Từ đề bài ta có :

\(\frac{13+n}{29+n}=\frac{5}{9}\)

\(\Rightarrow9\left(13+n\right)=5\left(29+n\right)\)

\(\Rightarrow117+9n=145+5n\)

\(\Rightarrow9n-5n=145-117\)

\(\Rightarrow4n=28\)

\(\Rightarrow n=7\)

Vậy n = 7

Thử lại : \(\frac{13+n}{29+n}=\frac{13+7}{29+7}=\frac{20}{36}=\frac{5}{9}\)(đúng)

n=7 nha bạn

BÀI 1:

\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(S=1+\frac{1}{1.2}+\frac{1}{2.2}+\frac{1}{2.4}+\frac{1}{4.4}+\frac{1}{4.8}\)

\(S=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\)

\(S=1+1-\frac{1}{8}\)

\(S=\frac{15}{8}\)

BÀI 2:

\(A=1.2+2.3+3.4+...+98.99\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+98.99.3\)

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+98.99.\left(100-97\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99\)

\(3A=\left(1.2.3+2.3.4+3.4.5+98.99.100\right)-\left(1.2.3+2.3.4+...+97.98.99\right)\)

\(3A=98.99.100\)

\(3A=970200\)

\(\Rightarrow A=970200:3\)

\(A=323400\)

CHÚC BN HỌC TỐT!!!
 

\(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{225}\right)\)

\(=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{224}{225}\)

\(=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot...\cdot\frac{14\cdot16}{15\cdot15}\)

\(=\frac{1\cdot3\cdot2\cdot4\cdot...\cdot14\cdot16}{2\cdot2\cdot3\cdot3\cdot...\cdot25\cdot25}\)

\(=\frac{\left(1\cdot2\cdot3\cdot...\cdot14\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot16\right)}{\left(2\cdot3\cdot4\cdot...\cdot15\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot15\right)}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot14}{2\cdot3\cdot4\cdot...\cdot15}\cdot\frac{3\cdot4\cdot5\cdot...\cdot16}{2\cdot3\cdot4\cdot...\cdot15}\)

\(=\frac{1}{15}\cdot\frac{16}{2}\)

\(=\frac{1}{15}\cdot8\)

\(=\frac{8}{15}\)

( 1- 1/4 ) . ( 1 - 1/9 ) . ( 1 - 1/16 ) ...( 1 - 1/225 ) 

= 3/4 . 8/9 . 15/16 ... 224/225

= 3 . 8 . 15 ... 224 / 4 . 9 . ..225 

= 3 . 2 . 4 . 3 . 5 ... 14 . 16 / 2 . 2 . 3 . 3 ... 15 . 15 

= ( 3 . 4 . 5 ... 16 ) . ( 2 . 3 . 4 ... 14 ) /  ( 2 . 3 ... 15 ) . ( 2 . 3 ... 15 ) 

=          16/ 2 . 15

=           `8/15

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

\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}.\frac{7}{6}.\frac{8}{7}\)

\(=\frac{3.4.5.6.7.8}{2.3.4.5.6.7}\)

\(=\frac{8}{2}\)

\(=4\)

a) \(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).\left(1+\frac{1}{5}\right).\left(1+\frac{1}{6}\right).\left(1+\frac{1}{7}\right)\)

= \(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}.\frac{7}{6}.\frac{8}{7}\)

= \(\frac{3.4.5.6.7.8}{2.3.4.5.6.7}\)

= \(4\)

\(\frac{x}{3}=\frac{48}{x}\)

\(\Rightarrow x×x=3.48\)

\(\Rightarrow x^2=144\)

\(\Rightarrow\)\(x^2\in\left\{12,-12\right\}\)

\(\frac{x}{3}=\frac{48}{x}\)

\(\Rightarrow x\cdot x=3\cdot48\)

\(\Rightarrow x^2=144\)

\(\Rightarrow x^2=\left(\pm12\right)^2\)

\(\Rightarrow x=\pm12\)

Gợi ý : 

a ) Tách số 19 ra 19 số 1 

Nhóm ở trên tử , mỗi số hạng cộng với 1 

=> ...

b )  Tách số 99 ở mẫu thành 99 số 1 

Nhóm ở dưới mẫu , mỗi số hạng cộng với 1 

=> ...

Chúc học tốt !!! 

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ...+ 1/19 - 1/20 

= ( 1 + 1/3 + 1/5 + ...+ 1/19 ) - ( 1/2 + 1/4 + ...+ 1/20 ) 

= ( 1 + 1/2 + 1/3 + 1/4 + ...+ 1/19 + 1/20 ) - 2 . ( 1/2 + 1/4 + ...+ 1/20 ) 

= ( 1 + 1/2 + 1/3 + ...+ 1/20 ) - ( 1 + 1/2 + ... + 1/10 ) 

=    1/11 + 1/12 + 1/13 + ...+ 1/20 ( Đpcm ) 
TK mk nha !!! 

\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{19}-\frac{1}{20}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{19}+\frac{1}{20}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{20}-1+\frac{1}{2}+....+\frac{1}{10}\)

\(=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\left(đpcm\right)\)

B=\(\frac{1.3}{2.2}\).\(\frac{2.4}{3.3}\)...\(\frac{\left(n-1\right)\left(n+1\right)}{nn}\)

B=\(\frac{\left[1.2...\left(n-1\right)\right]\left[3.4....\left(n+1\right)\right]}{\left(2.3...n\right)\left(2.3...n\right)}\)

B=\(\frac{1.\left(n+1\right)}{n.2}\)=\(\frac{n+1}{2n}\)