1/Tính : A = \(\dfrac{1}{3}-\dfrac{1}{18}-\dfrac{1}{54}-\dfrac{1}{108}-\dfrac{1}{270}-\dfrac{1}{378}\)
2/Tìm các số nguyên tố x, y sao cho : 51x + 26y = 2000
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H9
HT.Phong (9A5)
CTVHS
28 tháng 10 2023
3/ Ta có:
\(A=\dfrac{1-2x}{x+3}\)
\(A=\dfrac{-2x+1}{x+3}\)
\(A=\dfrac{-2x-6+7}{x+3}\)
\(A=\dfrac{-2\left(x+3\right)+7}{x+3}\)
\(A=-2+\dfrac{7}{x+3}\)
A nguyên khi \(\dfrac{7}{x+3}\) nguyên
⇒ 7 ⋮ \(x+3\)
\(\Rightarrow x+3\inƯ\left(7\right)\)
\(\Rightarrow x+3\in\left\{1;-1;7;-7\right\}\)
\(\Rightarrow x\in\left\{-2;-4;4;-10\right\}\)
H
10 tháng 4 2021
a) Trước hết ta chứng minh \(a^2-1=\left(a-1\right)\left(a+1\right)\text{tự chứng minh }\)
Áp dụng bổ đề trên ta có:
\(-A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right) =\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{100^2-1}{100^2}=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}=\dfrac{1\cdot2\cdot3^2\cdot...\cdot99^2\cdot100\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\dfrac{1\cdot101}{2\cdot100}>\dfrac{1}{2}\\ \Rightarrow A< -\dfrac{1}{2}\)
17 tháng 11 2023
\(\dfrac{1}{x-3}-\dfrac{1}{x}=\dfrac{x-\left(x-3\right)}{x\left(x-3\right)}=\dfrac{x-x+3}{x\left(x-3\right)}=\dfrac{3}{x\left(x-3\right)}\)
\(B=\dfrac{1}{x^2-3x}+\dfrac{1}{x^2-9x+18}+\dfrac{1}{x^2-15x+54}+\dfrac{1}{x^2-21x+108}\)
\(=\dfrac{1}{x\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-6\right)}+\dfrac{1}{\left(x-6\right)\left(x-9\right)}+\dfrac{1}{\left(x-9\right)\left(x-12\right)}\)
\(=\dfrac{1}{3}\left(\dfrac{3}{x\left(x-3\right)}+\dfrac{3}{\left(x-3\right)\left(x-6\right)}+\dfrac{3}{\left(x-6\right)\left(x-9\right)}+\dfrac{3}{\left(x-9\right)\left(x-12\right)}\right)\)
\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-6}-\dfrac{1}{x-6}+\dfrac{1}{x-9}-\dfrac{1}{x-9}+\dfrac{1}{x-12}\right)\)
\(=\dfrac{1}{3}\left(-\dfrac{1}{x}+\dfrac{1}{x-12}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{-\left(x-12\right)+x}{x\left(x-12\right)}\)
\(=\dfrac{4}{x\left(x-12\right)}\)
23 tháng 3 2021
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
20 tháng 2 2021
cái chỗ math processing error kia là \(3\left(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}\right)+\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)\ge\dfrac{985}{108}\)
11 tháng 2 2022
b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
28 tháng 1 2022
a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
| x | 1 | -1 | 2 | -2 | 5 | -5 | 10 | -10 |
| y | -10 | 10 | -5 | 5 | -2 | 2 | -1 | 1 |
c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 1 | 1 | -1 | 3 | -3 |
| y + 1 | 3 | -3 | 1 | -1 |
| x | 2 | 0 | 4 | -2 |
| y | 2 | -4 | 0 | -2 |
28 tháng 1 2022
b: =>xy=12
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)
Bài 1:
Ta có:
\(A=\dfrac{1}{3}-\dfrac{1}{18}-\dfrac{1}{54}-\dfrac{1}{108}-\dfrac{1}{270}-\dfrac{1}{378}\)
\(=\dfrac{1}{3}-\left(\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+\dfrac{1}{270}+\dfrac{1}{378}\right)\)
\(=\dfrac{1}{3}-\left(\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{18.21}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{18}-\dfrac{1}{21}\right)\)
\(=\dfrac{1}{3}-\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{21}\right)=\dfrac{1}{3}-\dfrac{1}{3}.\dfrac{6}{21}\)
\(=\dfrac{1}{3}-\dfrac{2}{21}=\dfrac{5}{21}\)
Vậy \(A=\dfrac{5}{21}\)
Bài 2:
Ta có: \(51x+26y=2000\)
Mà \(\left\{{}\begin{matrix}26y⋮2\\2000⋮2\end{matrix}\right.\) \(\Leftrightarrow51x⋮2\)
\(\left(51;2\right)=1\Rightarrow x⋮2\)
Mặt khác \(x\) là số nguyên tố nên \(x=2\)
Khi đó:
\(51.2+26y=2000\Leftrightarrow y=73\) (thỏa mãn)
Vậy các số nguyên tố \(\left(x,y\right)=\left(2;73\right)\)