Ta có: a/b + b/a = 41/20

=> BCNN(a;b) = 20 

=>BC(a;b) = {0;20;40;60;..........}

Tôi Mãi Mãi Yêu Lớp Tôi giari vows vaanr

Bạn nói đúng Nguyễn Huy Thắng ạ. Giải quá vớ vẩn

trình bày đầy đủ :

Ta có BĐT sau: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)( x,y >0 )

CM: \(\Leftrightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\)

Áp dụng bđt cô si cho 2 số dương x,y ta có:

\(x+y\ge2\sqrt{xy}\)

\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\)

\(\Rightarrow\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4\)( đúng )

Áp dụng bđt trên ta có: 

\(P=\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\ge\frac{4}{2\sqrt{2}}=\sqrt{2}\)

Dấu "=" xảy ra <=> \(a=b=\sqrt{2}\)

Vậy MIN P= \(\sqrt{2}\)\(a=b=\sqrt{2}\)

\(bđtcosi\)

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\ge\frac{4}{2\sqrt{2}}=\sqrt{2}\)

Dấu = xảy ra <=> a=b=\(\sqrt{2}\)

Min P=\(\sqrt{2}\)<=>a=b=\(\sqrt{2}\)

Ta có:

\(A=\frac{3}{2}+\frac{13}{12}+\frac{31}{30}+\frac{57}{56}+\frac{91}{90}\)

\(=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{12}\right)+\left(1+\frac{1}{30}\right)+\left(1+\frac{1}{56}\right)+\left(1+\frac{1}{90}\right)\)

\(=\left(1+1+1+1+1\right)+\left(\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)\)

\(=5+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right)\)

\(B=\frac{5}{6}+\frac{19}{20}+\frac{41}{42}+\frac{71}{72}+\frac{109}{110}\)

\(=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{110}\right)\)

\(=\left(1+1+1+1+1\right)-\left(\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\right)\)

\(=5-\left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\right)\)

=> A - B =\(\left[5+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right)\right]-\left[5-\left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\right)\right]\)

= \(5+\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}-5+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)

= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

= \(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{12}\right)+\left(1+\frac{1}{30}\right)+\left(1+\frac{1}{56}\right)+\left(1+\frac{1}{90}\right)\)

\(B=\left(1-\frac{1}{6}\right)+\left(1-\frac{19}{20}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{110}\right)\)

Mk gợi ý đến đây thôi , mk bí rồi đợi mk nghĩ đã!

mk sửa lại 1-1/20 chứ ko phải 1-19/20

\(A=\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)\)

\(B=\left(5-\frac{1}{6}-\frac{1}{20}-\frac{1}{42}-\frac{1}{72}-\frac{1}{110}\right)\)

\(A-B=\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)-\left(5-\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\right)\)

\(A-B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\)

\(A-B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)

\(A-B=1-\frac{1}{11}\)

\(A-B=\frac{10}{11}\)

\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)

\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)

\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)

\(\Leftrightarrow x+2=41\)

\(\Leftrightarrow x=41-2\)

\(\Leftrightarrow x=39\)

???????????????????????????????????????????????????????

99% LÀ 39

CÒN LAI LÀ ĐÁP ÁN KHÁC

ta có 20/39 > 14/39

22/27 > 22/29

18/43 < 18/41

=> 20/39+22/27+18/43 > 14/39+22/29+18/41

ta có 20/39 > 14/39

22/27 > 22/29

18/43 < 18/41

=> 20/39+22/27+18/43 > 14/39+22/29+18/41

   Bằng dấu lớn ( > ) bạn nhé!^-^

Chúc bạn học tốt nha!^-^

a/2 >hoặc = a/5 ( xảy ra giấu bằng với a=0)

b/3> hoặc = b/5 ( xảy randaaus bằng với a=0

Do đó : a/2 +b/3 = a/5 + b/5 chỉ trong trường hợp a=b=0

tìm các số tự nhiên a,b,c sao cho a^2 <=b;b^2<=c;c^2<=a

very easy

\(a)\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}=\frac{20}{41}\)

\(2\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{x\left(x+2\right)}\right)=2\cdot\frac{20}{41}\)

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{x\left(x+2\right)}=\frac{40}{41}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{40}{41}\)

\(1-\frac{1}{x+2}=\frac{40}{41}\)

\(\frac{x+1}{x+2}=\frac{40}{41}\)

\(\Leftrightarrow\hept{\begin{cases}x+1=40\\x+2=41\end{cases}\Leftrightarrow\hept{\begin{cases}x=40-1\\x=41-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=39\\x=39\end{cases}}}\)

Vậy x=39

\(b)|x+2016|\ge0\forall x;|x+2017|\ge0\forall x\)

\(\Leftrightarrow x+2016+x+2017+2018=3x\)

\(\Leftrightarrow2x+6051=3x\)

\(\Leftrightarrow6051=3x-2x\)

\(\Leftrightarrow6051=x\)

Vậy x=6051