`#3107.101107`

`99^{20}` và `9999^{10}`

Ta có:

\(99^{20}=99^{10}\cdot99^{10}\)

\(9999^{10}=99^{10}\cdot101^{10}\)

Vì \(99^{10}< 101^{10}\Rightarrow99^{10}\cdot99^{10}< 99^{10}\cdot101^{10}\)

\(\Rightarrow99^{20}< 9999^{10}.\)

99²⁰ và 9999¹⁰ phải không em?

Trung bình 1 em nặng :

 ( 36+38+40+34) :4 =37 (cân)

   đs: 37 cân

@kurumi cute Copy phải ghi Tk nhé!

Tk = Tham khảo

a: \(\dfrac{63^2-47^2}{215^2-105^2}=\dfrac{\left(63-47\right)\cdot\left(63+47\right)}{\left(215-105\right)\left(215+105\right)}\)

\(=\dfrac{16\cdot110}{110\cdot320}=\dfrac{16}{320}=\dfrac{1}{20}\)

b: \(\dfrac{437^2-363^2}{537^2-463^2}=\dfrac{\left(437-363\right)\left(437+363\right)}{\left(537-463\right)\left(537+463\right)}\)

\(=\dfrac{74\cdot800}{74\cdot1000}=\dfrac{800}{1000}=\dfrac{4}{5}\)

a: \(x\times8-10398=24194\)

=>\(x\times8=10398+24194=34592\)

=>x=34592:8=4324

b: \(583\times29-583\times x=500-65+24:3\)

=>\(583\times\left(29-x\right)=435+8=443\)

=>\(29-x=\dfrac{443}{583}\)

=>\(x=29-\dfrac{443}{583}=\dfrac{16464}{583}\)

\(x\cdot8-10398=24194\\ x\cdot8=35132\\ x=4391,5\\ \\ 583\cdot29-583\cdot x=500-65+24:3\\ 583\cdot\left(29-x\right)=443\\ 29-x=\dfrac{443}{583}\\ x=\dfrac{16464}{583}\)

11/8 : x - 1/3 = 2/5

11/8 : x = 2/5 + 1/3

11/8 : x = 6/15 + 5/15

11/8 : x = 11/15

x = 11/8 : 11/15

x = 11/8 x 15/11

x = 165/88

x = 15/8

\(\dfrac{11}{8}:x-\dfrac{1}{3}=\dfrac{2}{5}\)

=>\(\dfrac{11}{8}:x=\dfrac{2}{5}+\dfrac{1}{3}=\dfrac{11}{15}\)

=>\(x=\dfrac{11}{8}:\dfrac{11}{15}=\dfrac{15}{8}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

\(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{99\times101}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =\dfrac{1}{3}-\dfrac{1}{101}\\ =\dfrac{98}{303}\)

35,2<n+4<48,1

=>35,2-4<n<48,1-4

=>31,2<n<44,1

mà n là số tự nhiên

nên \(n\in\left\{32;33;...;43;44\right\}\)

\(35,2< n+4< 48,1\Leftrightarrow31,2< n< 44,1\)

Mà \(n\in\)N => n = 32;33;34;35;36;37;38;39;40;41;42;43;44

Bài 1: ĐKXĐ: \(x\ne-1\)

Để \(\dfrac{x+5}{2x+2}\) là số nguyên thì \(x+5⋮2x+2\)

=>\(2x+10⋮2x+2\)

=>\(2x+2+8⋮2x+2\)

=>\(8⋮2x+2\)

=>\(2x+2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

=>\(x\in\left\{-\dfrac{1}{2};-\dfrac{3}{2};0;-2;1;-3;3;-5\right\}\)

mà x nguyên

nên \(x\in\left\{0;-2;1;-3;3;-5\right\}\)

Bài 3:

Số đối của \(-\dfrac{4}{5}\) là \(\dfrac{4}{5}\)

Số đối của \(1\dfrac{1}{3}=\dfrac{4}{3}\) là \(-\dfrac{4}{3}\)

Biểu diễn:

12,26<3n<27,4

mà \(3n⋮3\)

nên \(3n\in\left\{15;18;21;24;27\right\}\)

=>\(n\in\left\{5;6;7;8;9\right\}\)

\(12,26< 3n< 27,4\Leftrightarrow4,086< n< 9,13\)

Mà \(n\in N\)=> n = 5;6;7;8;9

\(\dfrac{-5}{7}\cdot\dfrac{2}{11}+\dfrac{5}{7}\cdot\dfrac{9}{11}\)

\(=\dfrac{5}{7}\cdot\left(-\dfrac{2}{11}+\dfrac{9}{11}\right)\)

\(=\dfrac{5}{7}\cdot\dfrac{7}{11}=\dfrac{5}{11}\)

\(-\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}=\dfrac{5}{7}\left(-\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}.\left(\dfrac{7}{11}\right)=\dfrac{5}{11}\)