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a)\(\dfrac{-6}{11}:\left(\dfrac{3}{5}.\dfrac{4}{11}\right)=\dfrac{-5}{2}\)
b)\(\dfrac{7}{12}+\dfrac{5}{12}:6-\dfrac{14}{30}=\dfrac{67}{370}\)
c)\(\left(\dfrac{4}{5}+\dfrac{1}{2}\right):\left(\dfrac{3}{13}-\dfrac{8}{13}\right)=-\dfrac{169}{50}\)
d)\(\left(\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}\right):\left(\dfrac{5}{12}+1-\dfrac{7}{11}\right)=\dfrac{115}{103}\)
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
Bài 1:
a)
\(A=\dfrac{-5}{6}\cdot\dfrac{3}{10}\\ =\dfrac{\left(-5\right)\cdot3}{6\cdot10}\\ =\dfrac{-1}{4}\)
b)
\(B=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{12}\\ =\dfrac{4}{12}-\dfrac{3}{12}+\dfrac{1}{12}\\ =\dfrac{4-3+1}{12}\\ =\dfrac{1}{6}\)
Bài 2:
\(A=\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{4}{5}-\dfrac{14}{5}\right|:\dfrac{8}{3}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{-10}{5}\right|\cdot\dfrac{3}{8}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+2\cdot\dfrac{3}{8}\\ =\dfrac{-3}{4}+\dfrac{3}{4}\\ =0\)
\(B=\left(\dfrac{-1}{2}\right)^2:1\dfrac{3}{8}+25\%\cdot\dfrac{3}{11}\\ =\left(\dfrac{-1}{2}\right)^2:\dfrac{11}{8}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{1}{4}\cdot\dfrac{8}{11}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{8}{44}+\dfrac{9}{44}\\ =\dfrac{17}{44}\)
\(C=\dfrac{-8}{5}+0,6+\left|\dfrac{-1}{2}\right|+\dfrac{1}{2}\\ =\dfrac{-8}{5}+\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{1}{2}\\ =\left(\dfrac{-8}{5}+\dfrac{3}{8}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)\\ =\left(-1\right)+1\\ =0\)
\(D=\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}:\dfrac{13}{11}+1\dfrac{5}{9}\\ =\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}\cdot\dfrac{11}{13}+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot\left(\dfrac{2}{13}+\dfrac{11}{13}\right)+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot1+\dfrac{14}{9}\\ =\dfrac{-5}{9}+\dfrac{14}{9}\\ =1\)
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
bn ơi ko thì bn làm từng phép một cũng đc nhé các bn giải hộ mình mai mình phải nộp rồi
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
a) \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)
\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{37}{3}-\dfrac{17}{2}=\dfrac{74}{6}-\dfrac{51}{6}=\dfrac{23}{6}\)
b) \(3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+\dfrac{13}{6}.6=\dfrac{23}{6}+\dfrac{78}{6}=\dfrac{101}{6}\)
c) \(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}=-\dfrac{92}{14}=-\dfrac{46}{7}\)
d) \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}=\dfrac{9}{2}+\dfrac{1}{2}.\dfrac{2}{11}=\dfrac{9}{2}+\dfrac{1}{11}=\dfrac{99}{22}+\dfrac{2}{22}=\dfrac{101}{22}\)
a. \(12\dfrac{1}{3}-\left(3\dfrac{3}{4}+4\dfrac{3}{4}\right)=\dfrac{37}{3}-\left(\dfrac{15}{4}+\dfrac{19}{4}\right)\)
\(=\dfrac{37}{3}-\dfrac{34}{4}=\dfrac{23}{6}\)
\(b.3\dfrac{5}{6}+2\dfrac{1}{6}.6=\dfrac{23}{6}+13=\dfrac{101}{6}\)
\(c.3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}=\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}=\dfrac{20}{7}\)
d \(4\dfrac{1}{2}+\dfrac{1}{2}:5\dfrac{1}{2}\)
\(=\dfrac{9}{2}+\dfrac{1}{2}:\dfrac{11}{2}\)
\(=\dfrac{9}{2}+\dfrac{1}{11}\)
\(=\dfrac{101}{22}\)