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1/a,
-Ta có:
$B<1\Leftrightarrow B<\frac{10^{2005}+1+9}{10^{2006}+1+9}=\frac{10^{2005}+10}{10^{2006}+10}=\frac{10(10^{2004}+1)}{10(10^{2005}+1)}=\frac{10^{2004}+1}{10^{2005}+1}=A$
-Vậy: B<A
b,$A=1+(\frac{1}{2})^2+...+(\frac{1}{100})^2$
$\Leftrightarrow A=1+\frac{1}{2^2}+...+\frac{1}{100^2}$
$\Leftrightarrow A<1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}$
$\Leftrightarrow A<1+\frac{1}{1}-\frac{1}{2}+...+\frac{1}{99}-\frac{1}{100}$
$\Leftrightarrow A<1+1-\frac{1}{100}\Leftrightarrow A<2-\frac{1}{100}\Leftrightarrow A<2(đpcm)$
2,
a.
-Ta có:$\Rightarrow \frac{3x+7}{x-1}=\frac{3(x-1)+16}{x-1}=\frac{3(x-1)}{x-1}+\frac{16}{x-1}=3+\frac{16}{x-1}
-Để: 3x+7/x-1 nguyên
-Thì: $\frac{16}{x-1}$ nguyên
$\Rightarrow 16\vdots x-1\Leftrightarrow x-1\in Ư(16)\Leftrightarrow ....$
b, -Ta có:
$\frac{n-2}{n+5}=\frac{n+5-7}{n+5}=1-\frac{7}{n+5}$
-Để: n-2/n+5 nguyên
-Thì: \frac{7}{n+5} nguyên
$\Leftrightarrow 7\vdots n+5\Leftrightarrow n+5\in Ư(7)\Leftrightarrow ...$
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
a: Để A là phân số thì n-3<>0
hay n<>3
b: Để A là số nguyên thì \(n-3+4⋮n-3\)
\(\Leftrightarrow n-3\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{4;2;5;1;7;-1\right\}\)
c: Thay x=-1/2 vào A, ta được:
\(A=\dfrac{-\dfrac{1}{2}+1}{-\dfrac{1}{2}-3}=\dfrac{1}{2}:\dfrac{-7}{2}=-\dfrac{1}{7}\)
ta có :
A = \(\dfrac{5^{2020}+1}{5^{2020}+1}\)
B = \(\dfrac{5^{2019}+1}{5^{2020}+1}\)
\(\Leftrightarrow\) B < A
b.\(B=\dfrac{2n+5}{n+3}\)
\(B=\dfrac{n+n+3+3-1}{n+3}=\dfrac{n+3}{n+3}+\dfrac{n+3}{n+3}-\dfrac{1}{n+3}\)
\(B=1+1-\dfrac{1}{n+3}\)
Để B nguyên thì \(\dfrac{1}{n+3}\in Z\) hay \(n+3\in U\left(1\right)=\left\{\pm1\right\}\)
*n+3=1 => n=-2
*n+3=-1 => n= -4
Vậy \(n=\left\{-2;-4\right\}\) thì B có giá trị nguyên
18:
a: \(S=3\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)
=3*(1/2-1/4+1/4-1/6+...+1/98-1/100)
=3*49/100=147/100
b: Để A là số nguyên thì n-1 thuộc Ư(2)
=>n-1 thuộc {1;-1;2;-2}
=>n thuộc {2;0;3;-1}