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a) \(xy+3x=5y-2\)
\(\Leftrightarrow x\left(y+3\right)=5y-2\)
\(\Leftrightarrow x=\frac{5y-2}{y+3}\)
\(\Leftrightarrow x=\frac{5\left(y+3\right)-17}{y+3}\)
\(\Leftrightarrow x=5-\frac{17}{y+3}\)
Do x nguyên, y nguyên nên y+3 là Ư(17)
Ta có bảng:
| y+3 | -17 | -1 | 1 | 17 |
| y | -20 | -4 | -2 | 14 |
| x | 6 | 22 | -12 | 4 |
Vậy (x;y) là (6;-20);(22;-4);(-12;-2);(4;14)
b) \(\Leftrightarrow\frac{\frac{99.100.101}{3}}{100x^2+\frac{99.100}{2}}=\frac{6666}{131}\Rightarrow x=\pm4\)
Bài 1:
a) Ta có: \(\frac{2^8\cdot4\cdot13+2^7\cdot8\cdot65}{2^9\cdot39}\)
\(=\frac{2^8\cdot4\cdot13+2^8\cdot4\cdot13\cdot5}{2^9\cdot39}\)
\(=\frac{2^{10}\cdot13\left(1+5\right)}{2^9\cdot13\cdot3}=\frac{6}{3}=2\)
b) Đặt \(A=4+2^2+2^3+2^4+...+2^{20}\)
Ta có: \(A=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow2A=2^3+2^3+2^4+...+2^{21}\)
Ta có: \(2A-A=2^3+2^{21}-2^2-2^2=8+2^{21}-8=2^{21}\)
hay \(A=2^{21}\)
Vậy: \(4+2^2+2^3+2^4+...+2^{20}=2^{21}\)
a, \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]=178\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\) \(=-88\)
\(x+\dfrac{206}{100}=\dfrac{-5}{176}\)
\(x=\dfrac{-5}{176}-\dfrac{206}{100}\)
\(x=\dfrac{-9198}{4400}\)
a) \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left(1-\dfrac{1}{10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\dfrac{9}{10}.100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(90-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=90-89\)
\(\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=1\)
\(\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)=\dfrac{1}{2}\)
\(x+\dfrac{206}{100}=5\)
\(x=5-\dfrac{206}{100}\)
\(x=\dfrac{147}{50}\)
Vậy \(x=\dfrac{147}{50}\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)
\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)
\(x.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\left(1-\dfrac{1}{50}\right)=1\\ \Rightarrow x.\dfrac{49}{50}=1\\ \Rightarrow x=1:\dfrac{49}{50}\\ \Rightarrow x=\dfrac{50}{49}\)
Ta có công thức tổng quát của dãy số:
\(1\cdot2+2\cdot3+3\cdot4+...+n\left(n+1\right)=\dfrac{n\left(n+1\right)\left(n+2\right)}{3}\)
Trong đề bài ta có dãy số: \(1\cdot2+2\cdot3+...+99\cdot100\) có \(n=99\)
\(\Rightarrow1\cdot2+2\cdot3+...+99\cdot100=\dfrac{99\cdot\left(99+1\right)\cdot\left(99+2\right)}{3}=333300\)
Trở lại để bài:
\(\dfrac{1\cdot2+2\cdot3+3\cdot4+...+99\cdot100}{x^2+\left(x^2+1\right)+\left(x^2+2\right)+...+\left(x^2+99\right)}=50\dfrac{116}{131}\)
\(\Rightarrow\dfrac{333300}{x^2+x^2+1+x^2+2+...+x^2+99}=\dfrac{6666}{131}\)
\(\Rightarrow\dfrac{333300}{\left(x^2+x^2+...+x^2\right)+\left(1+2+3+...+99\right)}=\dfrac{6666}{131}\)
\(\Rightarrow\dfrac{333300}{100x^2+4950}=\dfrac{6666}{131}\)
\(\Rightarrow6666\left(100x^2+4950\right)=333300\cdot131\)
\(\Rightarrow666600x^2+32996700=43662300\)
\(\Rightarrow666600x^2=10665600\)
\(\Rightarrow x^2=\dfrac{10665600}{666600}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x^2=4^2\)
\(\Rightarrow x=\pm4\)