Sửa đề: A=5/2+5/6+...+5/2450

=5(1/2+1/6+...+1/2450)

=5(1-1/2+1/2-1/3+...+1/49-1/50)

=5*49/50<5

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

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

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

77/45

7/6

29/10

\(\dfrac{3}{5}+\dfrac{4}{3}\times\dfrac{5}{6}=\dfrac{3}{5}+\dfrac{10}{9}=\dfrac{27}{45}+\dfrac{50}{45}=\dfrac{77}{45}\)

\(2-\dfrac{7}{6}+\dfrac{1}{3}=\dfrac{12}{6}-\dfrac{7}{6}+\dfrac{2}{6}=\dfrac{7}{6}\)

\(\dfrac{7}{6}\times3-\dfrac{3}{5}=\dfrac{7}{2}-\dfrac{3}{5}=\dfrac{35}{10}-\dfrac{6}{10}=\dfrac{29}{10}\)

giải chi tiết đi

a: =11/11x3/7=3/7

b: =2/7x5/2=5/7

\(a,\left(\dfrac{6}{11}+\dfrac{5}{11}\right)\times\dfrac{3}{7}=\dfrac{11}{11}\times\dfrac{3}{7}=1\times\dfrac{3}{7}=\dfrac{3}{7}\\ b,\left(\dfrac{6}{7}-\dfrac{4}{7}\right):\dfrac{2}{5}=\dfrac{2}{7}\times\dfrac{5}{2}=\dfrac{5}{7}\)

a 11/11x3/7=3/7

b 2/7x5/2=5/7

Giải:

B=1,4 .15/49 - (4/5+2/3) :2 1/5

B=7/5.15/49 - 22/15 : 11/5

B=3/7 - 2/3

B=-5/21

Chúc bạn học tốt!

x:5/3=9/5+2/5

x:5/3=11/5

x=11/5.5/3

x=11/3

x : 5/3 = 9/5 - 2/5

x : 5/3 = 7/5

x = 7/5 x 5/3

x = 7/3

\(x:\dfrac{5}{3}+\dfrac{2}{5}=\dfrac{9}{5}\)

\(x:\dfrac{5}{3}=\dfrac{9}{5}-\dfrac{2}{5}\)

\(x:\dfrac{5}{3}=\dfrac{7}{5}\)

\(x=\dfrac{7}{5}\times\dfrac{5}{3}\)

\(x=\dfrac{35}{15}=\dfrac{7}{3}\)

a) -6 . (- \(\dfrac{2}{3}\) ) . 0,25 = 14 . 0,25 = 3,5 
b) -\(\dfrac{15}{4}\) . ( - \(\dfrac{7}{15}\) ) . ( - \(\dfrac{22}{25}\) ) = \(\dfrac{7}{4}\) . ( - \(\dfrac{22}{25}\) ) = - \(\dfrac{77}{50}\) = - 1,54
c) -\(\dfrac{21}{5}\) . ( - \(\dfrac{9}{11}\) ) . ( - \(\dfrac{11}{14}\) ) . \(\dfrac{2}{5}\) = \(\dfrac{189}{55}\) . ( - \(\dfrac{11}{14}\) ) . \(\dfrac{2}{5}\) = - \(\dfrac{297}{110}\) . \(\dfrac{2}{5}\) = - \(\dfrac{297}{275}\) 

Câu c nếu bn chx chắc chắn thì nên thử tính lại nhé

ok:

(2x+2⁴).5³=4.5⁵

(2x+2⁴)=4.5⁵:5³

(2x+2⁴)=4.5²

(2x+2⁴)=100

  2x=100-16

  2x=84

    x=84:2

    x=42

2x+24).53=4.55 (2x+24)=4.55:53 (2x+24)=4.52 (2x+24)=100 2x=100-16 2x=84 x=84:2 x=42

#include <bits/stdc++.h>

using namespace std;

long long n,ln,x,i;

int main()

{

cin>>n;

ln=LLONG_MIN;

for (i=1; i<=n; i++)

{

cin>>x;

ln=max(ln,x);

}

cout<<ln;

return 0;

}