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Ta có :
\(A=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}\)
\(A=\frac{1}{2^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< \frac{1}{2^2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\right)\)
\(A< \frac{1}{4}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)=\frac{1}{4}\left(1-\frac{1}{n}\right)\)
\(A< \frac{1}{4}-\frac{1}{4n}\)
Lại có \(n>0\) nên \(\frac{1}{4n}>0\)
\(\Rightarrow\)\(\frac{1}{4}-\frac{1}{4n}< \frac{1}{4}\)
Vậy \(A< \frac{1}{4}\)
\(\frac{a}{n+2}=\frac{b}{n+5}=\frac{c}{n+8}=k\Leftrightarrow a=nk+2k;b=nk=5k;c=nk+8k\)
\(\left(a+c\right)^2=\left(nk+2k+nk+8k\right)^2=4k^2\left(n+5\right)^2\) ( sai nhế)
\(4\left(a-b\right)\left(b-c\right)=4\left(nk+2k-nk-5k\right)\left(nk+5k-nk-8k\right)=4\left(-3k\right)\left(-3k\right)=36k^2\)
\(\left(a-c\right)^2=\left(nk+2k-nk-8k\right)^2=4\left(-6k\right)^2=36k^2\)
=> \(\left(a-c\right)^2=4\left(a-b\right)\left(b-c\right)\)
Tìm trước khi hỏi , google-sama chưa tính phí mà !
Câu hỏi của phạm minh anh - Toán lớp 7 - Học toán với OnlineMath
\(\frac{a}{b}\)= \(\frac{a\left(a+n\right)}{b\left(b+n\right)}\) = \(\frac{ab+an}{b^2+bn}\)
\(\frac{a+n}{b+n}\)= \(\frac{\left(a+n\right)b}{\left(b+n\right)b}\)= \(\frac{ab+nb}{b^2+bn}\)
Nếu a < b thì ab + an < ab + nb => \(\frac{a}{b}\)< \(\frac{a+n}{b+n}\)
Nếu a > b thì ab + an > ab + nb => \(\frac{a}{b}\)> \(\frac{a+n}{b+n}\)
Nếu a = b thì ab + an = ab + nb => \(\frac{a}{b}\)= \(\frac{a+n}{b+n}\)
1.
\(10x=|x+\dfrac{1}{10}|+|x+\dfrac{2}{10}|+...+|x+\dfrac{9}{10}| \ge 0\)
\(\Rightarrow x\ge0\)
\(pt\Leftrightarrow x+\frac{1}{10}+x+\frac{2}{10}+...+x+\frac{9}{10}=10x\)
\(\Leftrightarrow x=\frac{1}{10}+\frac{2}{10}+...+\frac{9}{10}=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{2}\)
4.
Áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{a}{b+3c}=\frac{b}{c+3a}=\frac{c}{a+3b}=\frac{a+b+c}{4\left(a+b+c\right)}=\frac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}4a=b+3c\left(1\right)\\4b=c+3a\left(2\right)\\4c=a+3b\left(3\right)\end{matrix}\right.\)
Từ \(\left(1\right);\left(2\right)\Rightarrow4a=b+3\left(4b-3a\right)\)
\(\Rightarrow12a=12b\Rightarrow a=b\left(4\right)\)
Từ \(\left(1\right);\left(3\right)\Rightarrow4c=a+3\left(4a-3c\right)\)
\(\Rightarrow12a=12c\Rightarrow a=c\left(5\right)\)
Từ \(\left(4\right);\left(5\right)\Rightarrow a=b=c\left(đpcm\right)\)
\(\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}.\frac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)n\left(n+1\right)}\)
\(=\frac{1}{2}\left[\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right]\)
Ta có đpcm.
Ta có
\(A=\frac{1}{14}+\frac{1}{29}+...+\frac{1}{n^2+\left(n+1\right)^2+\left(n+2\right)^2}+...+\frac{1}{1877}\)
\(=\frac{1}{1^2+2^2+3^2}+\frac{1}{2^2+3^2+4^2}+...+\frac{1}{n^2+\left(n+1\right)^2+\left(n+2\right)^2}+...+\frac{1}{24^2+25^2+26^2}\)
\(B=n^2+\left(n+1\right)^2+\left(n+2\right)^2=3n^2+6n+5\left(1\right)\)
+ Với \(n\ge1\)từ (1) ta có \(B\le3n^2+9n+6=3\left(n^2+3n+2\right)=3\left(n+1\right)\left(n+2\right)\)Từ đó
\(A>\frac{1}{3}\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\right)=\frac{1}{3}C\)
Với \(C=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}=\frac{1}{2}-\frac{1}{26}=\frac{6}{13}\)
\(\Rightarrow A>\frac{1}{3}\cdot\frac{6}{13}=\frac{2}{13}>0,15\)
+ Với \(n\ge1\)từ (1) ta có \(B>2n^2+6n+4=2\left(n^2+3n+2\right)=2\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A< \frac{1}{2}\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\right)=\frac{1}{2}C\)
Với \(C=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}=\frac{1}{2}-\frac{1}{26}=\frac{6}{13}\)
\(\Rightarrow A< \frac{1}{2}\cdot\frac{6}{13}=\frac{3}{13}< 0,25\)
Vậy \(0,15< A< 0,25\)
ta có \(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{n+a-n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
vậy \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)