lon hon 1 nha ban

sửa lại đề : Chứng tỏ rằng : A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}< 1\)

bài làm

A = \(\frac{1}{2!}+\frac{2}{3!}+...+\frac{2013}{2014!}\)

A = \(\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2014-1}{2014!}\)

A = \(1-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2014}{2014!}-\frac{1}{2014!}\)

A = \(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2013!}-\frac{1}{2014!}\)

A = \(1-\frac{1}{2014!}< 1\)

lớn hơn 1 nha.

\(n^2>\left(n-1\right)\left(n+1\right)\Rightarrow\frac{1}{n^2}< \frac{1}{\left(n-1\right)\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right).\) 

 Do đó:   \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2013^2}+\frac{1}{2014^2}< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2012.2014}+\frac{1}{2013.2015}=\) 

\(=\frac{1}{2}[1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2012}-\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2015}]=\) 

\(=\frac{1}{2}[1+\frac{1}{2}-\frac{1}{2014}-\frac{1}{2015}]=\frac{1}{2}[\frac{3}{2}-\frac{1}{2014}-\frac{1}{2015}]=\frac{3}{4}-\frac{1}{2}\left(\frac{1}{2014}+\frac{1}{2015}\right)< \frac{3}{4}.\)

\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}\)

\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)

\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4!}+...+\frac{1}{98!}-\frac{1}{100!}\)

\(=2-\frac{1}{99!}-\frac{1}{100!}< 2\)

Vậy \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\left(đpcm\right)\)

Đặt \(A=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)

\(\Rightarrow2^2A=2^2.\left(\frac{1}{2^2}-\frac{1}{2^4}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\right)\)

\(\Rightarrow4A=1-\frac{1}{2^2}+\frac{1}{2^4}-...-\frac{1}{2^{4n-2}}+\frac{1}{2^{4n}}-...-\frac{1}{2^{2002}}\)

\(\Rightarrow4A+A=\left(1-\frac{1}{2^2}+\frac{1}{2^4}-...-\frac{1}{2^{4n-2}}+\frac{1}{2^{4n}}-...-\frac{1}{2^{2002}}\right)+\left(\frac{1}{2^2}-\frac{1}{2^4}+...+\frac{1}{2^{4n-2}}-\frac{1}{2^{4n}}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\right)\)

\(\Rightarrow5A=1-\frac{1}{2^{2004}}\)

Vì \(1-\frac{1}{2^{2004}}< 1.\)

\(\Rightarrow5A< 1\)

\(\Rightarrow A< \frac{1}{5}=0,2\)

\(\Rightarrow A< 0,2\left(đpcm\right).\)

Chúc bạn học tốt!

Ta có:

\(x^2+y^2=1\Rightarrow\left(x^2+y^2\right)^2=1\)(1)

Thay (1) vào \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)ta có:

\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{x^4+2x^2y^2+y^4}{a+b}\)

\(\Leftrightarrow\left(x^4b+y^4a\right)\left(a+b\right)=\left(x^4+2x^2y^2+y^4\right).ab\)

\(\Leftrightarrow x^4ab+x^4b^2+y^4a^2+y^4ab=x^4ab+2x^2y^2ab+y^4ab\)

\(\Leftrightarrow x^4b^2+y^4a^2=2x^2y^2ab\)

\(\Leftrightarrow\left(x^2b\right)^2-2x^2y^2ab+\left(y^2a\right)^2=0\)

\(\Leftrightarrow\left(x^2b-y^2a\right)^2=0\)

\(\Leftrightarrow x^2b-y^2a=0\)

\(\Leftrightarrow x^2b=y^2a\)

\(\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)

\(\Rightarrow\left(\frac{x^2}{a}\right)^{1002}=\left(\frac{y^2}{b}\right)^{1002}=\left(\frac{1}{a+b}\right)^{1002}\)

\(\Rightarrow\frac{x^{2004}}{a^{1002}}=\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}\)

\(\Rightarrow\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{1}{\left(a+b\right)^{1002}}+\frac{1}{\left(a+b\right)^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\left(đpcm\right)\)

Chúc bạn học tốt!

\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)

\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)

\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)

Giải :

\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)

\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)

\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)

~~Học tốt~~

\(\text{Giải :}\)

\(\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{2}\frac{\text{5}}{\text{27}}-\text{10}\frac{\text{5}}{\text{6}}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\left(\text{1}\frac{\text{3}}{\text{7}}+\frac{\text{10}}{\text{3}}\right):\left(\text{12}\frac{\text{1}}{\text{3}}-\text{14}\frac{\text{2}}{\text{7}}\right)}=\frac{\left[\text{13}\frac{\text{1}}{\text{4}}-\left(\text{2}\frac{\text{5}}{\text{27}}+\text{10}\frac{\text{5}}{\text{6}}\right)\right].\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{100}}{\text{21}}:\frac{\text{-41}}{\text{21}}}\)

\(=\frac{\left(\text{13}\frac{\text{1}}{\text{4}}-\text{13}\frac{\text{1}}{54}\right).\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\frac{\text{25}}{\text{108}}.\text{230}\frac{\text{1}}{\text{25}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}\)

\(=\frac{\text{53}\frac{\text{1}}{\text{4}}+\text{46}\frac{\text{3}}{\text{4}}}{\frac{\text{-100}}{\text{41}}}=\frac{\text{100}}{\frac{-\text{100}}{\text{41}}}=\text{-41}\)

\(\text{~~Học tốt~~}\)

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

........

\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)

=> \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)

Đpcm 

b)B=1/4(1/2^2+1/3^2+...+1/n^2)=1/4*A<1/4

tìm 2 số 3:

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