11^5}{11^5}\cdot\frac{13^7}{13^8}=-\frac{1}{13}\)

c: \(2^{10}\cdot3^{10}-2^{10}\cdot3^9\)

\(=2^{10}\cdot3^9\cdot\left(3-1\right)\)

\(=2^{11}\cdot3^9\)

Ta có: \(\frac{2^{10}\cdot3^{10}-2^{10}\cdot3^9}{2^9\cdot3^{10}}\)

\(=\frac{2^{11}\cdot3^9}{2^{10}\cdot3^9}=2\)

d: \(5^{11}\cdot7^{12}+5^{11}\cdot7^{11}\)

\(=5^{11}\cdot7^{11}\left(7+1\right)=5^{11}\cdot7^{11}\cdot8\)

\(5^{12}\cdot7^{12}+9\cdot5^{11}\cdot7^{11}\)

\(=5^{11}\cdot7^{11}\left(5\cdot7+9\right)\)

\(=5^{11}\cdot7^{11}\left(35+9\right)=5^{11}\cdot7^{11}\cdot44\)

Ta có: \(\frac{5^{11}\cdot7^{12}+5^{11}\cdot7^{11}}{5^{12}\cdot7^{12}+9\cdot5^{11}\cdot7^{11}}\)

\(=\frac{5^{11}\cdot7^{11}\cdot8}{5^{11}\cdot7^{11}\cdot44}=\frac{8}{44}=\frac{2}{11}\)

a) (x - 3)(y - 3) = 9 = 1.9 = 3.3

Lập bảng:

Vậy ...

b) A = \(\frac{10^{19}+1}{10^{20}+1}\) => 10A = \(\frac{10^{20}+10}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)

B = \(\frac{10^{20}+1}{10^{21}+1}\) => 10B = \(\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)

Do \(10^{20}+1< 10^{21}+1\) => \(\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\) => 10A > 10B => A > B