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Em chưa học làm dạng này , em làm thử thôi nhá, sai xin chỉ dạy thêm nha
2 . \(\dfrac{n^7+n^2+1}{n^8+n+1}=\dfrac{n^7-n+n^2+n+1}{n^8-n^2+n^2+n+1}\)
\(=\dfrac{n\left(n^6-1\right)+n^2+n+1}{n^2\left(n^6-1\right)+n^2+n+1}=\dfrac{n\left(n^3+1\right)\left(n^3-1\right)+n^2+n+1}{n^2\left(n^3+1\right)\left(n^3-1\right)+n^2+n+1}\)\(=\dfrac{n\left(n^3+1\right)\left(n-1\right)\left(n^2+n+1\right)+n^2+n+1}{n^2\left(n^3+1\right)\left(n-1\right)\left(n^2+n+1\right)+n^2+n+1}\)
\(=\dfrac{\left(n^2+n+1\right)\left[\left(n^4+n\right)\left(n-1\right)\right]}{\left(n^2+n+1\right)\left[\left(n^5+n^2\right)\left(n-1\right)+1\right]}\)
\(=\dfrac{n^5-n^4+n^2-n}{n^6-n^5+n^3-n^2+1}=\dfrac{n^4\left(n-1\right)+n\left(n-1\right)}{n^5\left(n-1\right)+n^2\left(n-1\right)+1}\)
\(=\dfrac{\left(n-1\right)\left(n^4+n\right)}{\left(n-1\right)\left(n^5+n^2\right)+1}\)
Vậy ,với mọi số nguyên dương n thì phân thức trên sẽ không tối giản
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\(a^2+\left(a+1\right)^2=a^2+a^2+2a+1\\ =2a^2+2a+1>2a\left(a+1\right)\\ \Rightarrow\dfrac{1}{a^2+\left(a+1\right)^2}< \dfrac{1}{2a\left(a+1\right)}\)
\(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+...+\dfrac{1}{n^2+\left(n+1\right)^{^2}}\\ =\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+...+\dfrac{1}{n^2+\left(n+1\right)^2}\\ < \dfrac{1}{2.1.\left(1+2\right)}+\dfrac{1}{2.2\left(2+1\right)}+....+\dfrac{1}{2n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{n+1}\right)\\ =\dfrac{1}{2}\left(\dfrac{5}{6}-\dfrac{1}{n+1}\right)\\ =\dfrac{5}{12}-\dfrac{1}{2n+2}< \dfrac{5}{12}< \dfrac{9}{20}\)
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\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)
\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)
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a) Gọi ƯCLN(3n+1;5n+2) là d
ta có: 3n+1 chia hết cho d => 15n + 5 chia hết cho d
5n + 2 chia hết cho d => 15n + 6 chia hết cho d
=> 15n + 6 - 15n - 5 chia hết cho d
=> 1 chia hết cho d
=> 3n+1/5n+2 là phân số tối giản
gọi d là ƯC(3n + 1; 5n + 2) (d thuộc Z)
\(\Rightarrow\hept{\begin{cases}3x+1⋮d\\5n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}5\left(3n+1\right)⋮d\\3\left(5n+2\right)⋮d\end{cases}\Rightarrow\hept{\begin{cases}15n+5⋮d\\15n+6⋮d\end{cases}}}}\)
=> (15n + 5) - (15n + 6) ⋮ d
=> 15n + 5 - 15n - 6 ⋮ d
=> (15n - 15n) - (6 - 5) ⋮ d
=> 0 - 1 ⋮ d
=> 1 ⋮ d
=> d = 1 hoặc d = -1
vậy \(\frac{3n+1}{5n+2}\) là phân số tối giản với mọi n thuộc N
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C1:Biến đổi tương đương
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Leftrightarrow\dfrac{x}{xy}+\dfrac{y}{xy}\ge\dfrac{4}{x+y}\)
\(\Leftrightarrow\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2+y^2+2xy\ge4xy\Leftrightarrow x^2+y^2-2xy\ge0\Leftrightarrow\left(x-y\right)^2\ge0\)
C2:Dùng AM-GM
\(x+y\ge2\sqrt{xy}\);\(\dfrac{1}{x}+\dfrac{1}{y}\ge2\sqrt{\dfrac{1}{x}\cdot\dfrac{1}{y}}=2\sqrt{\dfrac{1}{xy}}\)
Nhân theo vế 2 BĐT
\(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\ge4\sqrt{xy\cdot\dfrac{1}{xy}}=4\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
C3:Dùng Cauchy-Schwarz (dạng Engel)
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{\left(1+1\right)^2}{x+y}=\dfrac{4}{x+y}\)
-3 cách trên đều có dấu "=" khi \(x=y\)
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Tk mình đi mọi người mình bị âm nè!
Ai tk mình mình tk lại cho!
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Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
\(\Rightarrow A< \dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+...+\dfrac{1}{n^2-1}\)
\(\Rightarrow2A< \dfrac{2}{1.3}+\dfrac{2}{2.4}+\dfrac{2}{3.5}+...+\dfrac{2}{\left(n-1\right)\left(n+1\right)}\)
\(\Rightarrow2A< 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n+1}\)
\(\Rightarrow2A< 1\)
\(\Rightarrow A< \dfrac{1}{2}< \dfrac{2}{3}\)