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c,
= \(\dfrac{5}{9}.\left(\dfrac{7}{13}+\dfrac{9}{13}+\dfrac{-3}{13}\right)\)
= \(\dfrac{5}{9}.1\)
= \(\dfrac{5}{9}\)
1) \(\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}-\dfrac{8}{9}+\dfrac{10}{9}-\dfrac{5}{6}+\dfrac{6}{7}-\dfrac{7}{8}+\dfrac{8}{9}\)
\(=\left(\dfrac{5}{6}-\dfrac{5}{6}\right)-\left(\dfrac{6}{7}+\dfrac{6}{7}\right)+\left(\dfrac{7}{8}-\dfrac{7}{8}\right)-\left(\dfrac{8}{9}+\dfrac{8}{9}\right)+\dfrac{10}{9}\)
\(=0-0+0-0+\dfrac{10}{9}\)
\(=\dfrac{10}{9}\)
2) \(\dfrac{1}{13}+\dfrac{16}{7}+\dfrac{3}{105}-\dfrac{9}{7}-\dfrac{-12}{13}\)
\(=\left(\dfrac{1}{13}-\left(-\dfrac{12}{13}\right)\right)+\left(\dfrac{16}{7}-\dfrac{9}{7}\right)+\dfrac{3}{105}\)
\(=1+1+\dfrac{3}{105}\)
\(=\dfrac{213}{105}=\dfrac{71}{35}\)
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
13: \(=\dfrac{4}{9}\cdot\left(-7\right)+\left(6+\dfrac{5}{9}\right)\cdot\left(-7\right)\)
\(=\left(-7\right)\left(\dfrac{4}{9}+6+\dfrac{5}{9}\right)=\left(-7\right)\cdot7=-49\)
14: \(=\left(\dfrac{-3}{4}+\dfrac{5}{13}\right)\cdot\dfrac{7}{2}-\left(\dfrac{9}{4}+\dfrac{8}{13}\right)\cdot\dfrac{7}{2}\)
\(=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)\)
\(=\dfrac{7}{2}\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\cdot\dfrac{-42}{13}=\dfrac{-147}{13}\)
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: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
a: \(=\dfrac{4\cdot2+4\cdot9}{55}+\dfrac{5}{6}=\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{49}{30}\)
b: \(=\dfrac{3}{2}\cdot\dfrac{3}{5}-\left(\dfrac{3}{7}+\dfrac{3}{20}\right)\cdot\dfrac{10}{3}\)
\(=\dfrac{9}{10}-\dfrac{81}{140}\cdot\dfrac{10}{3}\)
\(=\dfrac{9}{10}-\dfrac{27}{14}=\dfrac{-36}{35}\)
c: \(=15+\dfrac{3}{13}-3-\dfrac{4}{7}-8-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
d: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=5\)
B= (7 \(\dfrac{4}{9}\) + 3\(\dfrac{8}{13}\)) - 5\(\dfrac{4}{9}\)
B=(\(\dfrac{67}{9}\)+ \(\dfrac{47}{13}\)) - \(\dfrac{49}{9}\)
B=\(\dfrac{116}{9}\)-\(\dfrac{47}{13}\)
B=\(\dfrac{116-47}{117}\)
B=\(\dfrac{23}{39}\)
tui cũng không biết đúng hông nữa 
Đề bài : Áp dụng tính chất các phép tính và quy tắc dấu ngoặc để tính giá trị các biểu thức
a) A = 3/7
b) B = 73/13
c) C = 37/7
d) D = 12
ba câu a) ,b) ,c) bn đổi ra hỗn số giúp mk nha
tick cho tớ nha
A = \(\dfrac{1}{4}.\dfrac{7}{3}.12\)
= \(\dfrac{1.7.12}{4.3}\)
= \(7\)
@Nguyễn Thành Đăng
B = \(\dfrac{3}{8}.56.\dfrac{25}{7}.\left(-4\right)\)
= \(-\dfrac{3.56.25.4}{8.7}\)
= -3.100
= -300
@Nguyễn Thành Đăng
\(\dfrac{3}{4}\cdot\dfrac{7}{9}\cdot\dfrac{1}{9}\cdot\dfrac{7}{4}\)
\(=\dfrac{3\cdot7\cdot1\cdot7}{4\cdot9\cdot9\cdot4}=\dfrac{3\cdot7\cdot1\cdot7}{4\cdot3\cdot3\cdot9\cdot4}\)
\(=\dfrac{7\cdot1\cdot7}{4\cdot3\cdot9\cdot4}=\dfrac{49}{432}\)
\(\dfrac{6}{7}\cdot\dfrac{8}{13}-\dfrac{6}{9}\cdot\dfrac{9}{7}+\dfrac{5}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\cdot\dfrac{8}{13}-\dfrac{6}{7}+\dfrac{5}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\left(\dfrac{8}{13}-1+\dfrac{5}{13}\right)\)
\(=\dfrac{6}{7}\cdot0\)
\(=0\)
\(2\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\dfrac{2\cdot11\cdot3\cdot9}{4\cdot121}=\dfrac{2\cdot11\cdot3\cdot9}{2\cdot2\cdot11\cdot11}\)
\(=\dfrac{3\cdot9}{2\cdot11}=\dfrac{27}{22}\)
bạn ơi, đề bài là gì vậy ạ?