Tìm phân số tối giản \(\dfrac{a}{b}\) biết: \(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) + \(\dfrac{a}{b}\) = \(\dfrac{1}{2}\)
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H9
HT.Phong (9A5)
CTVHS
10 tháng 10 2023
a) Các phân số tối giản là: \(\dfrac{1}{5};\dfrac{7}{6};\dfrac{9}{19}\)
b) Ba phân số tối giản là: \(\dfrac{3}{2};\dfrac{5}{6};\dfrac{4}{9}\)
Ba phân số chưa tối giản là:
\(\dfrac{10}{18}=\dfrac{10:2}{18:2}=\dfrac{5}{9}\)
\(\dfrac{20}{50}=\dfrac{20:10}{50:10}=\dfrac{2}{5}\)
\(\dfrac{3}{12}=\dfrac{3:3}{12:3}=\dfrac{1}{4}\)
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17 tháng 1 2021
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3x^2+2}-\sqrt{4+x}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{3x^2-x-2}{\sqrt{3x^2+2}+\sqrt{4+x}}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{3x+2}{\left(x+1\right)\left(\sqrt{3x^2+2}+\sqrt{4+x}\right)}=\dfrac{5}{2.2\sqrt{5}}=\dfrac{\sqrt{5}}{4}\).
Từ đó a = 5; b = 4 nên a - b = 1.
9 tháng 9 2023
a: \(A=\dfrac{1}{\left(3-1\right)\left(3+1\right)}+\dfrac{1}{\left(5-1\right)\left(5+1\right)}+...+\dfrac{1}{\left(99-1\right)\left(99+1\right)}\)
\(=\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{98\cdot100}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{49}{100}=\dfrac{49}{200}\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2022
\(\sqrt{\dfrac{1}{4}+\dfrac{1}{\left(2n-1\right)^2}+\dfrac{1}{\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(2n-1\right)^2\left(2n+1\right)^2+4\left(2n-1\right)^2+4\left(2n+1\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)
\(=\sqrt{\dfrac{\left(4n^2-1\right)^2+4\left(4n^2-4n+1\right)+4\left(4n^2+4n+1\right)}{4\left(2n-1\right)^2\left(2n+1\right)^2}}\)
\(=\sqrt{\dfrac{16n^4+24n^2+9}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\sqrt{\dfrac{\left(4n^2+3\right)^2}{4\left(2n-1\right)^2\left(2n+1\right)^2}}=\dfrac{4n^2+3}{2\left(2n-1\right)\left(2n+1\right)}\)
\(=\dfrac{\left(4n^2-1\right)+4}{2\left(2n-1\right)\left(2n+1\right)}=\dfrac{1}{2}+\dfrac{2}{\left(2n-1\right)\left(2n+1\right)}\)
\(=\dfrac{1}{2}+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\)
Do đó:
\(P=\left(\dfrac{1}{2}+\dfrac{1}{1}-\dfrac{1}{3}\right)+\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{2}-\dfrac{1}{399}-\dfrac{1}{401}\right)\)
\(=\dfrac{1}{2}.200+1-\dfrac{1}{401}=\dfrac{40500}{401}\)
\(\Rightarrow Q=400\)
NV
Nguyễn Việt Dũng
CTVVIP
23 tháng 8 2023
a)
Phân số đã tối giản: \(\dfrac{2}{3}\); \(\dfrac{5}{17}\);\(\dfrac{1}{10}\)Phân số nào chưa tối giản: \(\dfrac{9}{21}\); \(\dfrac{10}{15}\); \(\dfrac{7}{14}\)
b) Rút gọn
\(\dfrac{21}{9}\) = \(\dfrac{7}{3}\)
\(\dfrac{10}{15}\) = \(\dfrac{2}{3}\)
\(\dfrac{7}{14}\) = \(\dfrac{1}{2}\)
11 tháng 12 2023
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3x+1}-2}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{3x+1-4}{\sqrt{3x+1}+2}\cdot\dfrac{1}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{3x-3}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+2\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{3}{\left(x+1\right)\left(\sqrt{3x+1}+2\right)}=\dfrac{3}{\left(1+1\right)\left(\sqrt{3+1}+2\right)}\)
\(=\dfrac{3}{2\cdot4}=\dfrac{3}{8}\)
=>a=3;b=8
=>a2+b=9+8=17
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 1 2021
\(\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-2\right)\left(3x+2\right)}+\dfrac{1}{\left(x-2\right)\left(x-10\right)}\right)=\lim\limits_{x\rightarrow2}\dfrac{1}{\left(x-2\right)}\left(\dfrac{x-10+3x+2}{\left(3x+2\right)\left(x-10\right)}\right)\)
\(=\lim\limits_{x\rightarrow2}\dfrac{4\left(x-2\right)}{\left(x-2\right)\left(3x+2\right)\left(x-10\right)}=\lim\limits_{x\rightarrow2}\dfrac{4}{\left(3x+2\right)\left(x-10\right)}=-\dfrac{1}{16}\)
\(\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{a}{b}=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{2}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{a}{b}=1-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{3}{3}-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{2}{3}\)
Vậy phân số tối giản \(\dfrac{a}{b}=\dfrac{2}{3}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) thì sao?