Đề là: \(Q=\left(1+\dfrac{a}{x}\right)\left(1+\dfrac{a}{y}\right)\left(1+\dfrac{a}{z}\right)\) đúng không em nhỉ?

Ta có:

\(Q=\left(1+\dfrac{x+y+z}{x}\right)\left(1+\dfrac{x+y+z}{y}\right)\left(1+\dfrac{x+y+z}{z}\right)\)

\(=\dfrac{\left(x+x+y+z\right)\left(x+y+y+z\right)\left(x+y+z+z\right)}{xyz}\)

\(Q\ge\dfrac{4\sqrt[4]{x^2yz}.4\sqrt[4]{xy^2z}.4\sqrt[4]{xyz^2}}{xyz}=\dfrac{64xyz}{xyz}=64\)

\(Q_{min}=64\) khi \(x=y=z=\dfrac{a}{3}\)

\(\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow z=\dfrac{5y}{6}\)

mà \(z-y=40\)

\(\Rightarrow\dfrac{5y}{6}-y=40\)

\(\Rightarrow-\dfrac{y}{6}=40\)

\(\Rightarrow y=-240\Rightarrow z=40+y=40-240=-200\)

\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow x=\dfrac{3y}{4}=\dfrac{3.\left(-240\right)}{4}=-180\)

Vậy \(\left\{{}\begin{matrix}x=-180\\y=-240\\z=-200\end{matrix}\right.\)

\(\hept{\begin{cases}x-y=-9\\y-z=-10\\z+x=11\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=y-9\\z=10+y\\10+y+y-9=11\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=y-9\\z=10+y\\2y=10\end{cases}\Leftrightarrow\hept{\begin{cases}x=-4\\z=15\\y=5\end{cases}}}\)

a)\(\left|x-2\right|+\left|-17\right|=\left|-24\right|\)

\(\left|x-2\right|+17=24\)

\(\Rightarrow\left|x-2\right|=7\)

\(\Rightarrow x-2=\hept{\begin{cases}7\\-7\end{cases}}\)

\(\Rightarrow x=\hept{\begin{cases}9\\-5\end{cases}}\)

\(b,\left|x\right|=x\)

Vậy \(x\in N\)

\(c,\left|x\right|+\left|y\right|+\left|z\right|=0\)

Mà \(\left|x\right|+\left|y\right|+\left|z\right|\ge0\)

\(\Rightarrow x=0;y=0;z=0\)

\(a)\)\(\left|x-2\right|+\left|-17\right|=\left|-24\right|\)

\(\Leftrightarrow\left|x-2\right|+17=24\)

\(\Leftrightarrow\left|x-2\right|=24-17\)

\(\Leftrightarrow\left|x-2\right|=7\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=7\\x-2=-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=9\\x=-5\end{cases}}\)

Vậy\(x\in\left\{9;-5\right\}\)

\(b)\)\(\left|x\right|=x\)

\(\Leftrightarrow x\ge0\)

Vậy\(x\ge0\)

\(c)\) Ta thấy: \(\left|x\right|\ge0\)

                   \(\left|y\right|\ge0\)               \(\left(\forall x;y;z\right)\)

                   \(\left|z\right|\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left|x\right|=0\\\left|y\right|=0\\\left|z\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)

Vậy \(x=y=z=0\)

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}\)(*)

\(=\frac{\left(y+z+1\right)+\left(x+z+2\right)+\left(x+y-3\right)}{x+y+z}\)(Dãy tỉ số bằng nhau)

\(=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

\(\Rightarrow\frac{1}{x+y+z}=2\Leftrightarrow x+y+z=\frac{1}{2}\)

Thay vào (*), ta có:

\(\frac{\left(\frac{1}{2}-x\right)+1}{x}=\frac{\left(\frac{1}{2}-y\right)+2}{y}=\frac{\left(\frac{1}{2}-z\right)-3}{z}=2\)

\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{2}-x\\2y=\frac{5}{2}-y\\2z=-\frac{5}{2}-z\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=\frac{3}{2}\\3y=\frac{5}{2}\\3z=-\frac{5}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{5}{6}\\z=-\frac{5}{6}\end{cases}}\)

Vậy \(x=\frac{1}{2};y=\frac{5}{6};z=-\frac{5}{6}.\)

Đặt  thì a + b + c = x + y + z