\(\frac{254.399-145}{254+399.253}=\frac{\left(253+1\right).399-145}{254+399.253}=\frac{253.399+399-145}{254+399.253}=\frac{253.399+254}{254+399.253}=1\)

bằng 1 đó

= 1 nha

Dịch ra là: Ta có: 3A = 3. (1 + 3 + 32 + 33 + ... + 399 + 3100) (1 + 3 + 32 + 33 + ... + 399 + 3100) 3A = 3 + 32 + 33 + ... + 3100 + 31013 + 32 + 33 + ... + 3100 + 3101 Suy ra: 3A - A = (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) (3 + 32 + 33 + ... + 3100 + 3101) - (1 + 3 + 32 + 33 + ... + 399 + 3100) ⇒⇒ A = 3101−123101−12 Vậy A = 3101−12

Mà đoạn 2A sai nhé bạn, sửa lại:

2A = 3101−13101−1 2A=-10001

A=-10001/2

A=-5000,5

Vậy A=-5000,5

101201/101201

ghép đầu cuối dăt chung 399 ra ngoài 

kq vẫn như cũ chắc chắn 100% 

202292.4291

C=\(\frac{1}{2}.\frac{2}{3}.......\frac{2016}{2017}\)

C= CÂU HỎI TƯƠNG TỰ 

=> đcpm 

\(A=\frac{254\cdot399-145}{254+399\cdot253}\)

\(A=\frac{\left(253+1\right)\cdot399-145}{254+399\cdot253}\)

\(A=\frac{253\cdot399+\left(399-145\right)}{254+399\cdot253}\)

\(A=\frac{253\cdot399+254}{254+399\cdot253}\)

\(A=1\)

\(B=\frac{5932+6001\cdot5931}{5932\cdot6001-69}\)

\(B=\frac{5932+6001\cdot5931}{\left(5931+1\right)\cdot6001-69}\)

\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+\left(6001-69\right)}\)

\(B=\frac{5932+6001\cdot5931}{5931\cdot6001+5932}\)

\(B=1\)

\(C=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\)

\(C=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2016}{2017}\)

\(C=\frac{1\cdot2\cdot3\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2017}\)

\(C=\frac{1}{2017}\)

Vừa đếm thêm 100 000 (10 000; 1000; 10; 1) vừa điền vào chỗ chấm

a) 300 000; 400 000; 500 000;600 000; 700 000; 800 000.

b) 350 000; 360 000; 370 000; 380 000; 390 000; 400 000

c) 399 000; 399 100; 399 200;399 3000; 399 400; 399 500

d) 399 940; 399 950; 399 960;399 970; 399 980; 399 990

e) 456 784; 456 785; 456 786; 456 787; 456 788; 456 789

Vừa đếm thêm 100 000 (10 000; 1000; 10; 1) vừa điền vào chỗ chấm

a) 300 000; 400 000; 500 000;600 000; 700 000; 800 000.

b) 350 000; 360 000; 370 000; 380 000; 390 000; 400 000

c) 399 000; 399 100; 399 200;399 3000; 399 400; 399 500

d) 399 940; 399 950; 399 960;399 970; 399 980; 399 990

e) 456 784; 456 785; 456 786; 456 787; 456 788; 456 789

300 000 ; 400 000 ; 500 000 ; 600 000 ; 700 000 ; 800 000 
350 000 ; 360 000 ; 370 000 ; 380 000 ; 390 000 ; 400 000
399 000 ; 399 100 ; 399 200 ; 399 300 ; 399 400 ; 399 500

399 940 ; 399 950 ; 399 960 ; 399 970 ; 399 980 ; 399 990
456 784 ; 456 785 ; 456 786 ; 456 787 ; 456 788 ; 456 789

Xét phân thức phụ sau:

Ta có: \(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\frac{1}{\sqrt{n}+\sqrt{n+1}}\right)\)

\(=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\frac{1}{\sqrt{n\left(n+1\right)}}\cdot\left(\sqrt{n+1}-\sqrt{n}\right)\)

\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)

Thay vào ta được:

\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(BT=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)

Đặt biểu thức đã cho là A

Tổng quát ta có: Với \(a\inℕ^∗\)ta có:

\(\frac{1}{\left(a+1\right)\sqrt{a}+a.\sqrt{a+1}}=\frac{\left(a+1\right)-a}{\sqrt{a}.\sqrt{a+1}.\left(\sqrt{a}+\sqrt{a+1}\right)}\)

\(=\frac{\left(\sqrt{a+1}-\sqrt{a}\right)\left(\sqrt{a+1}+\sqrt{a}\right)}{\sqrt{a}.\sqrt{a+1}.\left(\sqrt{a}+\sqrt{a+1}\right)}=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a}.\sqrt{a+1}}\)

\(=\frac{\sqrt{a+1}}{\sqrt{a}.\sqrt{a+1}}-\frac{\sqrt{a}}{\sqrt{a}.\sqrt{a+1}}=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}\)

Áp dụng kết quả trên ta có:

Với \(n=1\)\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\)

Với \(n=2\)\(\Rightarrow\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

Với \(n=3\)\(\Rightarrow\frac{1}{4\sqrt{3}+3\sqrt{4}}=\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}\)

.....................

Với \(n=399\)\(\Rightarrow\frac{1}{400\sqrt{399}+399\sqrt{400}}=\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(\Rightarrow A=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+......+\frac{1}{\sqrt{399}}-\frac{1}{\sqrt{400}}\)

\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{400}}=1-\frac{1}{20}=\frac{19}{20}\)

Ta có công thức tổng quát là \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Vậy \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{400\sqrt{399}+399\sqrt{400}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{399}}-\dfrac{1}{\sqrt{400}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{400}}=1-\dfrac{1}{20}=\dfrac{19}{20}\)

45 x 399 + 35 x 399 + 20 x 399 
= ( 45 + 35 + 20 ) x 399 
= 100 x 399 
= 39900

 = 399 x (45 + 35 + 20)

 = 399 x 100

 = 39 900

45 x 399 + 35 x 399 + 20 x 399 

= ( 45 + 35 + 20 ) x 399 

= 100 x 399 

= 39900