\(\dfrac{3}{5}+\dfrac{-1}{25}-\dfrac{8}{20}\)

\(=\dfrac{15}{25}+\dfrac{-1}{25}-\dfrac{8}{20}\)

\(=\dfrac{14}{25}-\dfrac{2}{5}\)

\(=\dfrac{14}{25}-\dfrac{10}{25}\)

\(=\dfrac{4}{25}\)

\(=\dfrac{15}{100}+\dfrac{-4}{100}-\dfrac{40}{100}\)

\(=\dfrac{-29}{100}\)

Số tấn gạo ngày đầu bán là: 360 x 1/5 = 72 tấn 

Số tấn gạo ngày thứ hai bán là: 360 - 72 - 213 = 75 tấn 

Số kg gạo còn lại là: 360 - 72 - 75 = 213 tấn gạo

\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{95\cdot98}\)

\(3A=\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{95\cdot98}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{98}\)

\(A=\dfrac{\dfrac{1}{2}-\dfrac{1}{98}}{3}=\dfrac{8}{49}\)

 Viết lại đề: \(\left\{{}\begin{matrix}u_1=\dfrac{1}{7}\\u_{n+1}=\dfrac{u_n\left(1-u_n^8\right)}{1+u_n}\end{matrix}\right.\)

 *Tính \(\lim\limits_{n\rightarrow+\infty}u_n\):

 Bằng quy nạp, dễ chứng minh được \(0< u_n< 1,\forall n=1,2,...\)

 Ta có \(u_{n+1}-u_n=\dfrac{-u_n^9-u_n^2}{1+u_n}< 0\) nên \(\left(u_n\right)\) là dãy giảm. Mà \(\left(u_n\right)\) bị chặn nên \(\left(u_n\right)\) có giới hạn hữu hạn.

 Đặt \(\lim\limits_{n\rightarrow+\infty}u_n=L\left(0\le L< 1\right)\) thì \(L=\dfrac{L\left(1-L^8\right)}{1+L}\)

 \(\Leftrightarrow\left[{}\begin{matrix}L=0\\\dfrac{1-L^8}{1+L}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}L=0\\1-L^8=1+L\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}L=0\\L=-1\end{matrix}\right.\)

\(\Rightarrow L=0\) \(\Rightarrow\lim\limits_{n\rightarrow+\infty}u_n=0\)

Bài 1 :

b) \(\left\{{}\begin{matrix}2x-3y=2\\-4x+6y=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=4\\-4x+6y=0\end{matrix}\right.\)

\(\Rightarrow0x+0y=4\) (vô lý)

\(\Rightarrow\) HPT vô nghiệm

Bài 2 :

\(\left\{{}\begin{matrix}2x-ay=b\\ax+by=1\end{matrix}\right.\)

Khi \(x=1;y=2\)

\(hpt\Leftrightarrow\left\{{}\begin{matrix}2.1-a.2=b\\a.1+b.2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2a+b=2\\a+2b=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4a+2b=4\\a+2b=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3a=3\\2b=1-a\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1\\2b=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\) thỏa mãn đề bài

Bạn tự vẽ đồ thị nhé.

Tổng của hai số là : \(100\times2=200\)

Tổng số phần bằng nhau là : \(2+3=5\left(phần\right)\)

Số bé là : \(200:5\times2=80\)

Số lớn là : \(200:5\times3=120\)

gấp gấp mn ui ~~~

anh chị giúp em với ạ

Ta cos

\(360=2^3\times3^5\times5\)

\(246=2\times3\times41\)

\(220=2^2\times5\times11\)

=>\(BCNN\left(360,246,220\right)=2^3=8\)

\(246=2\times3\times41\)

\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\\ =\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+\dfrac{3}{13.16}+\dfrac{3}{16.19}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\\ =1-\dfrac{1}{19}=\dfrac{18}{19}\)

\(\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{70}+\dfrac{3}{130}+\dfrac{3}{208}+\dfrac{3}{304}\)

\(=\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{10\times13}+\dfrac{3}{13\times16}+\dfrac{3}{16\times19}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{19}\)

\(=1-\dfrac{1}{19}\)

=\(\dfrac{18}{19}\)

\(a=\left(246-12\right):18=13\)

\(\dfrac{1}{6}+\dfrac{1}{2}\times\dfrac{1}{5}=\dfrac{1}{6}+\dfrac{1}{10}\\ =\dfrac{5}{30}+\dfrac{3}{30}=\dfrac{8}{30}=\dfrac{4}{15}\)

1/6 + 1/2 x 1/5 = 1/6 + 1/10 = 16/60 = 8/30 = 2/15