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Sửa đề: 2a-3b+4c=330

Ta có: \(\dfrac{a}{10}=\dfrac{b}{5}\)

\(\Leftrightarrow\dfrac{a}{20}=\dfrac{b}{10}\)(1)

Ta có: \(\dfrac{b}{2}=\dfrac{c}{5}\)

\(\Leftrightarrow\dfrac{b}{10}=\dfrac{c}{25}\)(2)

Từ (1) và (2) suy ra \(\dfrac{a}{20}=\dfrac{b}{10}=\dfrac{c}{25}\)

\(\Leftrightarrow\dfrac{2a}{40}=\dfrac{3b}{30}=\dfrac{4c}{100}\)

mà \(2a-3b+4c=330\)

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: 

\(\dfrac{2a}{40}=\dfrac{3b}{30}=\dfrac{4c}{100}=\dfrac{2a-3b+4c}{40-30+100}=\dfrac{330}{110}=3\)

Do đó: 

\(\left\{{}\begin{matrix}\dfrac{2a}{40}=3\\\dfrac{3b}{30}=3\\\dfrac{4c}{100}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a=120\\3b=90\\4c=300\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=60\\b=30\\c=75\end{matrix}\right.\)

Vậy: (a,b,c)=(60;30;75)

15 tháng 8 2023

\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{2}\)

\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)

Vậy \(B=3\)

8 tháng 5 2022

bài 1

a)\(=\dfrac{16}{40}+\dfrac{15}{40}=\dfrac{31}{40}\)

b)\(=\dfrac{7}{6}-\dfrac{4}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

c)\(=\dfrac{30}{9}=\dfrac{10}{3}\)

d)\(=\dfrac{8}{5}\times\dfrac{7}{4}=\dfrac{56}{20}=\dfrac{14}{5}\)

8 tháng 5 2022

bài 2

a)\(x=\dfrac{5}{6}-\dfrac{4}{5}=\dfrac{25}{30}-\dfrac{24}{30}=\dfrac{1}{30}\)

b)\(x=5\times\dfrac{10}{7}=\dfrac{50}{7}\)

bài 4 :

145 ×× 69 + 22 x  145 +145 x 8 + 145 

\(=145\times\left(69+22+8+1\right)=145\times100=14500\)

 

\(\Leftrightarrow\dfrac{a}{b}\left(\dfrac{7}{15}+\dfrac{8}{15}+10-1\right)=\dfrac{5}{7}\)

=>a/b=5/70=1/14

23 tháng 5 2022

1/14

20 tháng 9 2023

a, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)

              \(\dfrac{2}{5}\)\(x\)            = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)

              \(\dfrac{2}{5}\) \(x\)           = - \(\dfrac{3}{20}\)

                  \(x\)          =  - \(\dfrac{3}{20}\)\(\dfrac{2}{5}\)

                  \(x\)          = - \(\dfrac{3}{8}\)

20 tháng 9 2023

b, \(\dfrac{1}{3}\) +   \(\dfrac{1}{2}\)\(x\) = - \(\dfrac{1}{5}\)

            \(\dfrac{1}{2}\)\(x\) =  - \(\dfrac{1}{5}\) - \(\dfrac{1}{3}\)

             \(\dfrac{1}{2}\)\(x\) = - \(\dfrac{8}{15}\)

                   \(x\) =  \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))

                    \(x\) =  - \(\dfrac{15}{16}\)

30 tháng 9 2024

bài1  

a) \(\dfrac{7}{6}-\dfrac{13}{12}+\dfrac{3}{4}\) 

=\(\dfrac{14}{12}-\dfrac{13}{12}+\dfrac{9}{12}\) 

=\(\dfrac{1}{12}+\dfrac{9}{12}\) 

=\(\dfrac{10}{12}=\dfrac{5}{6}\)

30 tháng 9 2024

bài 1 

b)\(1\dfrac{1}{2}.(\dfrac{-4}{5})\) + \(\dfrac{3}{10}\) 

\(\dfrac{3}{2}.\left(-\dfrac{4}{5}\right)+\dfrac{3}{10}\) 

\(-\dfrac{6}{5}+\dfrac{3}{10}\) 

=\(-\dfrac{12}{10}+\dfrac{3}{10}\) 

=\(-\dfrac{9}{10}\) 

\(A=\left(-\dfrac{43}{51}\right)\left(-\dfrac{19}{80}\right)\)

=>A>0(1)

\(B=\left(-\dfrac{7}{13}\right)\left(-\dfrac{4}{65}\right)\left(-\dfrac{8}{21}\right)\)

=>B<0(2)

C\(=-\dfrac{5}{10}.\left(-\dfrac{4}{10}\right).....\left(\dfrac{4}{10}\right)\left(\dfrac{5}{10}\right)=0\)

=>C=0(3)

Từ 1;2;3 =>A>C>B

3 tháng 9 2017

\(A=\dfrac{-43}{51}.\dfrac{-19}{80}\Leftrightarrow A>0\left(1\right)\)

\(B=\left(\dfrac{-7}{13}\right).\left(-\dfrac{4}{65}\right).\left(\dfrac{-8}{31}\right)\Leftrightarrow B< 0\left(2\right)\)

\(C=\dfrac{-5}{10}.\dfrac{-4}{10}...........\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\Leftrightarrow C=0\left(3\right)\)

Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow A>C>B\)

24 tháng 8 2023

a) \(\dfrac{2}{5}=\dfrac{2\times5}{5\times2}=\dfrac{10}{25}\)

\(\dfrac{4}{7}=\dfrac{4\times2}{7\times2}=\dfrac{8}{14}\)

b) \(\dfrac{36}{40}=\dfrac{36:4}{40:4}=\dfrac{9}{10}\)

\(\dfrac{24}{32}=\dfrac{24:8}{32:8}=\dfrac{3}{4}\)

a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)

\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)

\(\Leftrightarrow-7x=\dfrac{10}{3}\)

hay \(x=-\dfrac{10}{21}\)

b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)

\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)

\(\Leftrightarrow85x+85=84x+168\)

\(\Leftrightarrow x=83\)