Ta có : BE = BC ( gt)

=> ∆BEC cân tại B 

=> E = BCE 

Mà ta thấy ABC = E + BCE ( góc ngoài ∆ bằng tổng 2 góc trong ko kề với nó)

Mà E = BCE ( cmt)

=> ABC = 2E 

Mà ABD = DBC ( BD là phân giác ABC )

=> E = BCE = ABD = DBC 

=> DBC = BCE 

Mà 2 góc này ở vị trí so le trong 

=> BD //EC ( dpcm)

a, abab = 101ab 

Ta có: ab.101 = ab . ( 100 + 1 ) = ab00 + ab = abab

b, aaabbb = a00b . 111

Ta có: a00b.111 = a00b . 100 + a00b . 10 + a00b  = a00b00 + a00b0 + a00b = aa0bb0 + a00b = aaabbb  

a) ab.101=ab.(100+1)=ab.100+ab=(10a+b).100+10a+b=1000a+100b+10a+b=abab

b) a00b.111=a00b.(100+1)=a00b.100+a00b.10+a00b=a00b00+a00b0+a00b=aa00bb0=a00b=aaabbb

Trả lời:

Từ 1 đến 100: 11 số 2

từ 100 => 200: 12 số

200=> 300: 100 số 

300=> 900: 66 số

900=> 999 :11 số => 11 + 12 + 100 + 66 = 189 số 2

lỡ đếm thiếu số nào thì xin lỗi nha:)

300 so nhe

giải chi tiết nha^^

Đổi thành phân số rồi tính bình thường nha bạn

Áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m}\left(m\ne0;m\in N\right)\)

Ta có: \(D=\frac{10^{76}+1}{10^{77}+1}< \frac{10^{76}+1+9}{10^{77}+1+9}=\frac{10^{76}+10}{10^{77}+10}=\frac{10.\left(10^{75}+1\right)}{10.\left(10^{76}+1\right)}=\frac{10^{75}+1}{10^{76}+1}=C\)

\(\Rightarrow D< C\)

\(\frac{-7}{12}:\frac{13}{6}+\frac{-7}{12}:\frac{13}{7}.\frac{2.|-8|}{3}\)

\(=\frac{-7}{12}.\frac{6}{13}+\frac{-7}{12}.\frac{7}{13}.\frac{2.8}{3}\)

\(=\frac{-7}{12}.\left(\frac{6}{13}+\frac{7}{13}.\frac{2.8}{3}\right)\)

\(=\frac{-7}{12}.\frac{10}{3}\)

\(=\frac{-35}{18}\)

\(\frac{-7}{12}:\frac{13}{6}+\frac{-7}{12}:\frac{13}{7}\times\frac{2\times\left|-8\right|}{3}\)

\(=\frac{-7}{12}\times\frac{6}{13}+\frac{-7}{12}\times\frac{7}{13}\times\frac{2\times8}{3}\)

\(=\frac{-7}{12}\times\left(\frac{6}{13}+\frac{7}{13}+\frac{2\times8}{3}\right)\)

\(=\frac{-7}{12}\times\frac{10}{3}\)

\(=\frac{-35}{18}\)

Rất vui khi giúp đc bạn.<3. Nếu có sai sót mong bạn bỏ qua

1) 2x - 378 = 122

=> 2x = 122 + 378

=> 2x = 500

=> x = 500 : 2

=> x = 250

2) 8x - 4x = 1208

=> 4x = 1208

=> x = 1208 : 4

=> x = 302

3) x - 382 = 159 : 3

=> x - 382 = 53

=>  x = 53 + 382

=> x = 435

4) 3x + 30 = 420

=> 3x = 420 - 30

=> 3x = 390

=> x = 390 : 3

=> x = 130

5) 12x  - 13x - 500 = 1500

=> -x = 1500 + 500

=> -x = 2000

=> x = -2000

1,\(2x-378=122\)

\(2x=122+378\)

\(2x=500\)

\(x=500:2\)

\(x=250\)

2,\(8x-4x=1208\)

\(x.\left(8-4\right)=1208\)

\(x.4=1208\)

\(x=1208:4\)

\(x=302\)

3,\(x-382=159:3\)

\(x-382=53\)

\(x=52+382\)

\(x=434\)

4,\(3x+30=420\)

\(3x=420-30\)

\(3x=390\)

\(x=390:3\)

\(x=130\)

5,\(12x-13x-500=1500\)

như câu 2 

a)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+..+\frac{1}{10.11}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

= \(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

b) Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)

= \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)

=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\)

Lấy 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)

              A  = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^7}\)

              A  = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^2}-...-\frac{1}{2^6}+\frac{1}{2^6}-\frac{1}{2^7}\)

             A   =\(1-\frac{1}{2^7}\)

Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}+\frac{1}{110}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(A=1-\frac{1}{11}\)

\(A=\frac{10}{11}\)

Đặt \(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\left(1\right)\)

\(2B=\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+\frac{2}{2^4}+\frac{2}{2^5}+\frac{2}{2^6}+\frac{2}{2^7}\)

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

Lấy \(\left(2\right)-\left(1\right)\)hay \(2B-B\)ta có:

\(2B-B=\left(1+\frac{1}{2}+...+\frac{1}{2^6}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)\)

\(\Rightarrow B=1-\frac{1}{2^7}\)

\(\Rightarrow B=\frac{2^7-1}{2^7}=\frac{128-1}{128}=\frac{127}{128}\)

HOK TOT