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\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)
\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
Vậy \(P=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-1,008\right):\dfrac{4}{7}:\left[\left(3\dfrac{1}{4}-6\dfrac{5}{9}\right):2\dfrac{2}{17}\right]\)
\(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right):\dfrac{4}{7}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right).\dfrac{36}{17}\right]\)
\(=-\dfrac{116}{125}.\dfrac{7}{4}:\left(-\dfrac{119}{36}.\dfrac{36}{17}\right)\)
\(=\dfrac{-29.7}{125}:\left(-7\right)=\dfrac{29}{125}\)
Vậy \(Q=\dfrac{29}{125}\)
\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)
\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)
\(=\dfrac{-1}{8}\)
\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)
\(=\dfrac{71}{35}\)
\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)
\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)
\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)
\(=\left(-3\right)+\left(\dfrac{-5}{2}\right)+\left(\dfrac{-7}{5}\right)+\dfrac{5}{7}\)
\(=\dfrac{-433}{70}\)
\(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)
\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{125}{100}+\dfrac{79}{63}\right)\)
\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{5}{4}+\dfrac{79}{63}\right)\)
\(=-5,13:\left(\dfrac{145}{28}-\dfrac{85}{36}+\dfrac{79}{63}\right)\)
\(=-5,13:\dfrac{57}{14}=-5,13:\dfrac{15}{57}\)
\(=\dfrac{-71,82}{57}=1,26\)
Vậy \(A=1,26\)
\(B=\left(3\dfrac{1}{3}.1,9+19,5:4\dfrac{1}{3}\right).\left(\dfrac{62}{75}-\dfrac{4}{25}\right)\)
\(=\left(\dfrac{10}{3}.1,9+19,5:\dfrac{13}{3}\right).\left(\dfrac{62-12}{75}\right)\)
\(=\left(\dfrac{19}{3}+\dfrac{58,5}{13}\right).\dfrac{50}{75}\)
\(=\left(\dfrac{19}{3}+4,5\right).\dfrac{2}{3}\)
\(=\dfrac{32,5}{3}.\dfrac{2}{3}=\dfrac{65}{9}=7\dfrac{2}{9}\)
Vậy \(B=7\dfrac{2}{9}\)
Bạn tính hai vế à.!? Hay tính vế thứ nhất rồi với vế thứ 2.!???
a: \(=\dfrac{-3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)=\dfrac{-3}{4}\cdot40=-30\)
b: \(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)
\(=\dfrac{35}{6}:\dfrac{-175+132}{42}+\dfrac{15}{2}\)
\(=\dfrac{35}{6}\cdot\dfrac{42}{-43}+\dfrac{15}{2}\)
\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}\)
\(=\dfrac{-70\cdot7+15\cdot43}{86}=\dfrac{155}{86}\)
c: \(=\dfrac{-7}{5}\left(4+\dfrac{5}{9}+5+\dfrac{4}{9}\right)=\dfrac{-7}{5}\cdot10=-14\)
d: \(=4+\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}\cdot\dfrac{64}{125}\cdot\dfrac{-8}{27}\right)\)
\(=\dfrac{89}{16}+25\cdot\dfrac{-32}{375}\)
\(=\dfrac{89}{16}-\dfrac{32}{15}=\dfrac{823}{240}\)
e: \(=\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
1, \(x\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)
2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)
Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)
\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)
Vậy, ...
b, \(\left|x-\dfrac{1}{3}\right|\ge0\)
Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)
\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)
Vậy, ...
1)
a)
\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2)
a)
\(\left|x+\dfrac{4}{6}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)
Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)
b)
\(\left|x-\dfrac{1}{3}\right|\ge0\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
\(A=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{4}+\dfrac{16}{5}\right)\)
\(A=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{4}-\dfrac{16}{5}\)
\(A=\left(5-6-2\right)\)\(-\left(\dfrac{3}{4}+\dfrac{7}{4}-\dfrac{5}{4}\right)\)\(+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)\)
\(A=-3\)\(-\dfrac{5}{4}\)\(+\dfrac{-7}{5}\)
\(A=\dfrac{-113}{20}\)
\(A=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{4}+\dfrac{16}{5}\right)\)
\(A=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{4}-\dfrac{16}{5}\)
\(A=-3-\dfrac{5}{4}+\dfrac{-7}{5}\)
\(A=-\dfrac{113}{20}\)