D

D

\(H=\left(-2\right)^3\cdot\dfrac{-1}{4}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)

\(H=-8\cdot\dfrac{-1}{4}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\cdot\dfrac{12}{5}\)

\(H=2+\left(-\dfrac{1}{2}\right)\cdot\dfrac{12}{5}\)

\(H=2+\left(-\dfrac{6}{5}\right)\)

\(H=\dfrac{4}{5}\)

Câu 1: A

bài đâu em :V ?

Bài 6:

a:

b:

c:

Bài 2: 

a: 3/8=81/216

5/27=40/216

b: -2/9=-50/225

4/25=36/225

c: 1/15=1/15

-6=-90/15

cảm ơn :3

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=3\sqrt{5}-1\\4x+\left(2\sqrt{5}+2\right)y=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-6y=6\sqrt{5}-2\\4x+\left(2\sqrt{5}+2\right)y=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(-8-2\sqrt{5}\right)y=6\sqrt{5}+2\\2x-3y=3\sqrt{5}-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1-\sqrt{5}\\x=\dfrac{3\sqrt{2}-3\sqrt{5}+2}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-6y=6\sqrt{5}-2\\3.\left(\sqrt{5}-1\right)x+6y=3-3\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(3\sqrt{5}+1\right)x=1+3\sqrt{5}\\y=\dfrac{3\sqrt{5}-1-2x}{-3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3\sqrt{5}-1-2.1}{-3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{-3.\left(1-\sqrt{5}\right)}{-3}=1-\sqrt{5}\end{matrix}\right.\\ \Rightarrow\left(x;y\right)=\left(1;1-\sqrt{5}\right)\)