\(=2-\left(\dfrac{5}{3}-\dfrac{7}{6}+\dfrac{9}{10}-...-\dfrac{19}{45}\right)\)

\(=2-2\left(\dfrac{5}{6}-\dfrac{7}{12}+\dfrac{9}{20}-...-\dfrac{19}{90}\right)\)

\(=2-2\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{5}-...-\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=2-2\cdot\dfrac{4}{10}=2-\dfrac{8}{10}=2-\dfrac{4}{5}=\dfrac{6}{5}\)

Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-5c}{2}=\dfrac{5\left(3a-2b\right)\left(2c-5a\right)}{5.5+3.3+}=\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\dfrac{5b-3c}{2}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3c}{5}\\a=\dfrac{2c}{5}\end{matrix}\right.\)

Mà \(a+b+c=-50\)

\(\Leftrightarrow\dfrac{2c}{5}+\dfrac{3c}{5}+c=-50\)

\(\Leftrightarrow2c=-50\)

\(\Leftrightarrow c=-25\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-10\\b=-15\end{matrix}\right.\)

Vậy ...

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}\leftrightarrow\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{5\left(3a-2b\right)}{25}=\dfrac{3\left(2c-5a\right)}{9}=\dfrac{2\left(5b-3c\right)}{4}=\dfrac{5\left(3a-2b\right)+3\left(2c-5a\right)+2\left(5b-3c\right)}{25+9+4}=0\)\(\Rightarrow\left\{{}\begin{matrix}3a-2b=0\\2c-5a=0\\5b-3c=0\end{matrix}\right.\)
⇔ 15a= 10b = 6c ⇔ \(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{\dfrac{1}{15}}=\dfrac{b}{\dfrac{1}{10}}=\dfrac{c}{\dfrac{1}{6}}=\dfrac{a+b+c}{\dfrac{1}{15}+\dfrac{1}{10}+\dfrac{1}{6}}=-\dfrac{50}{\dfrac{1}{3}}=-150\)
\(\Rightarrow\left\{{}\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)

đề sai

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\(\dfrac{3a-2b}{5}\)=\(\dfrac{2c-5a}{3}\)=\(\dfrac{5b-3c}{2}\)=\(\dfrac{15a-10b}{5}\)=\(\dfrac{6c-15a}{9}\)=\(\dfrac{10b-6c}{2}\)

Suy ra: \(\dfrac{15a-10b+6c-15a+10b-6c}{25+9+4}\)=\(\dfrac{0}{38}\)=0

Suy ra: 3a=2b\(\Leftrightarrow\)\(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)(1)

2c=5a\(\Leftrightarrow\)\(\dfrac{c}{5}\)=\(\dfrac{a}{2}\)(2)

Từ (1) và (2) suy ra: \(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{5}\)

Theo tính chất của dãy tỉ số bằng nhau ta được:

\(\dfrac{a}{2}\)=\(\dfrac{b}{3}\)=\(\dfrac{c}{5}\)=\(\dfrac{a+b+c}{2+3+5}\)=\(\dfrac{-50}{10}\)=-5

Tự làm nốt nha.

Đúng thì tick cho mk nha

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

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}=\dfrac{5.\left(3a-2b\right)+3.\left(2c-5a\right)}{5.5+3.3}=\dfrac{-10b+6c}{34}=\)

\(=\dfrac{-5b+3c}{17}\)

Do đó: \(\dfrac{5b-3c}{14}=\dfrac{-5b+3c}{2}\)

Suy ra: \(5b-3c=0\Rightarrow b=\dfrac{3}{5}c\) và \(a=\dfrac{2}{5}c\)

Lại có: \(a+b+c=-50\Rightarrow\dfrac{2}{5}c+\dfrac{3}{5}c+c=-50\Rightarrow c=-25\)

\(\Rightarrow b=\dfrac{3}{5}.\left(-25\right)=-15\)

và \(a=\dfrac{2}{5}.\left(-25\right)=-10\)

Vậy \(\left\{{}\begin{matrix}a=-10\\b=-15\\c=-25\end{matrix}\right.\)

Chúc bạn học tốt!!!

Theo t,c dãy tỉ số bằng nhau ta có :

\(\dfrac{3a-2b}{5}=\dfrac{2c-5a}{3}=\dfrac{5b-3c}{2}=\dfrac{5\left(3a-2b\right)\left(2c-5a\right)}{5.5+3.3}=\dfrac{-10b+6c}{34}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\dfrac{5b-3c}{2}=\dfrac{-5b+3c}{17}\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{3c}{5}\\a=\dfrac{2c}{5}\end{matrix}\right.\)

Mà \(a+b+c=-50\)

\(\Leftrightarrow\dfrac{2c}{5}+\dfrac{3c}{5}+c=-50\)

\(\Leftrightarrow2c=-50\)

\(\Leftrightarrow c=-25\)

\(\Leftrightarrow\left\{{}\begin{matrix}b=-15\\a=-10\end{matrix}\right.\)

Vậy ..

Một cách giải khác:

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)

\(\Leftrightarrow \frac{5(3a-2b)}{25}=\frac{3(2c-5a)}{9}=\frac{2(5b-3c)}{4}=\frac{5(3a-2b)+3(2c-5a)+2(5b-3c)}{25+9+4}=0\)

\(\Rightarrow \left\{\begin{matrix} 3a-2b=0\\ 2c-5a=0\\ 5b-3c=0\end{matrix}\right.\)

\(\Leftrightarrow 15a=10b=6c\Leftrightarrow \frac{a}{\frac{1}{15}}=\frac{b}{\frac{1}{10}}=\frac{c}{\frac{1}{6}}=\frac{a+b+c}{\frac{1}{15}+\frac{1}{10}+\frac{1}{6}}=\frac{-50}{\frac{1}{3}}=-150\)

(Áp dụng tính chất dãy tỉ số bằng nhau)

\(\Rightarrow \left\{\begin{matrix} a=-10\\ b=-15\\ c=-25\end{matrix}\right.\)

Lời giải:

\(\frac{3a-2b}{5}=\frac{2c-5a}{3}=\frac{5b-3c}{2}\)

\(\Leftrightarrow \frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}\)

Áp dụng TC dãy tỉ số bằng nhau:

\(\frac{15a-10b}{25}=\frac{6c-15a}{9}=\frac{10b-6c}{4}=\frac{15a-10b+6c-15a+10b-6c}{25+9+4}=0\)

\(\Rightarrow 15a=10b=6c\)

\(\Leftrightarrow \frac{a}{6}=\frac{b}{9}=\frac{c}{15}\)

Tiếp tục áp dụng TCDTSBN:

$\frac{a}{6}=\frac{b}{9}=\frac{c}{15}=\frac{a+b+c}{6+9+15}=\frac{50}{30}=\frac{5}{3}$

$\Rightarrow a=10; b=15; c=25$

bạn tham khảo tại: câu hỏi của nguyễn thị thanh mai- onlinemath

\(=\dfrac{85}{18}:\dfrac{85}{9}-\dfrac{136}{45}:\dfrac{136}{15}=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\)

Để tính tổng của biểu thức này, chúng ta cần thực hiện các phép cộng và trừ theo thứ tự từ trái sang phải.

\[4 + \frac{5}{6} - \frac{1}{9} \times \frac{1}{10} - \frac{7}{12} + \frac{1}{36} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{9} \times \frac{9}{5} + 1 - \frac{1}{3}\]

Đầu tiên, chúng ta sẽ làm các phép tính liên quan đến phân số:

\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{7}{12} + \frac{1}{36} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]

Tiếp theo, chúng ta sẽ tổng hợp các phân số:

\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{35}{90} + \frac{5}{180} - 3 - \frac{18}{90} + \frac{60}{180} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{35}{90} + \frac{5}{180} - 3 - \frac{2}{10} + \frac{10}{30} - \frac{2}{10} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{1}{90} - \frac{35}{90} + \frac{5}{180} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{36 + 35}{90} + \frac{5}{180} - 3 - \frac{1}{5} + \frac{2}{6} - \frac{1}{5} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - 3 - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]

Tiếp theo, chúng ta sẽ tính tổng các số nguyên:

\[= 4 - 3 + 1\]

Cuối cùng, chúng ta sẽ tổng hợp các phân số:

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{30}{90}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6} - \frac{71}{90} + \frac{5}{180} - \frac{18}{90} + \frac{30}{90} - \frac{18}{90} + 1 - \frac{1}{3}\]

\[= 4 + \frac{5}{6}

\(\dfrac{2}{3}+\dfrac{5}{6}+\dfrac{9}{10}+\dfrac{14}{15}+\dfrac{20}{21}+\dfrac{27}{28}+\dfrac{35}{36}+\dfrac{44}{45}\\ =\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{6}\right)+\left(1-\dfrac{1}{10}\right)+\left(1-\dfrac{1}{15}\right)+\left(1-\dfrac{1}{21}\right)+\left(1-\dfrac{1}{28}\right)+\left(1-\dfrac{1}{36}\right)+\left(1-\dfrac{1}{45}\right)\\ =8-\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\right)\\ =8-\left(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\\ =8-\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+\dfrac{2}{9.10}\right)\\ =8-2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =8-2\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=8-2.\dfrac{2}{5}=8-\dfrac{4}{5}=\dfrac{36}{5}\)

he nho

Kq = 36/5 nha bạn