Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
\(\dfrac{1}{5}+\dfrac{2}{11}< \dfrac{x}{55}< \dfrac{2}{5}+\dfrac{1}{5}\)
\(\dfrac{11+10}{55}< \dfrac{x}{55}< \dfrac{3}{5}\)
\(\dfrac{21}{55}< \dfrac{x}{55}< \dfrac{33}{55}\)
Vậy \(x\in\left\{22;23;24;...\right\}\)
2:
a: x=2,4-0,4=2
b: =>2x=-1,5+0,8=-0,7
=>x=-0,35
c: =>x-16=-15
=>x=1
\(\Leftrightarrow\left(11+\dfrac{7}{18}-9-\dfrac{13}{18}\right):x=\dfrac{5}{3}+\dfrac{35}{33}\cdot\dfrac{11}{7}=\dfrac{10}{3}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{10}{3}=\dfrac{1}{2}\)
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: \(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}=\dfrac{-39}{36}=\dfrac{-13}{12}\)
b: \(=\dfrac{11}{9}\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)=\dfrac{11}{9}\cdot\dfrac{3}{4}=\dfrac{11}{12}\)
c: \(=15+\dfrac{9}{7}+6+\dfrac{2}{3}-5-\dfrac{5}{9}\)
\(=16+\dfrac{88}{63}=\dfrac{1096}{63}\)
d: \(=\dfrac{5}{6}-\dfrac{1}{3}+\dfrac{2}{18}\)
\(=\dfrac{15-6+2}{18}=\dfrac{11}{18}\)
\(-\dfrac{1}{4}\times13\dfrac{9}{11}-0,25\times6\dfrac{2}{11}\\ =-\dfrac{1}{4}\times\dfrac{152}{11}+\left(-\dfrac{1}{4}\right)\times\dfrac{68}{11}\\ =-\dfrac{1}{4}\times\left(\dfrac{152}{11}+\dfrac{68}{11}\right)\\ =-\dfrac{1}{4}\times\dfrac{220}{11}\\ =-\dfrac{1}{4}\times20\\ =-5\)
mk ko ghi lại đề nhé.
= -1/4 x 152/11 - 25/100 x 68/11
= -1/4 x 152/11 - 1/4 x 68/11
= -1/4 x ( 152/11 + 68/11)
= -1/4 x 20
= -5