Bài 1:

\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)

\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)

\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)

\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)

\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)

Bài 2: 

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)

làm ơn giúp mình ,bạn nào làm nhanh đúng mình chọn cho nhanh nha mọi người

nhanh nha mọi người

T.T

Bài làm

2².4⁴.2⁶.2.2=2².2⁸.2⁶.2²=2¹⁸

2.2³.2⁷.2⁴=2¹⁵

3².3=3³=27

3⁴.3³.3=3⁸

3.3⁵.3⁴.3²=3¹²

3.3⁶.3⁷.3⁴=3¹⁸

4.4³=4⁴

4⁵.4.4⁷=4¹³

4.4³.4⁵.4⁶=4¹⁵

5.5⁶=5⁷

5.5⁴=5⁵

5.5².5⁴=5⁷

5.5³.5⁴.5⁶=5¹⁴

5.5⁶.5⁴.5³=5¹⁴

x.x²=x³

x.x³.x⁵=x⁹

a.a⁴=a⁵

a.a².a⁴.a⁵.a⁷=a¹⁹

a.a³.a⁵.a⁶=a¹⁵

a.a⁴.a.a⁵.a⁶.a=a¹⁸

x.x³.x⁴=x⁸

x⁵.x⁴.x.x⁷.x⁶=x²³

7.7²=7³

7.7².7⁴.7⁵=7¹²

7³.7⁵.7⁴.7=7¹³

7².7.7⁶.7⁶.7³=7¹⁸

10.10²=10³

10.100.10³=10⁶

10.100.10⁴.1000=10¹⁰

10².10⁴.100.1000=10¹¹

Tk nha !

\(A=4.4^1.4^3.4^5.....4^{57}=4^{1+3+5+...+57}=4^{\left[\left(\dfrac{57-1}{2}\right)+1\right]:2\left(57+1\right)}=4^{841}\)\(B=3+3^2+3^3+3^4+...+3^{100}\)

\(3B=3\left(3+3^2+3^3+3^4+...+3^{100}\right)\)

\(3B=3^2+3^3+3^4+3^5+...+3^{101}\)

\(3B-B=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4....+...+3^{100}\right)\)

\(2B=3^{101}-3\Leftrightarrow B=\dfrac{3^{101}-3}{2}\)

2)

Từ \(1\rightarrow9\) có: \(\left(9-1\right):1+1=9\)(chữ số)

Từ \(10\rightarrow99\) có:\(2\left[\left(99-10\right):1+1\right]=180\)(chữ số)

Từ \(100\rightarrow386\) có:\(3\left[\left(386-100\right):1+1\right]=816\)(chữ số)

Như vậy,Để đánh số trang từ \(1\rightarrow386\) thì cần:

\(9+180+816=1005\)(chữ số)

\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(2A=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

\(2A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(2A=\frac{1}{1}-\frac{1}{100}\)

\(2A=\frac{99}{100}\Rightarrow A=\frac{99}{100}:2\Rightarrow A=\frac{99}{200}\)

Câu B và C làm tương tự.

bạn Nhi làm sai rồi

\(\frac{2}{2\cdot3}\) sao có thể bằng \(\frac{1}{2}-\frac{1}{3}\) được

\(\frac{1}{2\cdot3}\) mới bằng \(\frac{1}{2}-\frac{1}{3}\)

kết quả là : \(\frac{49}{100}\)

`3x-16:2^3=31`

`=>3x-16:8=31`

`=>3x-2=31`

`=>3x=31+2`

`=>3x=33`

`=>x=11`

__

`2^10:2^8+3[4.7+3.4]`

`=2^2+3[4(3+7)]`

`=4+3[4.10]`

`=4+3.40`

`=4+120`

`=124`

__

`4^6:4^3-2^2 . 2^3`

`=4^3-2^5`

`=64-32`

`=32`

__

`141+2^5 . 2^4-3^1 . 3^2`

`=141+2^9-3^3`

`=141+512-9`

`=644`

__

`x+2^5:2^4=4.4^2`

`=>x+2=4^3`

`=>x=64-2`

`=>x=62`

a) \(P=\dfrac{1}{1.2}+\dfrac{2}{2.4}+\dfrac{3}{4.7}+...\dfrac{10}{46.56}\)

\(P=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...\dfrac{1}{46}-\dfrac{1}{56}\)

\(P=1-\dfrac{1}{56}\)

\(P=\dfrac{55}{56}\)

b) \(A=\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+...+\dfrac{3}{99.100}\)

\(A=3\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)

\(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=3\left(1-\dfrac{1}{100}\right)\)

\(A=3.\dfrac{99}{100}\)

\(A=\dfrac{297}{100}\)

c) \(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\)

\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(B=1-\dfrac{1}{103}\)

\(B=\dfrac{102}{103}\)

d) \(C=\dfrac{5}{1.4}+\dfrac{5}{4.7}+\dfrac{5}{7.10}+...+\dfrac{5}{100.103}\)

\(C=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{100.103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)

\(C=\dfrac{5}{3}.\dfrac{102}{103}\)

\(C=\dfrac{170}{103}\)

e) \(D=\dfrac{7}{1.5}+\dfrac{7}{5.9}+\dfrac{7}{9.13}+...+\dfrac{7}{101.105}\)

\(D=\dfrac{7}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{101.105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{101}-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}\left(1-\dfrac{1}{105}\right)\)

\(D=\dfrac{7}{4}.\dfrac{104}{105}\)

\(D=\dfrac{26}{15}\)