Cho A = \(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)
B = \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)
Tính A : B.
Mik đang cần gấp ai làm nhanh và đúng nhất mik tik cho 3 cái lun nha!
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1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10x9/10
=9/10x(1/2+2/3)+(3/4+4/5)+(5/6+6/7)+(7/8+8/9)
=9/10x(1/3+3/5+5/7+7/9)
9/10x(1/3+3/5)+(5/7+7/9)
=9/10x1/5+5/9
9/50+5/9
=10
Bn Long làm đúng rồi bn nguyễn kim arica cứ làm theo cách đó là được .
Bn nào thấy đúng thì ủng hộ nha .
1/ Ta có: \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{2^2}{4}=\frac{4}{4}=1\)
Dấu "=" xảy ra khi x=y=1
Máy mình bị lỗi nên ko nhìn được các bài tiếp theo
Chúc bạn học tốt :)
Ta có : x+y=2 => x=2-y. Thay vào bt ta đc : xy= (2-y).y = 2y -y^2
Vì y^2 >= 0 =>2y-y^2 nhỏ hơn hoặc bằng 0
a) \(A=\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.\left(2^2.5\right)^4}{5^{2^5}.\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{\left(5^{10}:5^8\right).\left(2^{10}:2^8\right)}=\frac{1}{5^2.2^2}=\frac{1}{25.4}=\frac{1}{100}\)
b) \(B=\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}\)\(=\frac{2^3+2^3}{1}=\frac{8+8}{1}=16\)
c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\)
\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{99}}\)
\(\Rightarrow2C-C=\left(1+\frac{1}{2}+\frac{1}{2^2}+.........+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\right)\)
\(\Rightarrow C=1-\frac{1}{2^{100}}\)
d) \(D=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{100}}\)
\(\Rightarrow5D=5+1+\frac{1}{5^2}+\frac{1}{5^3}+...........+\frac{1}{5^{101}}\)
\(\Rightarrow5D-D=\left(5+1+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{101}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+..........+\frac{1}{5^{100}}\right)\)
\(\Rightarrow4D=5-\frac{1}{5^{101}}\)
\(\Rightarrow D=\frac{5-\frac{1}{5^{101}}}{4}\)
a) \(A=\frac{5^4x20^4}{25^5x4^5}=\frac{5^4x\left(2^2x5\right)^4}{\left(5^2\right)^5x\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{5^2x2^2}=\frac{1}{25.4}=\frac{1}{100}\)
b) \(B=\frac{2^{30}x5^7+2^{13}x5^{27}}{2^{27}x5^7+2^{10}x5^{27}}=\frac{2^{13}.5^7.\left(2^{17}+5^{20}\right)}{2^{10}.5^7.\left(2^{17}+5^{20}\right)}=2^3=8\)
c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2C-C=1-\frac{1}{2^{100}}\)
\(C=1-\frac{1}{2^{100}}\)
phần d bn lm tương tự như phần c nha!
Ta có :
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\)\(\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\)\(\frac{2}{7}-\frac{2}{7}\)
\(=\)\(0\)
Chúc bạn học tốt ~
1/3xD=1/(2x4)+1/(4x6)+...+1/(98x100)
2/3xD=2/(2x4)+2/(4x6)+...+1/(98x100)
2/3xD= 1/2-1/4+1/4-1/6+...+1/98-1/100
2/3xD=1/2-1/100
2/3xD=49/100
D=147/200
\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{9}{1}+\frac{8}{2}+...+\frac{1}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10-1}{1}+\frac{10-2}{2}+...+\frac{10-9}{9}\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right].x=\frac{10}{1}-1+...+\frac{10}{9}-1\)
=> \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10-9+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}\)= \(\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)
=>\(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right]x=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)
=> \(x=10\)
b) Tương tự câu a
từ đề bài ta có \(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)
\(\frac{A}{B}=10\)