Cho C= \(\dfrac{5}{5.8.11}+\dfrac{5}{8.11.14}+...+\dfrac{5}{302.305.308}\). Chứng minh C<\(\dfrac{1}{48}\)
Cho B=5/5.8.11+5/8.11.14+...+5/302.305.308
Chứng minh B<1/48
\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)
\(\Rightarrow\frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)
\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)
\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{8.11}+...+\frac{1}{302.305}-\frac{1}{305.308}\)
\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)
Cho B=5/5.8.11+5/8.11.14+...+5/302.305.308
Chứng minh B<1/48
Lời giải:
\(B=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.205.308}\)
\(\Rightarrow \frac{6}{5}B=\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\)
\(=\frac{11-5}{5.8.11}+\frac{14-8}{8.11.14}+...+\frac{308-302}{302.305.308}\)
\(=\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}-\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\)
\(=\frac{1}{5.8}-\frac{1}{305.308}< \frac{1}{5.8}\)
\(\Rightarrow B< \frac{1}{40}.\frac{5}{6}\Leftrightarrow B< \frac{1}{48}\)
cho c=5/5.8.11+5/8.11.14+5/302.305.308
c/m : c <1/48
Cho C=\(\frac{5}{5.8.11}\)+ \(\frac{5}{8.11.14}\)+..............+\(\frac{5}{302.305.308}\)
Chứng minh C< \(\frac{1}{48}\)
Chứng minh :
1,C=\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}.C< \frac{3}{4}\)
2,D=\(\frac{1}{5^2}+\frac{1}{9^2}+...+\frac{1}{409^2}< \frac{1}{12}\)
3,E=\(\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}< \frac{1}{48}\)
Chứng minh rằng:
1)B=\(\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}< 100\)
2)C=\(\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)<\(\frac{1}{48}\)
3)D=\(\frac{11}{9}+\frac{18}{16}+\frac{27}{25}+...+\frac{1766}{1764}\)
\(40\frac{20}{43}< D< 40\frac{20}{21}\)
Cho D=5/5.8.11+5/11.14.17+...+5/302.305.308
CMR:D nho hon 1/48
Bạn chép đề sai rồi, mình sửa lại đề và làm luôn nhé :
Ta có :
\(D=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\frac{1}{40}-\frac{5}{6}.\frac{1}{305.308}\)
\(\Rightarrow D=\frac{1}{48}-\frac{5}{6.305.308}< \frac{1}{48}\) (đpcm )
cho a,b,c >0 . Chứng minh \(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\)
Áp dụng BĐT \(AM-GM\) ta có :
\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2\ge5\sqrt[5]{\dfrac{a^{15}b^4}{b^9}}=5\dfrac{a^3}{b}\)
\(\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+c^2+c^2\ge5\sqrt[5]{\dfrac{b^{15}c^4}{c^9}}=5\dfrac{b^3}{c}\)
\(\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+a^2+a^2\ge5\sqrt[5]{\dfrac{c^{15}a^4}{a^9}}=5\dfrac{c^3}{a}\)
Cộng từng vế của BĐT ta được :
\(3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(a^2+b^2+c^2\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
Tiếp tục áp dụng BĐT \(AM-GM\) ta lại có :
\(\dfrac{a^5}{b^3}+\dfrac{a^5}{b^3}+b^2+b^2+b^2\ge5\sqrt[5]{\dfrac{a^{10}b^6}{b^6}}=5a^2\)
\(\dfrac{b^5}{c^3}+\dfrac{b^5}{c^3}+c^2+c^2+c^2\ge5\sqrt[5]{\dfrac{b^{10}c^6}{c^6}}=5b^2\)
\(\dfrac{c^5}{a^3}+\dfrac{c^5}{a^3}+a^2+a^2+a^2\ge5\sqrt[5]{\dfrac{c^{10}a^6}{a^6}}=5c^2\)
Cộng vế theo vế ta được :
\(2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+3\left(a^2+b^2+c^2\right)\ge5\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge a^2+b^2+c^2\)
\(\Rightarrow3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge3\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)+2\left(a^2+b^2+c^2\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
\(\Leftrightarrow5\left(\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\right)\ge5\left(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\right)\)
\(\Leftrightarrow\dfrac{a^5}{b^3}+\dfrac{b^5}{c^3}+\dfrac{c^5}{a^3}\ge\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\left(đpcm\right)\)
C=5/58.11+5/8.11.14+3.6/4.5+...+5/302.305.308 CMR C<1/48