C= 2-\(\dfrac{5}{3}\)+\(\dfrac{7}{6}\)-\(\dfrac{9}{10}\)+\(\dfrac{11}{15}\)-\(\dfrac{13}{21}\)+\(\dfrac{15}{28}\)-\(\dfrac{17}{36}\)+\(\dfrac{19}{45}\)
tính C
tính
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
b)\(\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}+\dfrac{29}{42}:\dfrac{1}{28}-8\)
c)\(-1\dfrac{5}{7}.15+\dfrac{2}{7}\left(-15\right)+\left(-105\right).\left(\dfrac{2}{3}-\dfrac{4}{5}+\dfrac{1}{7}\right)\)
a)\(\dfrac{-10}{11}.\dfrac{8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}.\dfrac{-8}{9}+\dfrac{7}{18}.\dfrac{10}{11}\)
=\(\dfrac{10}{11}(\dfrac{-8}{9}+\dfrac{7}{18})\)
=\(\dfrac{10}{11}.\dfrac{-1}{2}\)
=\(\dfrac{-5}{11}\)
b;
B = \(\dfrac{3}{14}\) : \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\) : \(\dfrac{1}{28}\) - 8
B = (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{9}{42}\) - \(\dfrac{26}{42}\) + \(\dfrac{29}{42}\)) - 8
B = (\(\dfrac{-17}{42}\) + \(\dfrac{29}{42}\)) - 8
B = \(\dfrac{2}{7}\) - 8
B = \(\dfrac{2}{7}-\dfrac{56}{7}\)
B = - \(\dfrac{54}{7}\)
c; C = -1\(\dfrac{5}{7}\).15 + \(\dfrac{2}{7}\)(-15) + (-105).(\(\dfrac{2}{3}\) - \(\dfrac{4}{5}\) + \(\dfrac{1}{7}\))
C = - 15.(- 1 - \(\dfrac{5}{7}\) + \(\dfrac{2}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\) + \(1\))
C = -15.[(1 - 1) - (\(\dfrac{5}{7}\) - \(\dfrac{2}{7}\)) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15.[0 - \(\dfrac{3}{7}\) + \(\dfrac{14}{3}\) - \(\dfrac{28}{5}\)]
C = -15 . [- \(\dfrac{45}{105}\) + \(\dfrac{490}{105}\) - \(\dfrac{588}{105}\)]
C = -15. [ \(\dfrac{445}{105}\) - \(\dfrac{588}{105}\)]
C = - 15.(- \(\dfrac{143}{105}\))
C = \(\dfrac{143}{7}\)
Cho \(A=\dfrac{13}{25}+\dfrac{9}{10}-\dfrac{11}{15}+\dfrac{13}{21}-\dfrac{15}{28}+\dfrac{17}{36}-...+\dfrac{197}{4851}-\dfrac{199}{4950}\)
Chứng minh \(A>\dfrac{9}{10}\)
>; <; = ?
a) \(\dfrac{21}{23}\) ? \(\dfrac{19}{23}\) \(\dfrac{8}{5}\) ? \(\dfrac{49}{30}\) \(\dfrac{20}{36}\) ? \(\dfrac{5}{9}\)
b) \(\dfrac{11}{15}\) ? \(\dfrac{11}{17}\) \(\dfrac{26}{13}\) ? 2 3 ? \(\dfrac{16}{5}\)
c) \(\dfrac{8}{9}\) ? 1 1 ? \(\dfrac{31}{27}\) \(\dfrac{8}{9}\) ? \(\dfrac{31}{27}\)
a) \(\dfrac{21}{23}>\dfrac{19}{23}\)
\(\dfrac{8}{5}=\dfrac{49}{30}\)
\(\dfrac{23}{36}>\dfrac{5}{9}\)
b) \(\dfrac{11}{15}>\dfrac{11}{17}\)
\(\dfrac{26}{13}=2\)
\(3< \dfrac{16}{5}\)
c) \(\dfrac{8}{9}< 1\)
\(1< \dfrac{31}{27}\)
\(\dfrac{8}{9}< \dfrac{31}{27}\)
A = \(\dfrac{28}{25}\) + \(\dfrac{9}{10}\) - \(\dfrac{11}{15}\) + \(\dfrac{13}{21}\) - \(\dfrac{15}{28}\) + \(\dfrac{17}{26}\) - ... + \(\dfrac{197}{4851}\) - \(\dfrac{199}{4950}\)
\(\dfrac{2}{3}+\dfrac{5}{6}+\dfrac{9}{10}+\dfrac{14}{15}+\dfrac{20}{21}+\dfrac{27}{28}+\dfrac{35}{36}+\dfrac{44}{45}\) kết quả bằng bao nhiêu a
\(\dfrac{2}{3}+\dfrac{5}{6}+\dfrac{9}{10}+\dfrac{14}{15}+\dfrac{20}{21}+\dfrac{27}{28}+\dfrac{35}{36}+\dfrac{44}{45}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{10}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{21}\right)+\left(1-\dfrac{1}{28}\right)+\left(1-\dfrac{1}{36}\right)+\left(1-\dfrac{1}{45}\right)\\ =8-\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\right)\\ =8-\left(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ =8-\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+\dfrac{2}{9.10}\right)\\ =8-2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=8-2.\dfrac{2}{5}=8-\dfrac{4}{5}=\dfrac{36}{5}\)
Tính hợp lý: M=\(\dfrac{2}{3}+\dfrac{9}{10}-\dfrac{11}{15}_{ }+\dfrac{13}{21}-\dfrac{15}{28}+\dfrac{17}{36}-\dfrac{19}{45}+\dfrac{21}{55}-\dfrac{23}{66}\)
Giúp em nha!!!Bài 1:
a,\(|x-3|+|2-x|=0\)
b,\(\left(2-\dfrac{3}{4}x\right).\left(x+1\right)=0\)
bài 2:
a,A=\(\dfrac{\dfrac{-6}{7}+\dfrac{6}{13}-\dfrac{6}{29}}{\dfrac{9}{7}-\dfrac{9}{13}+\dfrac{9}{29}}\)
b,B=\(\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{0,25-\dfrac{5}{28}+\dfrac{5}{52}}\)
c,C=\(\dfrac{50-\dfrac{4}{15}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{\dfrac{65}{121}-\dfrac{26}{71}-13}\)
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a: \(=\dfrac{-6\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}{9\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}=\dfrac{-6}{9}=\dfrac{-2}{3}\)
b: \(=\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{\dfrac{10}{40}-\dfrac{10}{56}+\dfrac{10}{104}}\)
\(=\dfrac{\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}{\dfrac{10}{8}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{2}{3}:\dfrac{5}{4}=\dfrac{2}{3}\cdot\dfrac{4}{5}=\dfrac{8}{15}\)
c: \(=\dfrac{2\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}{4\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{13\left(\dfrac{5}{121}-\dfrac{2}{21}-1\right)}\)
=2/4:(-1)/13=2/4x(-13)=-13/2