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1)\(P=\frac{3n+2}{n-1}=\frac{3n-3+5}{n-1}=\frac{3n-3}{n-1}+\frac{5}{n-1}=3+\frac{5}{n-1}\)
Để \(P\in Z\Rightarrow3+\frac{5}{n-1}\in Z\Rightarrow\frac{5}{n-1}\in Z\Rightarrow n-1\inƯ\left(5\right)\)
| n - 1 | -5 | -1 | 1 | 5 |
| n | -4 | 0 | 2 | 6 |
Vậy để P nguyên thì \(n\in\left\{-4;0;2;6\right\}\)
2) \(\left(-1,5\right)^2:2\frac{1}{5}-3,15=2,25:2,2-3,15=4,95-3,15=1,8\)
\(1.\frac{\left(-3\right)^x}{81}=-27\Rightarrow\left(-3\right)^x\div\left(-3\right)^4=\left(-3\right)^3\)
\(\Rightarrow\left(-3\right)^x=\left(-3\right)^7\Rightarrow x=7\)
\(2.\sqrt{x-5}-4=5\Rightarrow\sqrt{x-5}=9\Rightarrow\sqrt{x-5}=\sqrt{81}\Rightarrow x-5=81\Rightarrow x=86\)
\(\)
Câu 1:
\(P=\frac{2n-1}{n-1}=\frac{2\left(n-1\right)+1}{n-1}=\frac{2\left(n-1\right)}{n-1}+\frac{1}{n-1}=2+\frac{1}{n-1}\in Z\)
\(\Rightarrow1⋮n-1\)
\(\Rightarrow n-1\inƯ\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow n\in\left\{2;0\right\}\)
Câu 2:
Từ \(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\Rightarrow\frac{a}{2}=\frac{2b}{3}=\frac{3c}{4}\Rightarrow\frac{a}{2}=\frac{b}{\frac{3}{2}}=\frac{c}{\frac{4}{3}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{2}=\frac{b}{\frac{3}{2}}=\frac{c}{\frac{4}{3}}=\frac{a-b}{2-\frac{3}{2}}=\frac{15}{\frac{1}{2}}=30\)
\(\Rightarrow\left\{\begin{matrix}\frac{a}{2}=30\Rightarrow a=30\cdot2=60\\\frac{b}{\frac{3}{2}}=30\Rightarrow b=30\cdot\frac{3}{2}=45\\\frac{c}{\frac{4}{3}}=30\Rightarrow c=30\cdot\frac{4}{3}=40\end{matrix}\right.\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
4. (3/4-81)(3^2/5-81)(3^3/6-81)....(3^6/9-81).....(3^2011/2014-81)
mà 3^6/9-81=0 => (3/4-81)(3^2/5-81)....(3^2011/2014-81)=0
a: để P là số nguyên thì \(3n-3+5⋮n-1\)
\(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{2;0;6;-4\right\}\)
b: Để Q là số nguyên thì \(3\left|n\right|-1+2⋮3\left|n\right|-1\)
\(\Leftrightarrow3\left|n\right|-1\in\left\{1;-1;2\right\}\)
\(\Leftrightarrow\left|n\right|\in\left\{0;1\right\}\)
hay \(n\in\left\{0;1;-1\right\}\)