Ta có : \(A=\frac{1}{1.101}+\frac{1}{2.202}+\frac{1}{3.103}+...+\frac{1}{10.110}\)

=\(\frac{1}{100}.\left(\frac{100}{1.101}+\frac{100}{2.102}+\frac{100}{3.103}+...+\frac{100}{10.110}\right)\)

= \(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{10}-\frac{1}{110}\right)\)

= \(\frac{1}{100}\cdot\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)

Lại có : B = \(\frac{1}{10}.\left(\frac{10}{1.11}+\frac{10}{2.12}+\frac{10}{3.13}+...+\frac{10}{100.110}\right)\)

= \(\frac{1}{10}\left(1-\frac{1}{11}+\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{110}\right)\)

= \(\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)\)

Khi đó \(A:B=\frac{A}{B}=\frac{\frac{1}{100}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}{\frac{1}{10}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}-\frac{1}{101}-\frac{1}{102}-...-\frac{1}{110}\right)}=\frac{1}{10}\)

An góp được 3/5 số tiền mua bóng . Suy ra Bình góp được 2/5 số tiền mua bóng:

a)Vậy quả bóng đó giá:

80 000 : 2/5 = 200 000 ( đồng)

b) Số tiền An góp là:

200 000 - 80 000 = 120 000 ( đồng)

Đ/s; a) 200 000 đồng

          b) 120 000 đồng

\(A=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+\frac{28}{\left(6.8\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(100.98\right)^2}\)

\(=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+\frac{8^2-6^2}{6^2.8^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)

\(=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+\frac{1}{6^2}-\frac{1}{8^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

\(=\frac{1}{2^2}-\frac{1}{100^2}< \frac{1}{4}\)

A = \(\frac{12}{\left(2.4\right)^2}\) +  \(\frac{20}{\left(4.6\right)^2}\) + ..........+ \(\frac{388}{\left(96.98\right)^2}\) +  \(\frac{396}{\left(98.100\right)^2}\)

   = \(\frac{16-4}{\left(2.4\right)^2}\)+ \(\frac{36-16}{\left(4.6\right)^2}\)+...........+ \(\frac{9604-9216}{\left(96.98\right)^2}\) + \(\frac{10000-9604}{\left(98.100\right)^2}\)

   = \(\frac{1}{2^2}\) - \(\frac{1}{4^2}\)+  \(\frac{1}{4^2}\)- \(\frac{1}{6^2}\) + ............+ \(\frac{1}{96^2}\) - \(\frac{1}{98^2}\) + \(\frac{1}{98^2}\) - \(\frac{1}{100^2}\)

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

   = \(\frac{1}{4}\) - \(\frac{1}{100^2}\) < \(\frac{1}{4}\)

   =) A < \(\frac{1}{4}\)

Chúc bạn học tốt nhé !

Ta có: a chia cho 2 dư 1 => a - 1 ⋮2

         a chia cho 3 dư 1 => a - 1 ⋮3

=> a - 1 ⋮6 => a -1 + 6.2 ⋮ 6 => a +11 ⋮ 6 (1)

Ta có: a  chia 5 dư 4 => a - 4 ⋮5 => a - 4 + 5.3 ⋮5  => a + 11 ⋮5 (2)

Ta có: a chia 7 dư 3 => a - 3 ⋮7 => a - 3 + 7.2 ⋮7 => a + 11 ⋮7 (3)

Từ (1) ; (2) ; (3) => a +11 ∈∈BC ( 6; 5; 7 ) 

Có: BCNN ( 6; 5; 7 ) = 210 

=> a + 11 ∈ BC ( 6; 5; 7 ) 

=> a ∈ { 199; 409 ;....}

Mà a là số tự nhiên nhỏ nhất nên a = 199.

sai òi bn oi:<

ko cs chia cho 6

T^T

\(\frac{16^3.3^{10}+144.6^9}{4^6.3^{12}+6^{11}}=\frac{\left(2^4\right)^3.3^{10}+2^4.3^2.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^{13}.3^{11}}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}.\left(1+2.3\right)}{2^{11}.3^{11}.\left(1+2.3\right)}=\frac{2}{3}\)

Với \(x\le0\)không thỏa mãn. 

Với \(x\ge1\)thì dễ thấy ta sẽ có \(z>y\).

\(\hept{\begin{cases}x+6=3^y\\8x+3=3^z\end{cases}}\Rightarrow8\left(x+6\right)-\left(8x+3\right)=45=8.3^y-3^z\)

\(\Leftrightarrow5.3^2=3^y\left(8-3^{z-y}\right)\)

\(\Rightarrow\hept{\begin{cases}y=2\\z-y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\z=3\end{cases}}\)

Suy ra \(x=3\).

Vậy ta có nghiệm \(\left(3,2,3\right)\).

Thk you

Đề sai bạn ơi

sao mà ai cũng đề sai vậy

13/6+5/6-7/12x8

26/12+10/12-7/12x8

19/2x8

152/2

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