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Sao bn học nhanh thế mới có mấy tuần học mà đã học toán chứng minh rồi thế!
Cố lên nha bn !
Chúc bn có câu trả lời thoả đáng để giải bài toán này nha!
Chúc bn mãi mãi học giỏi!
Uk, mk cảm ơn bn đã chúc mk học giỏi.
Nhưng mk đã học tới toán chứng minh vì mk học nâng cao, giỏi hơn bn nhiều! OK?
Trả lời thì trả lời đừng ở đó mà ns nhiều!
Tính nhanh :
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{2}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\left(\frac{20}{20}-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}\)
\(=\frac{19}{10}\)
= \(\left(1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}-\frac{1}{3}+...+\frac{1}{38}-\frac{1}{38}+\frac{1}{39}\right)\)
= 1 + \(1+\frac{1}{39}=\frac{40}{39}\)
chỗ " 1 + " phía trước là bỏ
ngay chỗ dấu bằng thứ hai
c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
1) \(\frac{2}{5}\cdot\frac{1}{2}+\frac{1}{3}=\frac{1}{5}+\frac{1}{3}=\frac{8}{15}\)
2)\(\left(\frac{5}{7}-\frac{2}{5}\right):\frac{11}{7}=\frac{5}{7}:\frac{11}{7}-\frac{2}{5}:\frac{11}{7}=\frac{5}{7}\cdot\frac{7}{11}-\frac{2}{5}\cdot\frac{7}{11}=\frac{5}{11}-\frac{14}{55}=\frac{1}{5}\)
3)\(\frac{1-\frac{2}{3}+\frac{1}{4}}{2-\frac{1}{3}+\frac{1}{6}}=\frac{12-8+3}{12}:\frac{12-2+1}{6}=\frac{7}{12}\cdot\frac{\frac{6}{11}7}{22}\)
*)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
=\(1-\frac{1}{6}\)
=\(\frac{6}{6}-\frac{1}{6}\)
\(=\frac{5}{6}\)
*)\(\frac{2003}{1.2}+\frac{2003}{2.3}+\frac{2003}{3.4}+...+\frac{2003}{2002.2003}\)
\(=\frac{2003}{1}-\frac{2003}{2}+\frac{2003}{2}-\frac{2003}{3}+\frac{2003}{3}-\frac{2003}{4}+...+\frac{2003}{2002}-\frac{2003}{2003}\)
\(=2003-1\)
\(=2002\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2}{5}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{5}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2}{5}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2}{5}\)
\(\frac{1}{x+1}=\frac{1}{10}\)
\(\Rightarrow x+1=10\)
\(\text{Vậy x = 9}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{2}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2}{5}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2}{5}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{10}\)
\(\Rightarrow x+1=10\)
\(\Rightarrow x=10-1\)
\(\Rightarrow x=9\)
Vẫy = 9
\(19\frac{1}{4}+\frac{1}{2}.2\frac{1}{3}+5,75-\frac{1}{6}+74\)
=\(\frac{77}{4}+\frac{1}{2}.\frac{7}{3}+\frac{575}{100}-\frac{1}{6}+74\)
= \(\frac{77}{4}+\frac{7}{6}+\frac{23}{4}-\frac{1}{6}+74\)
= \(\left(\frac{77}{4}+\frac{23}{4}\right)+\left(\frac{7}{6}-\frac{1}{6}\right)+74\)
= \(\frac{100}{4}+1+74\)
= 100
^^
Đáp án là 100