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HH
2
6 tháng 7 2017
\(\frac{3^{17}\cdot8^{11}}{27^{10}\cdot9^{15}}=\frac{3^{17}\cdot8^{11}}{3^{30}\cdot9^{15}}=\frac{8^{11}}{3^{13}\cdot9^{15}}=\frac{8^{11}}{3^{43}}\)
1 tháng 9 2023
a) \(L=4-8+12-16+20-24+...+220-224\)
\(\Rightarrow L=\left(-4\right)+\left(-4\right)+\left(-4\right)+...+\left(-4\right)\) (có 28 số -4)
\(\Rightarrow L=\left(-4\right).28=-112\)
c) \(O=6-12+18-24+30-36+354-360\)
\(\Rightarrow O=\left(-6\right)+\left(-6\right)+\left(-6\right)+...+\left(-6\right)\) (có 30 số -6)
\(\Rightarrow O=\left(-6\right).30=-180\)
e) \(P=3-6+9-12+15-18+...+147-150\)
\(\Rightarrow P=\left(-3\right)+\left(-3\right)+\left(-3\right)+...+\left(-3\right)\) (có 25 số -3)
\(\Rightarrow P=\left(-3\right).25=-75\)
2 tháng 9 2023
b)
S = 3 + 5 - 7 - 9 + 11 + 13 - 15 - 17 + ... + 243 + 245 - 247 - 249
S = (3 - 7) + (5 - 9) + ... + (243 - 247) + (245 - 249)
S = (-4) + (-4) + ... + (-4) + (-4)
Tổng trên có số số hạng là : [(249 - 3) : 2 + 1] : 2 = 62 (số hạng)
Suy ra S = (-4) x 62 = -248
d)
E = 2 - 4 + 6 - 8 + ... + 218 - 220
E = (2 - 4) + (6 - 8) + ... + (218 - 220)
E = (-2) + (-2) + ... + (-2)
Tổng trên có số số hạng là: [(220 - 2) : 2 + 1] : 2 = 55 (số hạng)
Suy ra E = (-2) x 55 = -110
10 tháng 1 2022
\(=\dfrac{2^{19}\cdot3^9-3\cdot3^8\cdot2^{18}\cdot5}{2^{19}\cdot3^9+2^{20}\cdot3^{10}}=\dfrac{-3^{10}\cdot2^{18}}{2^{19}\cdot3^9\cdot7}=-\dfrac{3}{14}\)
T
23 tháng 8 2023
\(B=x+\dfrac{0,2-0,375+\dfrac{5}{11}}{-0,3+\dfrac{9}{16}-\dfrac{15}{22}}\)
\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\left(\dfrac{3}{10}-\dfrac{9}{16}+\dfrac{15}{22}\right)}\)
\(=x+\dfrac{\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}}{-\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{3}{8}+\dfrac{5}{11}\right)}\)
\(=x+\dfrac{1}{-\dfrac{3}{2}}\)
\(=x+\dfrac{-2}{3}\)
Với \(x=-\dfrac{1}{3}\), ta được:
\(B=-\dfrac{1}{3}+\dfrac{-2}{3}=-\dfrac{3}{3}=-1\)
23 tháng 7 2016
\(H=\frac{4}{15}-\frac{23}{28}-\left(-\frac{23}{28}+\frac{-11}{15}-\frac{24}{27}\right)-\frac{2}{27}\)
\(H=\frac{4}{15}+\frac{-23}{28}+\frac{23}{28}+\frac{11}{15}+\frac{24}{27}+\frac{-2}{27}\)
\(H=\left(\frac{4}{15}+\frac{11}{15}\right)+\left(\frac{-23}{28}+\frac{23}{28}\right)+\left(\frac{24}{27}+\frac{-2}{27}\right)\)
\(H=1+0+\frac{22}{27}\)
\(H=1+\frac{22}{27}\)
\(H=\frac{27}{27}+\frac{22}{27}=\frac{49}{27}\)
23 tháng 7 2016
\(H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{24}{27}\right)-\frac{2}{27}\)
\(H=\frac{4}{15}-\frac{23}{28}+\frac{23}{28}+\frac{11}{15}+\frac{24}{27}-\frac{2}{27}\)
\(H=\left(\frac{4}{15}+\frac{11}{15}\right)+\left(\frac{-23}{28}+\frac{23}{28}\right)+\left(\frac{24}{27}-\frac{2}{27}\right)\)
\(H=1+0+\frac{22}{27}\)
\(H=\frac{49}{27}\)
18 tháng 9 2021
Bài 1:
\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)
\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)
Bài 2:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)
19 tháng 10 2021
\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)
\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)