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Giúp em bài toán này với :
Bài 3: Tìm x :
b) X x \(\frac{1}{2}\)+ \(\frac{3}{2}\)x X = \(\frac{4}{5}\)
\(x.\frac{1}{2}+\frac{3}{2}.x=\frac{4}{5}\)
\(\Rightarrow x\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)
\(\Rightarrow x.1=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}\)
a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2
=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)
=> \(-2x=\frac{-85}{4}\)
=> \(x=\frac{-85}{4}:\left(-2\right)\)
=> \(x=\frac{85}{8}\)
b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)
=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)
=> \(\frac{-2}{3}x=\frac{15}{29}\)
=> x = \(\frac{15}{29}:\frac{-2}{3}\)
=> x = \(\frac{-45}{58}\)
- \(x = {-b \pm \sqrt{b^2-4ac} \over 2Ahs\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(d,\left(1-\frac{3}{4}\right)\left(1+\frac{1}{3}\right):\left(1-\frac{1}{3}\right)\)
\(=\frac{1}{4}.\frac{4}{3}.\frac{3}{2}\)
\(=\frac{1}{2}\)
b)\(\left(2016.1017+2017.2018\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right)\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right).0\)
\(=0\)
\(\left(x+\frac{3}{4}\right)\times\frac{5}{7}=\frac{10}{9}\)
\(\Rightarrow x+\frac{3}{4}=\frac{10}{9}:\frac{5}{7}=\frac{10}{9}\times\frac{7}{5}=\frac{14}{9}\)
\(\Rightarrow x=\frac{14}{9}-\frac{3}{4}=\frac{56-27}{36}=\frac{29}{36}\)
\(\left(x+\frac{3}{4}\right)\times\frac{5}{7}=\frac{10}{9}\)
\(\Leftrightarrow x+\frac{3}{4}=\frac{14}{9}\)
\(\Rightarrow x=\frac{29}{36}\)
P/s tham khảo nha