Question

tính giá trị biểu thức A =\(2x^5-5y^3-4\) tại x,y thỏa mãn\(\left(x-1\right)^{20}+\left(y+2\right)^{30}=0\)

Answer 1

Ta có: \(\left(x-1\right)^{20}+\left(y+2\right)^{30}=0\)

\(\Leftrightarrow\left[\left(x-1\right)^{10}\right]^2+\left[\left(y+2\right)^{15}\right]^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^{10}=0\\\left(y+2\right)^{15}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Thay x=1, y = -2 vào biểu thức A ta được A= 38

Answer 2

Ta có \(\left(x-1\right)^{20}\ge0\);\(\left(y+2\right)^{30}\ge0\)

\(\Rightarrow\left(x-1\right)^{20}+\left(y+2\right)^{30}\ge0\)

Mà \(\left(x-1\right)^{20}+\left(y+2\right)^{30}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^{20}=0\\\left(y+2\right)^{30}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

Thay vào ta có \(A=2.1^5-5.\left(-2\right)^3-4=2+40-4=38\)