giải giúp mk vs

\(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).\left(1-\frac{1}{25}\right).\left(1-\frac{1}{36}\right)\)

\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.\frac{35}{36}=\frac{3.8.15.24.35}{4.9.16.25.36}=\frac{1.3.2.4.3.5.4.6.5.7}{2.2.3.3.4.4.5.5.6.6}\)

\(=\frac{\left(1.2.3.4.5\right).\left(3.4.5.6.7\right)}{\left(2.3.4.5.6\right).\left(2.3.4.5.6\right)}=\frac{1.7}{2.2}=\frac{7}{4}\)

S=1-1/4+1-1/9+...+1-1/x²

S=(1+1+1+...+1)-(1/4+1/9+...+1/x²)

Có (1/4+1/9+...+1/x²)<1/(1.2)+1/(2.3)+...+1/(x-1)x=1-1/x<1

=> (1/4+1/9+...+1/x²) ko là số nguyên

=>S ko là số nguyên

Tacó cho công thức tổng quát: A² - B² = (A+B).(A-B) 
A = (1-1/4)x(1-1/9)x(1-1/16)x(1-1/25)x(1-1/3... 
= (1+1/2) x (1-1/2) x (1+1/3) x (1-1/3) x...x (1+1/n) x (1-1/n) 
= (1+1/2) x (1+1/3) x (1+1/4) x ... x [1 + 1/(n-1) ] x (1 + 1/n) 
x (1-1/2) x (1-1/3) x (1-1/4) x ... x [1 - 1/(n-1) ] x (1 - 1/n) 
= 3/2 x 4/3 x 5/4 x ... x [ n/(n-1) ] x [ (n+1)/n ] 
x 1/2 x 2/3 x 3/4 x ... x [ (n-2)/(n-1) ] x [ (n-1)/n] 
               Vậy dãy A là:
A = 1/2 x 2/3 x 3/2 x 3/4 x 4/3 x 4/5 x 5/4 x .... x [ (n-2)x(n-1) ] x [ (n-1)/n] x [ n/(n-1)] x [ (n+1)/n] 
= 1/2 x 1 x 1 x 1 x ... x 1 x [(n+1)/n] 

giai bang cach lop 5

lời giải luôn nhé

\(\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times\left(1-\frac{1}{25}\right)\times\left(1-\frac{1}{36}\right)\)

\(=\)\(\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times\frac{35}{36}\)

\(=\)\(\frac{3\times8\times15\times24\times35}{4\times9\times16\times25\times36}\)

\(=\)\(\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\frac{\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}3\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times4\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times2\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times5\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times3\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times6\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times4\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times5\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times7}{4\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times3\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times3\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times8\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times2\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times5\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times5\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times6\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\text{​​}\times6}\)

\(=\)\(\frac{7\text{​​}\text{​​}\text{​​}\times4}{8\times6}\)

\(=\)\(\frac{7\times4}{4\times2\times6}\)

\(=\)\(\frac{7}{2\times6}\)

\(=\)\(\frac{7}{12}\)

\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\cdot...\cdot\left(1-\dfrac{1}{9801}\right)\)

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

\(=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{98}{99}\right)\cdot\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\right)\)

\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)

\(K=\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{9}\right)x\left(1-\frac{1}{16}\right)x\left(1-\frac{1}{25}\right)x\left(1-\frac{1}{36}\right)\)

\(K=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x\frac{24}{25}x\frac{35}{36}\)

tự rút gọn

thôi tui làm nốt cho

...

\(K=\frac{1x3}{2x2}x\frac{2x4}{3x3}x\frac{3x5}{4x4}x\frac{4x6}{5x5}x\frac{5x7}{6x6}\)

\(K=\frac{7}{2}\)

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