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Chứng minh: 1/2^3 + 1/3^3 + 1/4^3 + ... + 1/n^3 < 1/4 với n thuộc N, n ≥ 2 - Toán học Lớp 8 - Bài tập Toán học Lớp 8 - Giải bài tập Toán học Lớp 8 | Lazi.vn - Cộng đồng Tri thức & Giáo dục
\(A< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left(n-1\right)\cdot n\cdot\left(n+1\right)}\)
Nhận xét: mỗi số hạng tổng có dạng
\(\frac{1}{\left(n-1\right)\cdot n\cdot\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n\left(n-1\right)}-\frac{1}{n\left(n+1\right)}\right)\)
Từ đó suy ra: \(A< \frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+....+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{n\left(n+1\right)}\right)< \frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}\left(đpcm\right)\)
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
Linh ơi;Phương Anh đây bài này dễ mà học nhà thầy rùi cách giải nè:
Ta có:1/23 <1/1.2.3 ;1/33 <1/2.3.4;.....;1/n3<1/.(n-1).n.(n+1)
Suy ra Đề bài <1/1.2.3+1/2.3.4+1/3.4.5+....+1/(N-1).N.(N+1)
<1/1.2-1/2.3+1/2.3-1/3.4+...+1/N-1-1/N+1/N1/N+1
<1/2-1/n+1<1/4
Vậy........
\(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{99^2}\)
\(\Rightarrow M< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\)
\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\)
\(\Rightarrow M< 1-\frac{1}{99}< 1\)
Dễ thấy M > 0 nên 0 < M < 1
Vậy M không là số tự nhiên.
\(S=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
\(\Rightarrow S>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\) (50 số hạng \(\frac{1}{100}\))
\(\Rightarrow S>\frac{1}{100}.50=\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\left(đpcm\right)\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
Câu 2:
b) ĐKXĐ: \(x\ne-1\)
Để \(\frac{3x+5}{x+1}\) là số nguyên thì \(3x+5⋮x+1\)
\(\Leftrightarrow3x+3+2⋮x+1\)
mà \(3x+3⋮x+1\)
nên \(2⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(2\right)\)
\(\Leftrightarrow x+1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{0;-2;1;-3\right\}\)(tm)
Vậy: Khi \(x\in\left\{0;-2;1;-3\right\}\) thì \(\frac{3x+5}{x+1}\) là số nguyên
Câu 3:
a) ĐKXĐ: \(n\ne-3\)
Gọi \(d=ƯCLN\left(n+4;n+3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}n+4⋮d\\n+3⋮d\end{matrix}\right.\Leftrightarrow n+4-n-3⋮d\Leftrightarrow1⋮d\Leftrightarrow d=1\)
\(\LeftrightarrowƯCLN\left(n+4;n+3\right)=1\)
hay \(\frac{n+4}{n+3}\) là phân số tối giản(đpcm)
b) Gọi \(e=ƯCLN\left(n+2;2n+5\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}n+2⋮e\\2n+5⋮e\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2n+4⋮e\\2n+5⋮e\end{matrix}\right.\Leftrightarrow2n+4-2n-5⋮e\)
\(\Leftrightarrow-1⋮e\Leftrightarrow e=1\)
hay \(ƯCLN\left(n+2;2n+5\right)=1\)
\(\Leftrightarrow\frac{n+2}{2n+5}\) là phân số tối giản
c) Gọi \(f=ƯCLN\left(2n+1;3n+1\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2n+1⋮f\\3n+1⋮f\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6n+3⋮f\\6n+2⋮f\end{matrix}\right.\Leftrightarrow6n+3-6n-2⋮f\)
\(\Leftrightarrow1⋮f\Leftrightarrow f=1\)
\(\LeftrightarrowƯCLN\left(2n+1;3n+1\right)=1\)
hay \(\frac{2n+1}{3n+1}\) là phân số tối giản(đpcm)
Đặt A= \(\frac{3}{9.14}+\frac{3}{14.19}+...+\frac{3}{\left(5n+1\right).\left(5n+4\right)}\)
\(\Rightarrow A=3.\left(\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(=3.5.\frac{1}{5}.\left(\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(=\frac{3}{5}\left(\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{\left(5n-1\right)\left(5n+4\right)}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{5n-1}-\frac{1}{5n+4}\right)\)
\(=\frac{3}{5}\left(\frac{1}{9}-\frac{1}{5n+4}\right)\)
\(\Rightarrow\)\(A< \frac{3}{5}.\frac{1}{9}\)\(\Rightarrow A< \frac{1}{15}\)(đpcm)
ta có 1/23<1/1*2*3 1/33<1/2*3*4 1/43<1/3*4*5 .... 1/n3<1/(n-1)*n*(n+1)
Vậy=1/23+1/33+...+1/n3<1/1*2*3+1/2*3*4+.....1/(n-1)*n*(n+1)
Ta có 1/1*2*3 + 1/2*3*4 +...+ 1/(n-1)*n*(n+1)
=1/2*(1/1*2-1/2*3 + 1/2*3-1/3*4 +...+ 1/(n-1)*n-1/n*(n+1)
=1/2*(1/2- 1/6 + 1/6 -1/12+..........+1/(n-1)*n-1/n*(n+1)
=1/2*(1/2-1/n*(n+1))
=1/4-1/2n*(n+1)<1/4
Vì 1/2^3+1/3^3+..+1/n^3<1/4-1/2n*(n+1)<1/4
nên =>1/2^3+1/3^3+...+1/n^3<1/4
\(< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left(n-1\right).n}\)
\(< 2\cdot\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(< \frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{4\cdot5}-\frac{1}{5\cdot6}+...+\frac{2}{\left(n-1\right)\cdot n}\)
\(< \frac{1}{2}\cdot\left(\frac{1}{2}-\frac{2}{\left(n-1\right)\cdot n}\right)\)
\(< \frac{1}{4}-\frac{1}{\left(n-1\right)\cdot n}\)
ĐPCM
Ta có :
\(\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}<\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{4.5.6}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\)
\(=\frac{1}{1.2}-\frac{1}{n\left(n+1\right)}=\frac{1}{2}-\frac{1}{n\left(n+1\right)}\)
Vì n > 2 nên \(\frac{1}{n\left(n+1\right)}\le\frac{1}{6}\)
Do đó \(\frac{1}{2}-\frac{1}{n\left(n+1\right)}<\frac{1}{4}\)
=> ĐPCM