\(\frac{1+2+3+4+5+6+7+8+9.0}{1+2+3+4+5+6+7+8+9.0}\)

= 1

( Vì tử và mẫu đều giống nhau )

bang 1 day dung 100% luon

Đặt P = ... ( biểu thức đề bài ) 

Nhận xét: Với \(k\inℕ^∗\) ta có: 

\(\frac{k+2}{k!+\left(k+1\right)!+\left(k+2\right)!}=\frac{k+2}{k!+\left(k+1\right).k!+\left(k+2\right).k!}=\frac{k+2}{2.k!\left(k+2\right)}=\frac{1}{2.k!}\)

\(\Rightarrow\)\(P=\frac{1}{2.1!}+\frac{1}{2.2!}+...+\frac{1}{2.6!}=\frac{1}{2}\left(1+\frac{1}{2}+...+\frac{1}{720}\right)=...\)

bài 1b)

\(8\frac{1}{14}-6\frac37\)

C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)

C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)

\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)

bài 1 c)\(7-3\frac67\)

C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)

C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)

giúp mik với mik cần gấp mai mik phải nộp cho cô

very important !!!!!!

Câu a:

(4 5/37 - 3 4/5 + 8 15/29) - (3 5/37 - 6 14/29)

= (4 5/37 - 3 5/37) + (8 15/29 + 6 14/29) - 3 4/5

= 1 + 15 - 3 4/5

= 16 - 3 - 4/5

= 13 - 4/5

= 61/5

Câu b:

(2\(\frac56\) + 1\(\frac49\)) : (10\(\frac{1}{12}\) - 9\(\frac12\))

= (17/6 + 13/9) :(121/12 - 19/2)

= (51/18 + 26/18) :(121/12 - 114/2)

= 77/18 : 7/12

= 77/18 x 12/7

= 22/3

$A=1-$$3+5-7+...+2001-2003+2005$

$=\left(1-3\right)+\left(5-7\right)+...+\left(2001-2003\right)+2005$(Có 1002 cặp)

$=\left(-2\right).1002+2005$

$=-2004+2005$

$=1$

a)\(\frac{\frac{51}{2}}{\frac{13}{2}}\)=\(\frac{51}{13}\)

b)\(-\frac{5}{\frac{15}{32}}:\frac{\frac{23}{52}}{\frac{8}{3}}=-\frac{13312}{207}\)

\(4\frac{5}{6}+\frac{1}{6}.\left(8-2\frac{2}{3}\right)\)

\(=\frac{29}{6}+\frac{1}{6}.\left(8-\frac{8}{3}\right)\)

\(=\frac{29}{6}+\frac{1}{6}.\left(\frac{24}{3}-\frac{8}{3}\right)\)

\(=\frac{29}{6}+\frac{1}{6}.\frac{16}{3}\)

\(=\frac{29}{6}+\frac{8}{9}\)

=\(=\frac{103}{18}\)

\(4\frac{5}{6}+\frac{1}{6}.\left(8-2\frac{2}{3}\right)\)

\(=\frac{29}{6}+\frac{1}{6}.\left(8-\frac{8}{3}\right)\)

\(=\frac{29}{6}+\frac{1}{6}.\left(\frac{24}{3}-\frac{8}{3}\right)\)

\(=\frac{29}{6}+\frac{1}{6}.\frac{16}{3}\)

\(=\frac{29}{6}+\frac{8}{9}\)

\(=\frac{87}{18}+\frac{16}{18}\)

\(=\frac{103}{18}\)

Kết quả là 2870.

Giải nhanh bằng công thức tổng bình phương:

\(1^{2} + 2^{2} + \hdots + n^{2} = \frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)

Với \(n = 20\):

\(\frac{20 \cdot 21 \cdot 41}{6} = \frac{17220}{6} = 2870.\)

Ta có biểu thức:

\(1\times1+2\times2+3\times3+\ldots+20\times20=1^2+2^2+3^2+\ldots+20^2\)

Đây là tổng các số chính phương từ 1 đến 20.

Áp dụng công thức tổng bình phương:

\(1^2+2^2+3^2+\ldots+n^2=\frac{n \left(\right. n + 1 \left.\right) \left(\right. 2 n + 1 \left.\right)}{6}\)

Thay \(n = 20\):

\(\frac{20 \times 21 \times 41}{6} = \frac{17220}{6} = 2870\)