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\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\)\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}\)\(=\dfrac{1}{50}\)
Ta có:
P= \(\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
P= \(\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\left(1+1+...+1\right)\)(có 49 chữ số 1)
P= \(\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)
P= \(\dfrac{50}{49}+\dfrac{50}{48}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
P= \(50.\left(\dfrac{1}{50}+\dfrac{1}{49}+...+\dfrac{1}{2}\right)\)
⇒ \(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}}{50.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)}\)
⇒ \(\dfrac{S}{P}=\dfrac{1}{50}\)
Vậy \(\dfrac{S}{P}=\dfrac{1}{50}\)
P = 1/49+2/48+3/47+...+48/2+49/1
Cộng 1 váo mỗi p/s trong 48 p/s đầu , trừ p/s cuối đi 48 ta được
P=(1/49+1)+(2/48+1)+...+(48/2+1)+1
P= 50/49+50/48+....+50/2+50/50
Đưa ps cuối lên đầu
P=50/50+50/49+50/48+...+50/2
=50.(1/50+1/49+1/48+...+1/4+1/3+1/2)
=50S
=> S/P=1/50
\(Q=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{47}{3}+\dfrac{48}{2}+\dfrac{49}{1}\\ =\dfrac{1}{49}+1+\dfrac{2}{48}+1+\dfrac{3}{47}+1+...+\dfrac{47}{3}+1+\dfrac{48}{2}+1+1\\ =\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{3}+\dfrac{50}{2}+\dfrac{50}{50}\\ =50\cdot\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+...+\dfrac{1}{3}+\dfrac{1}{2}\right)\\ =50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
\(\dfrac{P}{Q}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}}{50\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{48}+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
\(49,=\dfrac{38}{5}-\left(\dfrac{19}{7}+\dfrac{28}{5}\right)\)
\(=\dfrac{38}{5}-\dfrac{19}{7}-\dfrac{28}{5}\)
\(=\left(\dfrac{38}{5}-\dfrac{28}{5}\right)-\dfrac{19}{7}\)
\(=2-\dfrac{19}{7}=-\dfrac{5}{7}\)
\(50,=\dfrac{25}{81}.\dfrac{15}{22}=\dfrac{125}{594}\)
mik xinloi ạ câu 50 mik viết sai đề nên mong bạn giải lại giúp ạ <3
Ta có : \(\left(x-1\right)^2+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{41.45}=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{100}\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{10}\\x-1=-\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
Vậy ...
a:
ĐKXĐ: x<>3
\(\dfrac{x-3}{2}=\dfrac{72}{x-3}\)
=>\(\left(x-3\right)^2=72\cdot2=144\)
=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\left(nhận\right)\\x=-9\left(nhận\right)\end{matrix}\right.\)
b: \(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\cdot x=\dfrac{1}{49}+\dfrac{2}{48}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)
=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)=\dfrac{50}{49}+\dfrac{50}{48}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
=>\(x\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
=>x=50