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\(-\frac{3}{6}=\frac{x}{-2}=-\frac{18}{y}=-\frac{z}{24}\)
Ta có :+) \(-\frac{3}{6}=\frac{x}{-2}\)
\(\Rightarrow x=\frac{\left(-3\right)\left(-2\right)}{6}\)
\(\Rightarrow x=1\)
+)\(-\frac{3}{6}=-\frac{18}{y}\)
\(\Rightarrow y=\frac{6.\left(-18\right)}{-3}\)
\(\Rightarrow y=36\)
+)\(-\frac{3}{6}=-\frac{z}{24}\)
\(\Rightarrow-z=\frac{\left(-3\right)24}{6}\)
\(\Rightarrow-z=-12\)
\(\Rightarrow z=12\)
Vậy........................
\(a)\frac{x}{8}=\frac{-30}{y}=\frac{-48}{32}\)
Rút gọn : \(\frac{-48}{32}=\frac{(-48):16}{32:16}=\frac{-3}{2}\)
* Ta có : \(\frac{x}{8}=\frac{-3}{2}\)
\(\Rightarrow x\cdot2=-3\cdot8\)
\(\Rightarrow x=\frac{-3\cdot8}{2}=-12\)
* Ta có : \(\frac{-30}{y}=\frac{-3}{2}\)
\(\Rightarrow-30\cdot2=-3\cdot y\)
\(\Rightarrow y=\frac{-30\cdot2}{-3}=20\)
Mấy bài kia làm tương tự
a) \(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{302\cdot305}\)
\(B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{302\cdot305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{302}-\frac{1}{305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{305}\right)=\frac{1}{3}\cdot\frac{303}{610}=\frac{101}{610}\)
b) \(C=\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+....+\frac{6}{202\cdot205}\)
\(C=2\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{202\cdot205}\right)=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\right)\)
\(=2\left(1-\frac{1}{205}\right)=2\cdot\frac{204}{205}=\frac{408}{205}\)
c) \(D=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{266\cdot271}\)
\(D=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{266\cdot271}\right)\)
\(D=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\right)=5\left(1-\frac{1}{271}\right)=5\cdot\frac{270}{271}=\frac{1350}{271}\)
d) \(E=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{5}{16}\cdot...\cdot\frac{9999}{10000}=\frac{3\cdot8\cdot15\cdot...\cdot9999}{4\cdot9\cdot16\cdot...\cdot10000}=\frac{3}{10000}\)
e) \(F=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
\(F=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{2500}\right)\)
\(F=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}=\frac{3\cdot8\cdot15\cdot...\cdot2499}{4\cdot9\cdot16\cdot...\cdot2500}=\frac{3}{2500}\)
a. \(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{302.305}\)
\(\Rightarrow3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{302.305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{302}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{303}{610}\)
\(\Rightarrow B=\frac{101}{610}\)
b. \(C=\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{202.205}\)
\(\Rightarrow\frac{1}{2}C=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{202.205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=\frac{204}{205}\)
\(\Rightarrow C=\frac{408}{205}\)
c. \(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{266.271}\)
\(\Rightarrow\frac{1}{5}D=\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{266.271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=\frac{270}{271}\)
\(\Rightarrow D=\frac{1350}{271}\)
\(\frac{30}{25}=\frac{6}{5}\)
\(\frac{70}{49}=\frac{10}{7}\)
\(\frac{-54}{36}=\frac{-3}{2}\)
\(\frac{30}{25}=\frac{30:5}{25:5}=\frac{6}{5}\)
\(\frac{70}{49}=\frac{70:7}{49:7}=\frac{10}{7}\)
\(-\frac{54}{36}=\frac{54:18}{36:18}=\frac{3}{2}\)
Giải :
\(\text{S}=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(\text{S}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)
\(\text{S}=1-\frac{1}{1000}=\frac{999}{1000}\)
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{998.999.1000}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{998.999}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{999000}\right)\)
\(=\frac{1}{2}.\frac{499499}{999000}\)
\(=\frac{499499}{1998000}\)
Study well ! >_<
Bài làm
1.
a) \(\frac{5.7}{14.3}=\frac{5.7}{2.7.3}=\frac{5}{6}\)
b) \(\frac{3.55}{22.9}=\frac{3.5.11}{2.11.3.3}=\frac{5}{6}\)
c) \(\frac{10.63}{15.21}=\frac{2.5.7.3.3}{3.5.7.3}=2\)
d) \(\frac{3.11.13}{26.15}=\frac{3.11.13}{2.13.3.5}=\frac{11}{10}\)
2.
a) \(\frac{7.29+7.14}{43.5-43.11}=\frac{7\left(29+14\right)}{43\left(5-11\right)}=\frac{7.43}{43.-6}=-\frac{7}{6}\)
b) \(\frac{19.15-7.14}{38}=\frac{285-98}{38}=\frac{187}{38}\)
c) \(\frac{53.19-19.2}{7-60}=\frac{19\left(53-2\right)}{7-60}=\frac{19.51}{-53}=-\frac{969}{53}\)
d) \(\frac{120}{24.13-24.2}=\frac{120}{24\left(13-2\right)}=\frac{24.5}{24.11}=\frac{5}{11}\)