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\(A=\dfrac{2^4.3^3+2^3.3^4}{2^5.3^4-2^6.3^3}=\dfrac{2^3.3^3.\left(2+3\right)}{2^5.3^3.\left(3-2\right)}=\dfrac{2^3.3^3.5}{2^5.3^3.1}\)
\(=\dfrac{5}{2^2}=\dfrac{5}{4}\)
\(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)}=\dfrac{1}{50}\)
Bài 1: Tìm số đối.
- Số đối của \(\dfrac{1}{2}\) là \(-\dfrac{1}{2}\)
- Số đối của \(-\dfrac{3}{4}\) là \(\dfrac{3}{4}\)
- Số đối của \(\dfrac{7}{-12}\) là \(\dfrac{7}{12}\)
Bài 2: Thu gọn:
\(\dfrac{2^4.3^3-2^4.3^3}{2^5.3^4-2^6.3^3}=\dfrac{0}{2^5.3^4-2^6.3^3}=0\)
Ta có: \(P=\dfrac{1}{49}+\dfrac{2}{48}+\dfrac{3}{47}+...+\dfrac{48}{2}+\dfrac{49}{1}\)
\(P=\left(1+\dfrac{1}{49}\right)+\left(1+\dfrac{2}{48}\right)+\left(1+\dfrac{3}{47}\right)+...+\left(1+\dfrac{48}{2}\right)+1\)
\(P=\dfrac{50}{49}+\dfrac{50}{48}+\dfrac{50}{47}+...+\dfrac{50}{2}+\dfrac{50}{50}\)
\(P=50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(\Rightarrow\)\(\dfrac{S}{P}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}}{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)}\)\(=\dfrac{1}{50}\)
\(\dfrac{-2}{3}.x+\dfrac{1}{5}=\dfrac{3}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{1}{5}\)
\(\dfrac{-2}{3}.x=\dfrac{3}{10}-\dfrac{2}{10}\)
\(\dfrac{-2}{3}.x=\dfrac{1}{10}\)
\(x=\dfrac{1}{10}:\dfrac{-2}{3}\)
\(x=\dfrac{1}{10}.\dfrac{-3}{2}\)
\(x=\dfrac{-3}{20}\)
Đặt \(A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\)
\(\Rightarrow3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)
\(\Rightarrow3A-A=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{100}}\right)\)
\(\Rightarrow2A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\dfrac{100}{3^{100}}\)
Đặt \(B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3B=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3B-B=\left(3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2B=3-\dfrac{1}{3^{99}}\Rightarrow B=\dfrac{3}{2}-\dfrac{1}{3^{99}.2}\)
Do đó: \(2A=\dfrac{3}{2}-\dfrac{1}{3^{99}.2}-\dfrac{100}{3^{100}}=\dfrac{3^{101}}{3^{100}.2}-\dfrac{3}{3^{100}.2}-\dfrac{200}{3^{100}.2}=\dfrac{3^{101}-203}{3^{100}.2}\Rightarrow A=\dfrac{3^{101}-203}{3^{100}.4}\)
Vậy...
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