Ta có: \(\left|3x-5\right|\ge0\forall x\)

\(\left(2y+5\right)^{20}\ge0\forall y\)

\(\left(4z-3\right)^{206}\ge0\forall z\)

Do đó: \(\left|3x-5\right|+\left(2y+5\right)^{20}+\left(4z-3\right)^{206}\ge0\forall x,y,z\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{3};y=-\dfrac{5}{2};z=\dfrac{3}{4}\)

la

\(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\2y+5=0\\4z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4}\end{matrix}\right.\)

Sửa đề \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4x-3\right)^{20}\le0\)

Mà \(\left|3x-5\right|\ge0\);\(\left(2y+5\right)^{208}\ge0;\left(4x-3\right)^{20}\ge0\)

Do đó \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}=0\)

\(\Rightarrow\left\{{}\begin{matrix}3x-5=0\\2y+5=0\\4z-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4}\end{matrix}\right.\)

a) x/3 = y/2 = z/5 = 2y/4 = 2y- z/4-5 = -3/-1 = 3

x/3 = 3 suy ra x=9         ;        y/2 = 3 suy ra y=6         ;           z/5 = 3 suy ra z=15

 Vậy x=3 ; y=6 ; z=15

b) x/2 = y/2 suy ra x/6 = y/15 (nhân vs 3)           ;             y/3 = z/7 suy ra y/15 = z/35 (nhân vs 5) . Suy ra x/6 = y/15 = z/35

x/6 = y/15 = z/35 = 2x/12 = 3y/45 = 2x+ 3y- z/ 12+ 45- 35 = 22/22 =1

x/6 = 1 suy ra x=6 ; y/15 = 1 suy ra y=15 ; z/35 = 1 suy ra =35

  Vậy x=6 ; y=15 ; z= 35

Ta có: \(\left|3x-5\right|+\left(2y+5\right)^2+\left(4z-3\right)^{20}\ge0\)với \(\forall x;y;z\)

Mà \(\left|3x-5\right|+\left(2y+5\right)^2+\left(4z-3\right)^{20}\le0\)

\(\Rightarrow\left|3x-5\right|+\left(2y+5\right)^2+\left(4z-3\right)^{20}=0\)

\(\Rightarrow\hept{\begin{cases}3x-5=0\\2y+5=0\\4z-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=\frac{-5}{2}\\x=\frac{3}{4}\end{cases}}}\)

Vậy \(x=\frac{5}{3};y=\frac{-2}{5};z=\frac{3}{4}\)

                                                                Bài giải

\(\left|3x-5\right|+\left(2y+5\right)^{2008}+\left(4z-3\right)^{2006}\le0\)

Mà \(\hept{\begin{cases}\left|3x-5\right|\ge0\\\left(2y+5\right)^{2008}\ge0\\\left(4z-3\right)^{2006}\ge0\end{cases}}\) \(\Rightarrow\) Chỉ xảy ra trường hợp : \(\left|3x-5\right|+\left(2y+5\right)^{2008}+\left(4z-3\right)^{2006}=0\)

\(\Rightarrow\hept{\begin{cases}\left|3x-5\right|=0\\\left(2y+5\right)^{2008}=0\\\left(4z-3\right)^{2006}=0\end{cases}}\)            \(\Rightarrow\hept{\begin{cases}3x-5=0\\2y+5=0\\4z-3=0\end{cases}}\)         \(\Rightarrow\hept{\begin{cases}3x=5\\2y=-5\\4z=3\end{cases}}\)          \(\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{5}{2}\\x=\frac{3}{4}\end{cases}}\)

\(\Rightarrow\text{ }x=\frac{5}{3}\text{ , }y=-\frac{5}{2}\text{ , }z=\frac{3}{4}\)

\(\frac{2x}{3}=\frac{2y}{4}=\frac{4z}{5}=\frac{2x+2y+4z}{3+4+5}=\frac{2\left(x+y+z\right)}{12}=\frac{98}{12}=\frac{49}{6}\)

=> x = 49/4

=> y = 49/3

=. z = 245/24