Bài 1: Rút gọn

\(A=\dfrac{-56\cdot49+\left(-49\right)\cdot44}{73\cdot14+\left(-14\right)\cdot\left(-27\right)}\)

\(=\dfrac{49\cdot\left(-56-44\right)}{14\cdot\left(73+27\right)}\)

\(=\dfrac{-49\cdot100}{14\cdot100}=\dfrac{-7}{2}\)

Có cần ghi lời giải luôn ko bạn.

cần...........thanks bạn trước nhé

a 25 phần 41

b 9

c 1 phần 12

d 12

e 14 phần 15

f 24 phần 7

a 25/41 

b 9

c 1/12

d 12

e 14/15

f24/7

bạn ngọc hân đúng rồi

sửa đề: \(B=5+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}\)

Ta có: \(A=\frac23+\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}+\frac{194}{195}\)

\(=1-\frac13+1-\frac{1}{15}+1-\frac{1}{35}+1-\frac{1}{63}+1-\frac{1}{99}+1-\frac{1}{143}+1-\frac{1}{195}\)

\(=7-\left(\frac13+\frac{1}{15}+\cdots+\frac{1}{195}\right)\)

\(=7-\frac12\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\cdots+\frac{2}{13\cdot15}\right)\)

\(=7-\frac12\left(1-\frac13+\frac13-\frac15+\cdots+\frac{1}{13}-\frac{1}{15}\right)=7-\frac12\left(1-\frac{1}{15}\right)\)

\(=7-\frac12\cdot\frac{14}{15}=7-\frac{7}{15}=\frac{98}{15}\) >6

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{7^2}<\frac{1}{6\cdot7}=\frac16-\frac17\)

Do đó; \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}<1-\frac12+\frac12-\frac13+\cdots+\frac16-\frac17=1-\frac17<1\)

=>\(5+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{7^2}<5+1=6\)

=>B<6

mà A>6

nên B<A

a: \(A=\left(1-\frac49\right)\left(1-\frac{4}{25}\right)\left(1-\frac{4}{49}\right)\cdot\ldots\cdot\left(1-\frac{4}{\left(2n+1\right)^2}\right)\)

\(=\left(1-\frac23\right)\cdot\left(1-\frac25\right)\cdot\ldots\cdot\left(1-\frac{2}{2n+1}\right)\left(1+\frac23\right)\left(1+\frac25\right)\cdot\ldots\cdot\left(1+\frac{2}{2n+1}\right)\)

\(=\frac13\cdot\frac35\cdot\ldots\cdot\frac{2n-1}{2n+1}\cdot\frac53\cdot\frac75\cdot\ldots\cdot\frac{2n+3}{2n+1}\)

\(=\frac{1}{2n+1}\cdot\frac{2n+3}{3}=\frac{2n+3}{3\left(2n+1\right)}\)

b: Ta có công thức tổng quát:

\(1+\frac{1}{n^2-1}\)

\(=\frac{n^2-1+1}{n^2-1}=\frac{n^2}{n^2-1}=\frac{n\cdot n}{\left(n-1\right)\left(n+1\right)}\)

\(B=\left(1+\frac13\right)\left(1+\frac18\right)\cdot\ldots\left(1+\frac{1}{n^2-1}\right)\)

\(=\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{n^2-1}\right)\)

\(=\frac{2\cdot2}{\left(2-1\right)\left(2+1\right)}\cdot\frac{3\cdot3}{\left(3-1\right)\left(3+1\right)}\cdot\ldots\cdot\frac{n\cdot n}{\left(n-1\right)\left(n+1\right)}\)

\(=\frac{2\cdot3\cdot\ldots\cdot n}{1\cdot2\cdot\ldots\cdot\left(n-1\right)}\cdot\frac{2\cdot3\cdot\ldots\cdot n}{3\cdot4\cdot\ldots\cdot\left(n+1\right)}=\frac{n}{1}\cdot\frac{2}{n+1}=\frac{2n}{n+1}\)

c: Ta có công thức tổng quát:

\(1-\frac{1}{1+2+\cdots+n}\)

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

\(=1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

\(C=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\cdot\ldots\cdot\left(1-\frac{1}{1+2+\cdots+n}\right)\)

\(=\frac{\left(2+2\right)\left(2-1\right)}{2\left(2+1\right)}\cdot\frac{\left(3+2\right)\left(3-1\right)}{3\left(3+1\right)}\cdot\ldots\cdot\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)

\(=\frac{4\cdot5\cdot\ldots\cdot\left(n+2\right)}{3\cdot4\cdot\ldots\cdot\left(n+1\right)}\cdot\frac{1\cdot2\cdot\ldots\cdot\left(n-1\right)}{2\cdot3\cdot\ldots\cdot n}=\frac{n+2}{3}\cdot\frac{1}{n}=\frac{n+2}{3n}\)

a) \(\dfrac{16}{24}-\dfrac{1}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

b ) \(\dfrac{4}{5}-\dfrac{12}{60}=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{4}{5}\)

c ) \(\dfrac{35}{50}-\dfrac{3}{10}=\dfrac{7}{10}-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

d ) \(\dfrac{21}{28}-\dfrac{1}{4}=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)

1) Ta có: \(\dfrac{1}{7}x^2y^3\cdot\left(-\dfrac{14}{3}xy^2\right)\cdot\left(-\dfrac{1}{2}xy\right)\left(x^2y^4\right)\)

\(=\left(-\dfrac{1}{7}\cdot\dfrac{14}{3}\cdot\dfrac{-1}{2}\right)\left(x^2y^3\cdot xy^2\cdot xy\cdot x^2y^4\right)\)

\(=\dfrac{1}{3}x^6y^{10}\)

2) Ta có: \(\left(3xy\right)^2\cdot\left(-\dfrac{1}{2}x^3y^2\right)\)

\(=9xy^2\cdot\dfrac{-1}{2}x^3y^2\)

\(=-\dfrac{9}{2}x^4y^4\)

3) Ta có: \(\left(-\dfrac{1}{4}x^2y\right)^2\cdot\left(\dfrac{2}{3}xy^4\right)^3\)

\(=\dfrac{1}{16}x^4y^2\cdot\dfrac{8}{27}x^3y^{12}\)

\(=\dfrac{1}{54}x^7y^{14}\)