AH la duong cao cua cac hinh tam giac nao?

Viet ten day tuong ung cua hinh tam giac.

​​​​​ A B H D C

\(P=5+5^2+...+5^{101}+5^{102}\)

\(P=5\left(1+5\right)+...+5^{101}\left(1+5\right)\)

\(P=5\cdot6+...+5^{101}\cdot6\)

\(P=6\cdot\left(5+...+5^{101}\right)⋮6\left(đpcm\right)\)

C/m tương tự khi chứng minh chia hết cho 31 ( nhóm 3 số với nhau )

Ta có :

\(3a+2b⋮17\)

\(\Rightarrow9\left(3a+2b\right)⋮17\)

\(\Rightarrow27a+18b⋮17\)

\(\Rightarrow\left(17a+17b\right)+\left(10a+b\right)⋮17\)

\(\Rightarrow10a+b⋮17\)(1)

Ta có :

\(10a+b⋮17\)

\(\Rightarrow2\left(10a+b\right)⋮17\)

\(\Rightarrow20a+2b⋮17\)

\(\Rightarrow17a+3a+2b⋮17\)

\(\Rightarrow3a+2b⋮17\)(2)

Từ (1) và (2) \(\Rightarrow3a+2b⋮17\Leftrightarrow10a+b⋮17\)(đpcm)

_Chúc bạn học tốt_

Ta có

1/a+1=1a/a(a+1)

=>1/a+1 + 1/a(a+1) = 1a/a(a+1) + 1/a(a+1) = 1a+1/a(a+1) =1.(a+1)/a.(a+1)=1/a => dpcm

a) ( nhân chéo nha bạn )

\(\frac{7}{x}=\frac{3}{12}=\frac{1}{4}\)

=> x = 7 . 4

=> x = 28

b)

\(x+x\cdot\frac{1}{4}:\frac{2}{7}+x:\frac{2}{9}=255\)

\(x+x\cdot\frac{7}{8}+x\cdot\frac{9}{2}=255\)

\(x\cdot\left(1+\frac{7}{8}+\frac{9}{2}\right)=255\)

\(x\cdot\frac{51}{8}=255\)

\(x=40\)

\(A=\frac{3}{2}\times\left(\frac{1}{13\times11}+\frac{1}{13\times15}+\frac{1}{15\times17}+.....+\frac{1}{97\times99}\right)\)

\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+......+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{99}\right)\)

\(A=\frac{3}{2}\times\frac{8}{99}\)

\(A=\frac{4}{33}\)

b] \(\frac{A}{5}=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)

\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)

\(\frac{A}{5}=\frac{1}{31}-\frac{1}{57}\)

\(\Rightarrow A=5\left(\frac{1}{31}-\frac{1}{57}\right)=\frac{130}{1767}\)

c] Ta đặt \(\left(8n+5,6n+4\right)=d\)

\(\Rightarrow\frac{8n+5\div d}{6n+4\div d}\Rightarrow4\times\left(6n+4\right)-3\times\left(8n+5\right)=\left(24n+16\right)-\left(24n+15\right):d\)\(\Rightarrow d=1\)

Vậy \(\frac{8n+5}{6n+4}\)là phân số tối giản

nhanh gium minh dang gap, cam on

Bài 1 mk ko hiểu đề cho lắm 

Bài 2 : 

Đặt \(A=\frac{x+4}{x-2}+\frac{2x-5}{x-2}\)

Ta có : 

\(\frac{x+4}{x-2}+\frac{2x-5}{x-2}=\frac{x+4+2x-5}{x-2}=\frac{3x-1}{x-2}=\frac{3x-6+5}{x-2}=\frac{3\left(x-2\right)}{x-2}+\frac{5}{x-2}=3+\frac{5}{x-2}\)

Để \(A\) là số nguyên thì \(\frac{5}{x-2}\) phải là số nguyên \(\Rightarrow\) \(5⋮\left(x-2\right)\) \(\Rightarrow\) \(\left(x-2\right)\inƯ\left(5\right)\)

Mà \(Ư\left(5\right)=\left\{1;-1;5;-5\right\}\)

Do đó : 

Vậy \(x\in\left\{-3;1;3;7\right\}\) thì A là số nguyên 

Chúc bạn học tốt ~

a) -2 /3 x + 1/5 = 3/10

 -2/3x =1/10 

 x = -3/20 

 vậy x = -3/20

b) 25/9 - 12/13x = 7/

12/13x = 2

x = 13/6

c) (x) - 3/4 =5/3 

(x) = 29/12

x = 29/12 ; -29/-12

 d)  x = 11/2

\(S=\frac{2016}{2.3:2}+\frac{2016}{3.4:2}+...+\frac{2016}{2015.2016:2}\)

\(S=\frac{4032}{2.3}+\frac{4032}{3.4}+...+\frac{4032}{2015.2016}\)

\(S=4032\left[\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right]\)

\(S=4032\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right]\)

\(S=4032\left[\frac{1}{2}-\frac{1}{2016}\right]=4032\cdot\frac{1007}{2016}\)

\(S=2014\)

S = \(2016+\frac{2016}{1+2}+\frac{2016}{1+2+3+}+...+\frac{2016}{1+2+3+...+2015}\)

S = \(2016+\left(\frac{2016}{1+2}+\frac{2016}{1+2+3}+...+\frac{2016}{1+2+3+...+2015}\right)\)

S = \(2016+2016.\left(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2015}\right)\)

đặt A = \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2015}\)

A = \(\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+...+\frac{1}{\left(1+2015\right).2015:2}\)

A = \(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2015.2016}\)

A = \(2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+...+2.\left(\frac{1}{2015}-\frac{1}{2016}\right)\)

A = \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)

A = \(2.\left(\frac{1}{2}-\frac{1}{2016}\right)\)

A = \(2.\frac{1007}{2016}=\frac{1007}{1008}\)

Thay A vào ta được :

S = \(2016+2016.\frac{1007}{1008}\)

S = \(2016.\left(1+\frac{1007}{1008}\right)\)

S = \(2016.\frac{2015}{1008}\)

S = \(4030\)