\(C=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\)

\(D=x-4+\sqrt{16-8x+x^2}=x-4+\left|4-x\right|=x-4+x-4=2x-8\)

\(A=-x-z\left(x-y\right)+y=-x-xz+zy+y=-x\left(1+z\right)+y\left(1+z\right)=\left(1+z\right)\left(y-x\right)\)

\(\left(x-6\right)^3=\left(x-6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-6=0\\x-6=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=7\end{matrix}\right.\)

\(\dfrac{9}{10};\dfrac{4}{5}=\dfrac{8}{10};\dfrac{3}{2}=\dfrac{15}{10};\dfrac{6}{5}=\dfrac{12}{10};\dfrac{1}{2}=\dfrac{5}{10}\)

-> Sắp xếp: \(\dfrac{3}{2};\dfrac{6}{5};\dfrac{9}{10};\dfrac{4}{5};\dfrac{1}{2}\)

\(\Rightarrow\dfrac{-6}{x}=\dfrac{1}{2}\)

\(\Rightarrow x=-12\)

Số kẹo họ có là: 20 x 6 = 120 (cái)

Nếu chia mỗi túi 24 cái thì được 120 : 24 = 5 (túi)

\(5^x.5^2=5^{40}\)

\(\Rightarrow5^x=5^{40}:5^2=5^{38}\)

=> x = 38

Có: \(\left\{{}\begin{matrix}\left|x-\dfrac{1}{3}\right|\ge0\\\left|7y-1\right|\ge0\end{matrix}\right.\)

=> \(\left|x-\dfrac{1}{3}\right|+\left|7y-1\right|-1\ge-1\)

Dấu = xảy ra khi x = 1/3 và y = 1/7