Question

[Tran Le Hoang Vu]

 các ban làm giùm mình câu 9,10

Answer 1

[Nguyễn Lê Phước Thịnh]

9: \(A=\dfrac{\dfrac{1}{4}-5\cdot\left(\dfrac{3}{2}\right)^2}{10\dfrac{5}{9}+\left(-\dfrac{2}{3}\right)^2}=\dfrac{\dfrac{1}{4}-5\cdot\dfrac{9}{4}}{10+\dfrac{5}{9}+\dfrac{4}{9}}\)

\(=\dfrac{\dfrac{1}{4}-\dfrac{45}{4}}{10+1}=\dfrac{-44}{4}:11=-\dfrac{44}{44}=-1\)

\(B=\dfrac{5}{12}\cdot3,7-\dfrac{5}{12}\cdot6,7=\dfrac{5}{12}\cdot\left(3,7-6,7\right)\)

\(=\dfrac{5}{12}\cdot\left(-3\right)=-\dfrac{5}{4}\)

\(A-B=\left(-1\right)-\left(-\dfrac{5}{4}\right)=-1+\dfrac{5}{4}=\dfrac{1}{4}\)

10: \(P=\left(6,8;1,36-\dfrac{29}{3}:\dfrac{58}{9}\right):\dfrac{0.27^3}{0.09^3\cdot2}\)

\(=\left(5-\dfrac{29}{3}\cdot\dfrac{9}{58}\right):\dfrac{\left(0,3\right)^6\cdot3^3}{0,3^6\cdot2}\)

\(=\left(5-\dfrac{3}{2}\right):\dfrac{3^3}{2}=\dfrac{7}{2}\cdot\dfrac{2}{27}=\dfrac{7}{27}\)

\(P+\dfrac{1}{27}=\dfrac{7}{27}+\dfrac{1}{27}=\dfrac{8}{27}=\left(\dfrac{2}{3}\right)^3\)

=>\(P+\dfrac{1}{27}\) là bình phương của một số hữu tỉ