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E=1/1.3+1/3.6+1/6.9+.............+1/20.23
<=>E=1/1-1/3+1/3-1/6+1/6-1/9+...........+1/20-1/23
<=>E=1/1-1/23
<=>E=22/23
Kb và k mk nha mn.
Cho hai phan so 1/n va 1/n+1 (n thuoc z)chung to rang h cua hai phan so nay bang hieu cua chung
3.25.8+4.37.6+2.38.12
=24.25+24.37+24.38
=24.(25+37+38)
=24.100
=2400
chúc bạn học tốt :)!
3.25.8+4.37.6+2.38.12
=24.25+24.37+24.38
=24.(25+37+38)
=24.100=2400
mik nha
Ta có : \(\frac{-32}{27}-\left(3x-\frac{7}{9}\right)^3=\) \(\frac{-24}{27}\)
<=> \(-\left(3x-\frac{7}{9}\right)^3=\left(\frac{2}{3}\right)^3\)
<=> \(-\left(3x-\frac{7}{9}\right)=\left(\frac{2}{3}\right)\)
<=> x= \(\frac{1}{27}\) -------chỗ này tắt
A=212x35-46x92:(22x3)6+84x35
=212x35-212x34:212x36+212x35
=212(35-34+35):212x36=212x405:(212x36)=1x\(\dfrac{5}{9}\)=\(\dfrac{5}{9}\).
\(\frac{4}{3.6}+\frac{4}{6.9}+...+\frac{4}{12.15}\)
\(=\frac{4\left(\frac{3}{3.6}+\frac{3}{6.9}+...+\frac{3}{12.15}\right)}{3}\)
\(=\frac{4\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{12}-\frac{1}{15}\right)}{3}\)
\(=\frac{4\left(\frac{1}{3}-\frac{1}{15}\right)}{3}\)
\(=\frac{\frac{16}{15}}{3}=\frac{48}{15}\)
\(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6}=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6}=\frac{2^{12}\left(3^5-3^4\right)}{2^{12}.3^6}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^6}=\frac{2^{12}.3^4.2}{2^{12}.3^6}=\frac{2^{13}.3^4}{2^{12}.3^6}=\frac{2}{3^2}=\frac{2}{9}\)
Ta có:\(\frac{4}{9}=\frac{4\cdot6}{9\cdot6}=\frac{24}{54}\) ; \(\frac{6+9}{6\cdot9}=\frac{15}{54}\) ; \(\frac{2}{3}=\frac{2\cdot18}{3\cdot18}=\frac{36}{54}\)
=>\(\frac{15}{54}<\frac{24}{54}<\frac{36}{54}\)
=>\(\frac{6+9}{6\cdot9}<\frac{4}{9}<\frac{2}{3}\)
6+6.9+6.92+6.93+..............+6.999
\(=6.\left(1+9+9^2+9^3+...........+9^{99}\right)\)
Đặt A=1+9+92+93+........+999
\(\Rightarrow9A=9+9^2+9^3+9^4+.....+9^{100}\)
\(\Rightarrow9A-A=\left(9+9^2+9^3+9^4+.....+9^{100}\right)-\left(1+9+9^2+9^3+............+9^{99}\right)\)
\(\Rightarrow8A=9^{100}-1\)
\(\Rightarrow A=\frac{9^{100}-1}{8}\)
+)Thay A vào \(6.\left(1+9+9^2+9^3+...........+9^{99}\right)\)được:
\(=6.\frac{9^{100}-1}{8}=\frac{54^{100}-6}{8}\)
Chúc bn học tốt