a)\(\left|2x-3y\right|+\left|2y-4z\right|=0\)

\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\forall x;y\\\left|2y-4z\right|\ge0\forall y;z\end{matrix}\right.\) \(\Rightarrow\left|2x-3y\right|+\left|2y-4z\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|2y-4z\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=3y\\2y=4z\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=\dfrac{y}{2}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=\dfrac{y}{4}\\\dfrac{y}{4}=\dfrac{z}{2}\end{matrix}\right.\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{2}=\dfrac{x+y+z}{6+4+2}=\dfrac{7}{12}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{12}.6=\dfrac{7}{2}\\y=\dfrac{7}{12}.4=\dfrac{7}{3}\\z=\dfrac{7}{12}.2=\dfrac{7}{6}\end{matrix}\right.\)

b)\(\left|x-2\right|+\left|x-3\right|+\left|x-4\right|=0\)

\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|x-3\right|\ge0\\\left|x-4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|x-2\right|+\left|x-3\right|+\left|x-4\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-2\right|=0\\\left|x-3\right|=0\\\left|x-4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=3\\x=4\end{matrix}\right.\)

Vì \(2\ne3\ne4\) nên \(x\in\varnothing\)

c)

\(\left|x+1\right|+\left|x+2\right|+...+\left|x+8\right|+\left|x+9\right|\)

Với mọi \(x\ge0\) ta có:

\(\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x+2\right|=x+2\\\left|x+8\right|=x+8\\\left|x+9\right|=x+9\end{matrix}\right.\)\(\Leftrightarrow x+1+x+2+...+x+8+x+9=x-1\)

\(\Leftrightarrow9x+90=x-1\)

\(\Leftrightarrow9x=x-89\)

\(\Leftrightarrow-8x=89\)

\(\Leftrightarrow x=\dfrac{89}{-8}\left(KTM\right)\)

Với mọi \(x< 0\) ta có:

\(\left\{{}\begin{matrix}x+1=-x-1\\x+2=-x-2\\x+8=-x-8\\x+9=-x-9\end{matrix}\right.\) \(\Leftrightarrow\left(-x-1\right)+\left(-x-2\right)+...+\left(-x-8\right)+\left(-x-9\right)=x-1\)

\(\Leftrightarrow-9x-90=x-1\)

\(\Leftrightarrow-9x=x+89\)

\(\Leftrightarrow-10x=89\)

\(\Leftrightarrow x=\dfrac{89}{-10}\left(TM\right)\)

d)\(\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|=0\)

\(\left\{{}\begin{matrix}\left|2x-3y\right|\ge0\\ \left|5y-2z\right|\ge0\\ \left|2z-6\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow\left|2x-3y\right|+\left|5y-2z\right|+\left|2z-6\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|2x-3y\right|=0\\\left|5y-2z\right|=0\\\left|2z-6\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}z=3\\y=\dfrac{6}{5}\\x=\dfrac{9}{5}\end{matrix}\right.\)

a) x=36,y=24,z=18                                                                                           b)x=12,y=8,z=24

cứuuuuuuuuuuuu

\(x^2+3y^2+2z^2-2x+12y+4z+15=0\)

\(x^2-2x+1+\left(\sqrt{3}y\right)^2+2.6.y+\left(2\sqrt{3}\right)^2+\left(\sqrt{2}z\right)^2+2.2.z+\left(\sqrt{2}\right)^2=0\)

\(\left(x-1\right)^2+\left(\sqrt{3}y+2\sqrt{3}\right)^2+\left(\sqrt{2}z+\sqrt{2}\right)^2=0\)

\(\Rightarrow x=1;y=-2;z=-1\)

<=>(x²-2x+1)+(3y²+12y+12)+(2z²+4z+2)=0

<=>(x-1)²+3(y+2)²+2(z+1)²=0

Vì \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\3\left(y+2\right)^2\ge0\\2\left(z+1\right)^2\ge0\end{cases}\Rightarrow\left(x-1\right)^2+3\left(y+2\right)^2+2\left(z+1\right)^2\ge0}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-2\\z=-1\end{cases}}}\)