18/53 lớn hơn nha bn

chuk bn học tốt

Cách giải:
26/78=13/36 
13/36=13x53/36x53=689/1708 
18/53=18x36/53x36=648/1708 
=>26/78>18/53
Chúc bạn học tốt!

=2018.2018/2019.2019

=1.1/1.1

=1/1

1/1=444444/444444

vì 888887>4444444=>888887/444444>4444444/444444

\(\frac{x+1}{19}+\frac{x+2}{18}=\frac{x+3}{17}+\frac{x+4}{16}\)

\(\Rightarrow\left(\frac{x+1}{19}+1\right)+\left(\frac{x+2}{18}+1\right)=\left(\frac{x+3}{17}+1\right)+\left(\frac{x+4}{16}+1\right)\)

\(\Rightarrow\frac{x+20}{19}+\frac{x+20}{18}=\frac{x+20}{17}+\frac{x+20}{16}\)

\(\Rightarrow\frac{x+20}{19}+\frac{x+20}{18}-\frac{x+20}{17}-\frac{x+20}{16}=0\)

\(\Rightarrow\left(x+20\right).\left(\frac{1}{19}+\frac{1}{18}-\frac{1}{17}-\frac{1}{16}\right)=0\)

Vì : \(\frac{1}{19}+\frac{1}{18}-\frac{1}{17}-\frac{1}{16}\ne0\)nên \(x+20=0\)

Suy ra : \(x=0-20=-20\)

Vậy \(x=-20\)

-20 nha 

tk mk nha

hok tốt

theo bai ra ta co : 

   5n+2:2n-3

=>2(5n+2):2n-3

=>10n+4:2n-3 (1)

mà 2n-3:2n-3

=>5(2n-3):2n-3

=>10n-15:2n-3 (2)

 tu (1) va (2) suy ra:

(10n+4)-(10n-15):(2n-3)

hay (10n+4-10n+15):(2n-3)

=>19:2n-3

=>2n-3 thuoc uoc cua 19

=>2n-3 thuoc {1;-1;19;-19}

=>2n thuoc {4;2;22;-16}

=>n thuoc {2;1;11;-8}

vay n thuoc {2;1;11;-8}

voi lai day la tim n chu ko phai x :)

\(A=\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{96^2}+\frac{1}{98^2}\)

\(A< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}+\frac{1}{97.99}\)

\(A< \frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}+\frac{1}{99}\)

\(A< 1-\frac{1}{99}\)

\(A< \frac{98}{99}\)

gọi đó là A đi.

Ta có:

1/13+1/14+1/14< 1/12+1/12+1/12=3/12=1/4

1/61+1/62+1/63< 1/60+1/60+1/60=3/60=1/20

=> 1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/5+1/4+1/20=1/2

=>A< 1/2 (ĐPCM)

ta có 
1 = 1 
1/2 + 1/3 < 1/2 + 1/2 = 1 
1/4 + 1/5 + .. + 1/7 < 1/4 +..+ 1/4 = 4/4 = 1 
1/8 + 1/9 + .. + 1/15 < 1/8 + .. + 1/8 = 8/8 = 1 
tương tự 
1/16 +1/17 + .. + 1/31 < 1 
1/32 + 1/33 + .. + 1/63 < 1 
=> cộng lại => B < 6

Trả lời

Ta có: \(1=1\)

\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)

\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)

\(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}< \frac{1}{32}+\frac{1}{32}+...+\frac{1}{32}=1\)

\(\Rightarrow B< 1+1+1+1+1+1\)

\(\Rightarrow B< 6\)

\(a,\frac{7}{25}\cdot\frac{39}{-14}\cdot\frac{50}{78}=\frac{7}{25}\cdot-\frac{39}{14}\cdot\frac{25}{39}=-\frac{7}{14}=-\frac{1}{2}\)

\(b,\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)

\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}=\frac{1}{4}+\frac{1}{2}=\frac{6}{8}=\frac{3}{4}\)

Còn nữa:

7/19.8/11+3/11.7/19+-12/19

a) \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)

\(3F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)

\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)

\(3F=\frac{1}{3}-\frac{1}{33}\)

\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(F=\frac{1}{3}.\frac{1}{3}-\frac{1}{3}.\frac{1}{33}=\frac{1}{9}-\frac{1}{99}=\frac{11}{99}-\frac{1}{99}=\frac{10}{99}\)

b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)

\(A=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)

\(A=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)

\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\left(\frac{7}{70}-\frac{1}{70}\right)=7.\frac{6}{70}\)

\(A=\frac{7.6}{70}=\frac{1.6}{10}=\frac{1.3}{5}=\frac{3}{5}\)

a, \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)

\(F=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)

\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)

\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)

\(F=\frac{1}{3}-\frac{10}{33}\)

\(F=\frac{10}{99}\)

Đặt : \(A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)

Do \(\frac{32}{99}>32\%\)nên \(A>32\%\left(đpcm\right)\)

7/15=1/5+4/15