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25 tháng 10 2022
a: \(=7\cdot\dfrac{6}{7}-5+\dfrac{3\sqrt{2}}{2}=1+\dfrac{3}{2}\sqrt{2}\)
b: \(=-\dfrac{8}{7}-\dfrac{3}{5}\cdot\dfrac{5}{8}+\dfrac{1}{2}=\dfrac{-16+7}{14}-\dfrac{3}{8}=\dfrac{-9}{14}-\dfrac{3}{8}\)
\(=\dfrac{-72-42}{112}=\dfrac{-114}{112}=-\dfrac{57}{56}\)
c: \(=20\sqrt{5}-\dfrac{1}{4}\cdot\dfrac{4}{3}+\dfrac{3}{2}=20\sqrt{5}+\dfrac{3}{2}-\dfrac{1}{3}=20\sqrt{5}+\dfrac{7}{6}\)
31 tháng 7 2022
1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)
2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)
c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)
25 tháng 10 2018
\(a,-\dfrac{3}{5}.y=\dfrac{21}{10}\)
\(y=\dfrac{21}{10}:\dfrac{-3}{5}=\dfrac{-7}{2}\)
\(b,y:\dfrac{3}{8}=-1\dfrac{31}{33}\)
\(y=-1\dfrac{31}{33}.\dfrac{3}{8}=\dfrac{-8}{11}\)
Vậy \(y=-\dfrac{8}{11}\)
\(c,1\dfrac{2}{5}.y+\dfrac{3}{7}=-\dfrac{4}{5}\)
\(\Rightarrow1\dfrac{2}{5}y=-\dfrac{4}{5}-\dfrac{3}{7}=\dfrac{-43}{35}\)
\(\Rightarrow y=\dfrac{-43}{35}:1\dfrac{2}{5}=\dfrac{-43}{49}\)
\(d,-\dfrac{11}{12}.y+0,25=\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{11}{12}.y=\dfrac{5}{6}-0,25=\dfrac{7}{12}\)
\(\Rightarrow y=\dfrac{7}{12}:\dfrac{-11}{12}=\dfrac{-7}{11}\)
11 tháng 9 2017
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
TH
9 tháng 3 2017
a,
\(\dfrac{\left(3^3\right)^{15}.5^3.\left(2^3\right)^4}{\left(5^2\right)^2.\left(3^4\right)^{11}.2^{11}}=\dfrac{3^{45}.5^3.2^{12}}{5^4.3^{44}.2^{11}}=\dfrac{6}{5}\)
b, \(\left(-\dfrac{14}{25}\right)^2.\dfrac{125}{49}+\left(-3\dfrac{11}{36}\right).2\dfrac{2}{17}=\dfrac{4}{5}.\left(-7\right)=-\dfrac{28}{5}\)
c, \(\dfrac{1}{3}-2.1=-\dfrac{5}{3}\)
VN
24 tháng 7 2018
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
HS
24 tháng 7 2018
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
5 tháng 8 2018
Ta có:
\(B=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\\ =\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}=\dfrac{72}{5}\)
Vậy B = \(\dfrac{72}{5}\)
5 x ( \(\dfrac{-5}{2}\) )2 + \(\dfrac{2}{15}\) x \(\sqrt{\dfrac{9}{4}}\) - ( -2022)0 + | -0,25|
= 5 x \(\dfrac{25}{4}\) + \(\dfrac{2}{15}\) x \(\dfrac{3}{2}\) - 1 + 0,25
= \(\dfrac{125}{4}\) + \(\dfrac{1}{5}\) - ( 1 - 0,25)
= 31,25 + 0,2 - 0,75
= 31,45 - 0,75
= 30,7