\(\frac{1^2}{1.2}\)\(\frac{2^2}{2.3}\)\(\frac{3^2}{3.4}\).......\(\frac{10^2}{10.11}\)=..........
Đáp án : \(\frac{1}{11}\) các bạn giúp mình cách làm nha
A=\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)
bn nào lm đc mk tick cx dễ đúng ko Vio vòng 2 thôi:D
A= (1x2x3x...x10)/(1x2x3x...x10)x(1x2x3x...x10)/(2x3x4x...x11)
A=1x 1/11=1/11
bạn nhớ nha
\(\frac{1^2}{1.2}\) . \(\frac{2^2}{2.3}\) . \(\frac{3^2}{3.4}\) ..... \(\frac{10^2}{10.11}\)
Hãy rút gọn biểu thức:
M=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\)..........=\(\frac{10^2}{10.11}\)
\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{10^2}{10.11}\)
\(M=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}......\frac{10.10}{10.11}\)
\(M=\frac{1.2.3.....10}{1.2.3....10}.\frac{1.2.3.....10}{2.3.4.....11}\)
\(M=1.\frac{1}{11}\)
\(M=\frac{1}{11}\)
Tính nhanh :
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}\)
\(B=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}...-\frac{1}{3.2}-\frac{1}{2.1}\)
giúp mk với nha các bạn
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{9^2}{9.10}\)
\(A=\frac{1.1.2.2.3.3...9.9}{1.2.2.3.3.4...9.10}\)
\(A=\frac{1}{10}\)
\(B=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(B=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-\frac{1}{98}+\frac{1}{98}-\frac{1}{97}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(B=\frac{1}{99}-\left(\frac{1}{99}-1\right)\)
\(B=\frac{1}{99}-\frac{1}{99}+1\)
\(B=1\)
sorry nha Thiên Sứ đội lốt Ác Quỷ mk 5 - 6
Hãy tính giúp mình nha
B=\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}....\frac{99^2}{99.100}\)
\(=\frac{1.2.3.....99}{1.2.3.....98}.\frac{1.2.3......99}{2.3.4.5....100}\)
\(=99.\frac{1}{100}\)
\(=\frac{99}{100}\)
\(B=\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}\)
\(B=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.....\frac{99.99}{99.100}\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{99}{100}\)
\(B=\frac{1}{100}\)
tính:
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{20.21}\)
giúp mik nha,ai làm nhanh và rõ ràng chinh xác thì mik sẽ k 3 cái luôn nha!!
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\)
\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{20.21}\right)\)
\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{20}-\frac{1}{21}\right)\)
\(=2\left(1-\frac{1}{21}\right)=2.\frac{20}{21}=\frac{40}{21}\)
A=\(\frac{1^2}{1.2}\)x\(\frac{2^2}{2.3}\)x\(\frac{3^2}{3.4}\)x\(\frac{4^2}{4.5}\)
giúp mình và giải chi tiết nha - thanks
\(A=\frac{1\cdot1}{1\cdot2}\cdot\frac{2\cdot2}{2\cdot3}\cdot\frac{3\cdot3}{3\cdot4}\cdot\frac{4\cdot4}{4\cdot5}=\frac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot3\cdot4}\cdot\frac{1\cdot2\cdot3\cdot4}{2\cdot3\cdot4\cdot5}=\frac{1}{5}\)
A= 1/2 * 2/3 * 3/4 * 4/5
= 1*2*3*4/2*3*4*5
= 1/5
A = \(\frac{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5}\)= \(\frac{\left(1\cdot2\cdot3\cdot4\right)\cdot\left(1\cdot2\cdot3\cdot4\right)}{\left(1\cdot2\cdot3\cdot4\right)\cdot\left(2\cdot3\cdot4\cdot5\right)}\)= \(\frac{1}{5}\)
Bạn k cho mik nhé.
Tìm x
\(X-\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-....-\frac{1}{98.99}=\frac{1}{100}+\)1/99.100
Mình đang cần gấp giúp mình với ai làm đúng tớ sẽ tích cho bạn đó, nêu cả cách giải giúp mình nhé!
\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{100}+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow x-\frac{98}{99}=\frac{1}{99}\Leftrightarrow x=1\)
Tìm X
(\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...........+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)) .100 -[\(\frac{5}{2}\):(X+\(\frac{266}{100}\))]:\(\frac{1}{2}\)=89
chấm là nhân các bạn giúp mình nha mình đang cần gấp
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\Rightarrow x+\frac{266}{100}=5\Rightarrow x=\frac{117}{50}\)
Vậy x = 117/50
Ta có:
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right).100\\ =\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100\)
\(=\left(1-\frac{1}{10}\right).100\)
\(=\frac{9}{10}.100\)
= 90
Khi đó đề bài sẽ thành : \(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\)
\(\Rightarrow x+\frac{266}{100}=5\)
\(\Rightarrow x=\frac{117}{50}\)
Vậy \(x=\frac{117}{50}\)