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Ta có : \(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
\(\Leftrightarrow2\cdot\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}=\dfrac{x}{98}\)
\(\Leftrightarrow\dfrac{1}{3}-\dfrac{1}{21}=\dfrac{x}{98}\)
\(\Leftrightarrow\dfrac{2}{7}=\dfrac{x}{98}\Rightarrow x=28\)
Vậy $x=28$
1) \(x:\dfrac{2}{3}=150\)
\(\Leftrightarrow x=150.\dfrac{2}{3}\)
\(\Leftrightarrow x=100\).
2) \(\dfrac{35}{9}:x=\dfrac{35}{6}\)
\(\Leftrightarrow x=\dfrac{35}{9}:\dfrac{35}{6}\)
\(\Leftrightarrow x=\dfrac{2}{3}\).
3) \(\dfrac{49}{7}:x=\dfrac{49}{5}\)
\(\Leftrightarrow x=\dfrac{49}{7}:\dfrac{49}{5}\)
\(\Leftrightarrow x=\dfrac{5}{7}\).
4) \(1-\left\{5\dfrac{4}{9}+x-7\dfrac{7}{18}\right\}:15\dfrac{3}{5}=0\)
\(\Leftrightarrow1-\left\{\dfrac{49}{9}+x-\dfrac{133}{18}\right\}:\dfrac{78}{5}=0\)
\(\Leftrightarrow\left\{\dfrac{-35}{18}+x\right\}:\dfrac{78}{5}=1-0\)
\(\Rightarrow\dfrac{-35}{18}+x=1.\dfrac{78}{5}\)
\(\Leftrightarrow\dfrac{-35}{18}+x=\dfrac{78}{5}\)
\(\Rightarrow x=\dfrac{1579}{90}\).
Gọi a,b,c.. cho dễ nhé.Thớt vui tính quá, dấu phẩy cũng không viết hộ con dân =)))
a, \(x:\dfrac{2}{3}=150\)
\(\Leftrightarrow x=150.\dfrac{2}{3}\)
\(\Leftrightarrow x=100\)
Vậy...
b, \(\dfrac{35}{9}:x=\dfrac{35}{6}\)
\(\Leftrightarrow x=\dfrac{35}{9}:\dfrac{35}{6}\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy...
c, \(\dfrac{49}{7}:x=\dfrac{49}{5}\)
\(\Leftrightarrow x=\dfrac{49}{7}:\dfrac{49}{5}\)
\(\Leftrightarrow x=\dfrac{5}{7}\) Vậy...
d, \(1-\left\{5\dfrac{4}{9}+x-7\dfrac{7}{18}\right\}:15\dfrac{3}{4}=0\)
\(\Leftrightarrow1-\left\{\dfrac{49}{9}+x-\dfrac{133}{18}\right\}:\dfrac{63}{4}=0\)
\(\Leftrightarrow1-\left\{x-\dfrac{35}{18}\right\}:\dfrac{63}{4}=0\)
\(\Leftrightarrow1-\left(\dfrac{\left(18x-35\right).4}{18.63}\right)=0\)
\(\Leftrightarrow1-\left(\dfrac{72x-140}{1134}\right)=0\)
\(\Leftrightarrow1-\dfrac{72x-140}{1134}=0\)
\(\Leftrightarrow\dfrac{1134-72x+140}{1134}=0\)
\(\Leftrightarrow1274-72x=0\)
\(\Leftrightarrow72x=1274\)
\(\Leftrightarrow x=\dfrac{637}{36}\)
Vậy...
a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)
b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)
c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)
d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)
e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)
f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)
13)\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}\right)+\dfrac{45}{8}\left(\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}+\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left[\left(\dfrac{4}{15}+\dfrac{11}{15}\right)-\left(\dfrac{7}{8}-\dfrac{9}{8}\right)\right]\)
=\(\dfrac{45}{8}.\dfrac{5}{4}\)=\(\dfrac{225}{32}\)
14)\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right):\dfrac{7}{15}\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right).\dfrac{15}{7}\)
=\(\dfrac{15}{7}\left[\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right)\right]\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}-\dfrac{2}{5}+\dfrac{27}{4}\right)\)
=\(\dfrac{15}{7}.\dfrac{35}{8}\)=\(\dfrac{75}{8}\)
\(A=\dfrac{14}{8.11}+\dfrac{14}{11.14}+\dfrac{14}{14.17}+.....+\dfrac{14}{197.200}\)
\(A=\dfrac{14}{3}\left(\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{197}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\left(\dfrac{1}{8}-\dfrac{1}{200}\right)\)
\(A=\dfrac{14}{3}.\dfrac{24}{200}=\dfrac{28}{25}\)
\(B=\dfrac{7}{15}+\dfrac{7}{35}+\dfrac{7}{63}+...+\dfrac{7}{399}\)
\(B=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+.....\dfrac{7}{19.21}\)
\(B=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{19}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)\)
\(B=\dfrac{7}{2}.\dfrac{6}{21}=1\)
a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)
\(\Rightarrow-4< x< \dfrac{-3}{10}\)
\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)
\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)
b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)
\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)
\(\Rightarrow x=\varnothing\)
c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)
\(\Rightarrow x\in\left\{1;2\right\}\)
+) Với \(x=1\)
\(\Rightarrow y\in\left\{1;2\right\}\)
+) Với \(x=2\)
\(\Rightarrow y=2\)
Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).
a: \(=\dfrac{-3}{7}+\dfrac{-9}{35}-\dfrac{2}{5}\)
\(=\dfrac{-15-9-14}{35}=\dfrac{-38}{35}\)
b: \(=\left(\dfrac{15}{24}-\dfrac{7}{12}\right)\cdot\dfrac{-12}{7}\)
\(=\dfrac{15-14}{24}\cdot\dfrac{-12}{7}=\dfrac{1}{24}\cdot\dfrac{-12}{7}=\dfrac{-1}{14}\)
c: \(=\dfrac{7}{5}\cdot\dfrac{15}{19}\cdot\dfrac{-8}{15}+\dfrac{7}{15}\)
\(=\dfrac{-56}{95}+\dfrac{7}{15}\)
\(=\dfrac{-7}{57}\)
Giải:
\(\dfrac{4}{15}+\dfrac{4}{35}+\dfrac{4}{63}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
\(\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{19.21}=\dfrac{x}{49}\)
\(2.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{19.21}\right)=\dfrac{x}{49}\)
\(2.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)=\dfrac{x}{49}\)
\(2.\left(\dfrac{1}{3}-\dfrac{1}{21}\right)=\dfrac{x}{49}\)
\(2.\dfrac{2}{7}=\dfrac{x}{49}\)
\(\dfrac{4}{7}=\dfrac{x}{49}\)
\(\Rightarrow x=\dfrac{4.49}{7}=28\)
Chúc bạn học tốt!
\(\dfrac{4}{15}+\dfrac{4}{35}+...+\dfrac{4}{399}=\dfrac{x}{49}\)
2 . \(\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{399}=\dfrac{x}{49}\)
2 . \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{19.21}=\dfrac{x}{49}\)
2 . ( \(\dfrac{1}{3}-\dfrac{1}{21}\) ) = \(\dfrac{x}{49}\)
2 . \(\dfrac{2}{7}\) = \(\dfrac{x}{49}\)
=> \(\dfrac{4}{7}=\dfrac{x}{49}\)
=> \(\dfrac{21}{49}=\dfrac{x}{49}\)
=> \(x=21\)
Vậy \(x=21\)