Vì \(\hept{\begin{cases}\left|3-2x\right|\text{≥ }0\\\left|4-5y\right|\text{≥ }0\\\left|5x-3y+z\right|\text{≥ }0\end{cases}\Rightarrow\left|3-2x\right|+\left|4-5y\right|+\left|5x-3y+z\right|\text{≥ }0}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|3-2x\right|=0\\\left|4-5y\right|=0\\\left|5x-3y+z\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{4}{5}\\z=\frac{51}{10}\end{cases}}}\)

\(\dfrac{4x-3y}{5}=\dfrac{5y-4z}{3}=\dfrac{3z-5x}{4}\)

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

=>\(\left\{{}\begin{matrix}3\left(4x-3y\right)=5\left(5y-4z\right)\\4\left(4x-3y\right)=5\left(3z-5x\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-9y-25y+20z=0\\16x-12y-15z+25x=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}12x-34y+20z=0\\41x-12y-15z=0\end{matrix}\right.\)

mà x-y+z=200 nên ta có hệ phương trình:

\(\left\{{}\begin{matrix}12x-34y+20z=0\\41x-12y-15z=0\\x-y+z=200\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}36x-102y+60z=0\\164x-48y-60z=0\\60x-60y+60z=12000\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}200x-150y=0\\-24x-42y=-12000\\x-y+z=200\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x-3y=0\\4x+7y=2000\\x-y+z=200\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-10y=-2000\\4x-3y=0\\x-y+z=200\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=200\\4x=3y\\x-y+z=200\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=200\\x=\dfrac{3}{4}y=150\\150-200+z=200\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=200\\x=150\\z=250\end{matrix}\right.\)

a) \(x^4+4x^2-5=x^4+4x^2+4-9=\left(x^2+2\right)^2-3^2\)

\(\left(x^2+2-3\right)\left(x^2+2+3\right)\)

b) \(-x-y^2+x^2-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)\(=\left(x+y\right)\left(x-y-1\right)\)

c) \(x\left(x+y\right)-5x-5y=x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)

d) \(x^2-5x+5y-y^2=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-5\right)\)

e) \(5x^3-5x^2y-10x^2+10xy=5x^2\left(x-y\right)-10x\left(x-y\right)\)

\(=5\left(x-y\right)\left(x^2-2x\right)\)

f) \(27x^3-8y^3=\left(3x\right)^3-\left(2y\right)^3=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)

x(x+y)-5x-5y  

=x(x+y) -5(x+y)

=(x+y)(x-5)

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