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\(E=\dfrac{98:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right)\cdot\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}\\ E=\dfrac{98}{\dfrac{3}{5}}+\dfrac{\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\\ E=\dfrac{490}{3}+\dfrac{\dfrac{7}{4}}{7}=\dfrac{490}{3}+\dfrac{1}{4}=\dfrac{1963}{12}\)
bạn ơi chỗ kia mik nhìn hơi loạn tí bạn giải thích giúp mik với
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}-\dfrac{1213}{100}=2\cdot\left[\left(x-\dfrac{10}{7}\right)\cdot\dfrac{49}{50}+\dfrac{2}{5}\right]\)
\(\Leftrightarrow\left(\dfrac{13}{4}-x\right)\cdot\dfrac{101}{25}=\dfrac{49}{25}\left(x-\dfrac{10}{7}\right)+\dfrac{4}{5}+\dfrac{1213}{100}\)
\(\Leftrightarrow\dfrac{1313}{100}-\dfrac{101}{25}x=\dfrac{49}{25}x-\dfrac{490}{175}+\dfrac{1293}{100}\)
=>-6x=13/5
hay x=-13/30
\(\left|\dfrac{2}{3}-1\right|-\dfrac{5}{2}.\sqrt[]{\dfrac{4}{25}}=\left|-\dfrac{1}{3}\right|-\dfrac{5}{2}.\dfrac{2}{5}=\dfrac{1}{3}-1=-\dfrac{2}{3}\)
a: =>\(-\dfrac{6+x}{2}-\dfrac{3}{2}=2\)
=>-x-6-3=4
=>-x-9=4
=>-x=5
hay x=-5
b: =>(x+1)2=16
=>x+1=4 hoặc x+1=-4
=>x=3 hoặc x=-5
c: \(\Leftrightarrow\left(\dfrac{x-2}{27}-1\right)+\left(\dfrac{x-3}{26}-1\right)+\left(\dfrac{x-4}{25}-1\right)+\left(\dfrac{x-5}{24}-1\right)+\left(\dfrac{x-44}{5}+3\right)=0\)
=>x-29=0
hay x=29
\(a.\dfrac{-4}{7}-\dfrac{5}{13}\times\dfrac{-39}{25}+\dfrac{-1}{42}:\dfrac{-5}{6}\)
\(=\dfrac{-4}{7}+\dfrac{3}{5}+\dfrac{1}{35}\) \(=\dfrac{1}{35}+\dfrac{1}{35}=\dfrac{2}{35}\)
\(b.\dfrac{2}{9}\times\left[\dfrac{4}{5}:\left(\dfrac{1}{5}-\dfrac{2}{15}\right)+1\dfrac{2}{3}\right]-\dfrac{-5}{27}\)
\(=\dfrac{2}{9}\times\left[\dfrac{4}{5}:\dfrac{1}{15}+\dfrac{5}{3}\right]-\dfrac{-5}{27}\)
\(=\dfrac{2}{9}\times\left(12+\dfrac{5}{3}\right)-\dfrac{-5}{27}\)
\(=\dfrac{2}{9}\times\dfrac{41}{3}-\dfrac{-5}{27}=\dfrac{82}{27}-\dfrac{-5}{27}=\dfrac{29}{9}\)
\(a,\dfrac{x-1}{x+5}=\dfrac{6}{7}\\ \Leftrightarrow\left(x-1\right).7=6\left(x+5\right)\\ \Rightarrow7x-7=6x+30\\ \Rightarrow7x-6x=7+30\\ \Rightarrow x=37\)
Vậy \(x=37\)
\(b,\dfrac{x^2}{6}=\dfrac{24}{25}\\ \Leftrightarrow x^2.25=24.6\\ \Rightarrow x^2.5^2=144\\ \Rightarrow\left(5x\right)^2=144\\ \Rightarrow\left(5x\right)^2=\left(\pm12\right)^2\\ \Rightarrow\left\{{}\begin{matrix}5x=12\\5x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)
Vậy \(x=\pm\dfrac{12}{5}\)
\(\dfrac{6}{5}\cdot\sqrt{\dfrac{25}{16}}-\left(\dfrac{3}{4}\right)^2:0,25\\ =\dfrac{6}{5}\cdot\dfrac{5}{4}-\dfrac{9}{16}\cdot4\\ =\dfrac{3}{2}-\dfrac{9}{4}\\ =-\dfrac{3}{4}\)
\(\dfrac{6}{5}\cdot\sqrt{\dfrac{25}{16}}-\left(\dfrac{3}{4}\right)^2:0,25\)
\(=\dfrac{6}{5}\cdot\dfrac{5}{4}-\dfrac{9}{16}:\dfrac{1}{4}\)
\(=\dfrac{6\cdot5}{5\cdot4}-\dfrac{9\cdot4}{16}\)
\(=\dfrac{6}{4}-\dfrac{9}{4}\)
\(=\dfrac{3}{4}\)
\(a,=-5+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}=-5-1+\dfrac{1}{5}=-\dfrac{29}{5}\\ b,\Rightarrow\left|\dfrac{1}{2}-x\right|=\dfrac{28}{5}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}-x=\dfrac{28}{5}\\x-\dfrac{1}{2}=\dfrac{28}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{51}{10}\\x=\dfrac{61}{10}\end{matrix}\right.\)
\(\left(x-\dfrac{1}{5}\right)^2=\dfrac{4}{25}\Leftrightarrow\left(x-\dfrac{1}{5}\right)^2=\left(\dfrac{2}{5}\right)^2\)
\(\Leftrightarrow x-\dfrac{1}{5}=\dfrac{2}{5}\Leftrightarrow x=\dfrac{3}{5}\)
pt \(\Leftrightarrow\)\(\left[{}\begin{matrix}x-\dfrac{1}{5}=\dfrac{2}{5}\\x-\dfrac{1}{5}=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)