Question

Tìm giá trị của các đa thức sau:

a, \(A=3x^4+5x^2y^2+2y^4+2x^2\) . Biết: \(x^2+y^2=0\)

b, \(B=x^6-20x^5+20x^4-20x^3+20x^2-20x+20\) . Biết: \(x=19\)

c, \(C=\left(1+\frac{x}{y}\right).\left(1+\frac{y}{z}\right).\left(1+\frac{z}{x}\right)\) . Biết: \(x+y+z=0\) và \(x,y,z\ne0\)

Answer 1

a/ \(x^2+y^2=0\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\) \(\Rightarrow A=0\)

b/ Do \(x=19\Rightarrow20=x+1\)

\(B=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+20\)

\(B=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+20\)

\(B=20-x=20-19=1\)

c/ \(x+y+z=0\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)

\(C=\frac{\left(x+y\right)}{y}.\frac{\left(y+z\right)}{z}.\frac{\left(x+z\right)}{x}=\frac{-z}{y}.\frac{-x}{z}.\frac{-y}{x}=\frac{-xyz}{xyz}=-1\)