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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: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>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...
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
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....
\(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\left(\dfrac{5}{3}\right)^6:\left(\dfrac{625}{343}\right)^2\)
\(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{5^6}{7^6}:\dfrac{5^8}{7^6}=\dfrac{1}{25}\)
1: =>|1/4x^2+1/45|=1/20
=>1/4x^2+1/45=1/20 hoặc 1/4x^2+1/45=-1/20
=>1/4x^2=1/36
=>x^2=1/36:1/4=1/9
=>x=1/3 hoặc x=-1/3
2: =(x^2-3)(x^2-2x)
=x(x-2)(x^2-3)
\(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}\)
\(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-\dfrac{1}{8}-\dfrac{27}{16}}{-\dfrac{29}{16}}\)
\(=\dfrac{-\dfrac{29}{16}}{-\dfrac{29}{16}}=1\)
Chúc bạn học tốt!!!
a) ta có : \(\left(x-\dfrac{1}{3}\right).\left(x+\dfrac{2}{3}\right)>0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\\x+\dfrac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{1}{3}\\x>\dfrac{-2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{1}{3}\\x< \dfrac{-2}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< \dfrac{-2}{3}\end{matrix}\right.\) vậy \(x>\dfrac{1}{3}\) hoặc \(x< \dfrac{-2}{3}\)
b) \(\left(x+\dfrac{3}{5}\right).\left(x+1\right)< 0\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\\x+1>0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{-3}{5}\\x< -1\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{-3}{5}\\x>-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\-1< x< \dfrac{-3}{5}\end{matrix}\right.\) vậy \(-1< x< \dfrac{-3}{5}\)
\(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{3}\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{1}{3}>0\Rightarrow x>\dfrac{1}{3}\\x+\dfrac{2}{3}>0\Rightarrow x>-\dfrac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{1}{3}< 0\Rightarrow x< \dfrac{1}{3}\\x+\dfrac{2}{3}< 0\Rightarrow x< -\dfrac{2}{3}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x>-\dfrac{2}{3}\) hoặc \(x< \dfrac{1}{3}\)
\(\left(x+\dfrac{3}{5}\right)\left(x+1\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\dfrac{3}{5}< 0\Rightarrow x< -\dfrac{3}{5}\\x+1>0\Rightarrow x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x+\dfrac{3}{5}>0\Rightarrow x>-\dfrac{3}{5}\\x+1< 0\Rightarrow x< -1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1< x< -\dfrac{3}{5}\)
\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)
\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
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 \(Q=\dfrac{29}{125}\)
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 = \(\dfrac{72}{5}\)