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\(A=2.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\right)\)
\(A=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+....+\dfrac{3}{95.98}\right)\)
\(A=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)
\(A=\dfrac{2}{3}\dfrac{24}{49}=\dfrac{16}{49}\)
Ta có: A=\(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}\)
\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}\right)\)\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\)\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)
\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{49}{98}-\dfrac{1}{98}\right)\)
\(\Rightarrow A=\dfrac{3}{2}.\dfrac{48}{98}\)
\(\Rightarrow A=\dfrac{3.2.2.12}{2.2.49}\)
\(\Rightarrow A=\dfrac{36}{49}\)
a) Để phân số \(\dfrac{12}{n}\) có giá trị nguyên thì :
\(12⋮n\)
\(\Leftrightarrow n\inƯ\left(12\right)\)
\(\Leftrightarrow n\in\left\{-1;1;-12;12;-2;2;-6;6;-3;3;-4;4\right\}\)
Vậy \(n\in\left\{-1;1;-12;12;-2;2-6;6;-3;3;-4;4\right\}\) là giá trị cần tìm
b) Để phân số \(\dfrac{15}{n-2}\) có giá trị nguyên thì :
\(15⋮n-2\)
\(\Leftrightarrow x-2\inƯ\left(15\right)\)
Tới đây tự lập bảng zồi làm típ!
c) Để phân số \(\dfrac{8}{n+1}\) có giá trị nguyên thì :
\(8⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(8\right)\)
Lập bảng rồi làm nhs!
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
a)\(\dfrac{-1}{4}\cdot13\dfrac{9}{11}-0,25\cdot6\dfrac{2}{11}\)
\(=\dfrac{-1}{4}\cdot\dfrac{152}{11}-\dfrac{1}{4}\cdot\dfrac{68}{11}\)
=\(\dfrac{1}{4}\cdot\left(\dfrac{-152}{11}-\dfrac{68}{11}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{-220}{11}=-5\)
\(\)\(\left|x-1,5\right|+\left|2,5-x\right|=0\)
Với mọi \(x\in R\) thì:
\(\left\{{}\begin{matrix}\left|x-1,5\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix} \left|x-1,5\right|=0\\ \left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\)
Khi đó không tồn tại giá trị x
\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\sqrt{\dfrac{1}{6}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+\sqrt{\dfrac{1}{6}}\\x=\dfrac{1}{2}-\sqrt{\dfrac{1}{6}}\end{matrix}\right.\)
\(\sqrt{\dfrac{1}{6}=?}\)
mk ko hiểu Linh Nguyễn
mk chưa hk đến căn
a)
ta có:
\(\left\{{}\begin{matrix}\dfrac{b-a}{b-a}=1..\forall a\ne b\\\dfrac{b-a}{a.b}=\dfrac{1}{a}-\dfrac{1}{b}..\forall a,b\ne0\end{matrix}\right.\)(*)
\(A=\dfrac{1}{2.5}+\dfrac{1}{5.8}+..+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
\(\left\{{}\begin{matrix}a=3n-1\\b=3n+2\end{matrix}\right.\)\(\Rightarrow b-a=3..\forall n\)
Thay (*) vào dãy A
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-....+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)=\dfrac{1}{3}\left(\dfrac{3n+2-2}{2.\left(3n+2\right)}\right)=\dfrac{n}{6n+4}=VP\rightarrow dpcm\)
B) tương tự
\(A=\left(5+\dfrac{3}{8}+18+\dfrac{1}{2}-\dfrac{5}{24}\right):\left(16+\dfrac{2}{3}\right)\)
\(A=\left(5+8+\dfrac{3}{8}+\dfrac{1}{2}-\dfrac{5}{24}\right):\dfrac{50}{3}\)
\(A=\left(23+\dfrac{2}{3}\right).\dfrac{3}{50}\)
\(A=\dfrac{71}{3}.\dfrac{3}{50}\)
\(A=\dfrac{71}{50}=1\dfrac{21}{50}\)