\(\dfrac{3}{17}\)x\(\dfrac{6}{29}\)-\(\d...">
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\(\dfrac{3}{17}\cdot\dfrac{6}{29}-\dfrac{35}{29}+2022\dfrac{3}{17}\)

\(=\dfrac{3}{17}\cdot\dfrac{6}{29}+2022+\dfrac{3}{17}-\dfrac{35}{29}\)

\(=\dfrac{3}{17}\left(\dfrac{6}{29}+1\right)-\dfrac{35}{29}+2022\)

\(=\dfrac{3}{17}\cdot\dfrac{35}{29}-\dfrac{35}{29}+2022\)

\(=\dfrac{35}{29}\left(\dfrac{3}{17}-1\right)+2022\)

\(=\dfrac{35}{29}\cdot\dfrac{-14}{17}+2022\)

\(=\dfrac{-490}{493}+2022=\dfrac{996356}{493}\)

20 tháng 7 2018

a) \(\dfrac{3}{4}+\dfrac{3}{5}-\dfrac{18}{60}\) ( MTC: 60)

= \(\dfrac{3.15}{4.15}+\dfrac{3.12}{5.12}-\dfrac{18}{60}\)

= \(\dfrac{45}{60}+\dfrac{36}{60}-\dfrac{18}{60}\)

= \(\dfrac{45+36-18}{60}\)=\(\dfrac{63}{60}=\dfrac{21}{20}\)

b)\(\dfrac{17}{8}-\dfrac{11}{6}-\dfrac{2}{9}\) (MTC:72)

=\(\dfrac{17.9}{8.9}-\dfrac{11.12}{6.12}-\dfrac{2.8}{9.8}\)

= \(\dfrac{153}{72}-\dfrac{132}{72}-\dfrac{16}{72}\)

=\(\dfrac{153-132-16}{72}\)

=\(\dfrac{5}{72}\)

c)\(\dfrac{23}{29}+\dfrac{5}{11}+\dfrac{17}{11}\) (MTC:319)

= \(\dfrac{23.11}{29.11}+\dfrac{5.29}{11.29}+\dfrac{17.29}{11.29}\)

=\(\dfrac{253}{319}+\dfrac{145}{319}+\dfrac{493}{319}\)

=\(\dfrac{253+145+493}{319}\)=\(\dfrac{891}{319}=\dfrac{81}{29}\)

20 tháng 7 2018

c) \(\dfrac{20}{45}+\dfrac{14}{35}+\dfrac{32}{44}\)

= \(\dfrac{4}{9}+\dfrac{2}{5}+\dfrac{8}{11}\)(Rút gọn b/thức)(MTC:495)

=\(\dfrac{4.55}{9.55}+\dfrac{2.99}{5.99}+\dfrac{8.45}{11.45}\)

=\(\dfrac{220}{495}+\dfrac{198}{495}+\dfrac{360}{495}\)

=\(\dfrac{220+198+360}{495}\)=\(\dfrac{778}{495}\)

e)\(17\dfrac{25}{27}+3\dfrac{7}{2}\)

= \(\dfrac{484}{27}+\dfrac{13}{2}\) (MTC:54)

=\(\dfrac{484.2}{27.2}+\dfrac{13.27}{2.27}\)

\(=\dfrac{968}{54}+\dfrac{351}{54}\)

=\(\dfrac{968+351}{54}=\dfrac{1319}{54}\)

27 tháng 3 2017

A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)

A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)

A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)

A=\(\dfrac{7}{24}\)

B=\(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)

B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)

B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)

B=\(1+\left(-1\right)+\left(-1\right)=-1\)

C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)

C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)

C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)

D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)

D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)

3 tháng 8 2017

a,

\(\dfrac{13}{17}=1-\dfrac{4}{17}\\ \dfrac{25}{29}=1-\dfrac{4}{29}\\ \dfrac{4}{17}>\dfrac{4}{29}\Rightarrow1-\dfrac{4}{17}< 1-\dfrac{4}{29}\Leftrightarrow\dfrac{13}{17}< \dfrac{25}{29}\)

Vậy \(\dfrac{13}{17}< \dfrac{25}{29}\)

b,

\(\dfrac{59}{101}>\dfrac{56}{101}>\dfrac{56}{105}\\ \Rightarrow\dfrac{59}{101}>\dfrac{56}{105}\)

Vậy \(\dfrac{59}{101}>\dfrac{56}{105}\)

c,

\(\dfrac{14}{55}>\dfrac{14}{56}=\dfrac{1}{4}=\dfrac{20}{80}>\dfrac{20}{83}\)

Vậy \(\dfrac{14}{55}>\dfrac{20}{83}\)

3 tháng 8 2017

d,

\(\dfrac{13}{57}< \dfrac{13}{39}=\dfrac{1}{3}=\dfrac{29}{87}< \dfrac{29}{73}\)

Vậy \(\dfrac{13}{57}< \dfrac{29}{73}\)

e,

\(\dfrac{17}{21}=\dfrac{17\cdot101}{21\cdot101}=\dfrac{1717}{2121}\)

Vậy \(\dfrac{17}{21}=\dfrac{1717}{2121}\)

3 tháng 8 2017

3. a) Ta có : 13.29 = 377
25.17 = 425
=> \(\dfrac{13}{17}< \dfrac{25}{29}\)
b) Ta có : 59.105 > 56.101
=> \(\dfrac{59}{101}>\dfrac{56}{105}\)
c) Ta có : 14.83 = 1162
20.55 = 1100
=> \(\dfrac{14}{55}>\dfrac{20}{83}\)
d) Ta có : 13.73 = 949
29.57 = 1653
=> \(\dfrac{13}{57}< \dfrac{29}{73}\)
e) Ta có : \(\dfrac{1717}{2121}=\dfrac{17}{21}\)
=> \(\dfrac{17}{21}=\dfrac{1717}{2121}\)
@Đặng Vũ Hoài Anh

3 tháng 8 2017

4. Gọi các phân số cần tìm có dạng \(\dfrac{x}{3}\)
Ta có : \(\dfrac{-1}{2}< \dfrac{x}{3}< \dfrac{1}{2}\)

=> \(\dfrac{-3}{6}< \dfrac{2x}{6}< \dfrac{3}{6}\)

=> -3 < 2x < 3
=> 2x = -2; 0; 2
=> x = -1; 0; 1 (thỏa mãn)
@Đặng Vũ Hoài Anh

16 tháng 5 2017

5\(\dfrac{8}{17}\):x + (-\(\dfrac{1}{17}\)) : x + 3\(\dfrac{1}{17}\) : 17\(\dfrac{1}{3}\)= \(\dfrac{4}{17}\)

\(\dfrac{93}{17}\).\(\dfrac{1}{x}\) + (-\(\dfrac{1}{17}\)) .\(\dfrac{1}{x}\) +\(\dfrac{3}{17}\)= \(\dfrac{4}{17}\)

\(\dfrac{1}{x}\).\(\dfrac{92}{17}\)=\(\dfrac{1}{17}\)

\(\dfrac{1}{x}\)=\(\dfrac{1}{17}\):\(\dfrac{92}{17}\)
x= 92
16 tháng 5 2017

\(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+...+\(\dfrac{1}{x.\left(x+3\right)}\)=\(\dfrac{6}{19}\)

3(\(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+...+\(\dfrac{1}{x.\left(x+3\right)}\))=3.\(\dfrac{6}{19}\)
\(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{18}{19}\)
1-\(\dfrac{1}{x+3}\)=\(\dfrac{18}{19}\)
\(\dfrac{1}{x+3}\)=\(\dfrac{1}{19}\)
x+3 =19
x=19-3
x=17
23 tháng 7 2017

Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OKleuleu

23 tháng 7 2017

Mình khuyen bạn phải suy nghĩ kĩ bài trước khi đăng lên nhé!!hum

9 tháng 4 2018

a,\(\dfrac{3}{7}\).\(\dfrac{14}{5}\)=\(\dfrac{6}{5}\)

b,\(\dfrac{35}{9}\).\(\dfrac{81}{7}\)=45

c,\(\dfrac{28}{17}\).\(\dfrac{68}{14}\)=8

d,\(\dfrac{35}{46}\).\(\dfrac{23}{105}\)=\(\dfrac{1}{6}\)

e,\(\dfrac{12}{5}\):\(\dfrac{16}{15}\)=\(\dfrac{12}{5}\).\(\dfrac{15}{16}\)=\(\dfrac{9}{4}\)

i,\(\dfrac{9}{8}\):\(\dfrac{6}{5}\)=\(\dfrac{9}{8}\).\(\dfrac{5}{6}\)=\(\dfrac{15}{16}\)

5 tháng 8 2018

2)

S = \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{43.46}\)

S = 3 . (\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{43.46}\))

S = 1 . (\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{43.46}\))

S = 1 . (\(1-\dfrac{1}{4}+...+\dfrac{1}{43}-\dfrac{1}{46}\))

S = 1 . (\(1-\dfrac{1}{46}\))

S = 1 . \(\dfrac{45}{46}\)

S = \(\dfrac{45}{46}\)

=> \(\dfrac{45}{46}\) < 1

4 tháng 8 2018

bài 2:tính hợp lý

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: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)