Giả sử a, b là các số thực sao cho

D

- Hướng dẫn giải

Đáp án D.

Ta có:

$$\begin{gathered} \left\{\begin{array}{l}\log \left(x+y\right)=z\\ \log \left(x^2+y^2\right)=z+1\end{array}\right. \\ \iff \left\{\begin{array}{l}x+y={10}^z+\\ x^2+y^2={10}^{z+1}={10.10}^z\end{array}\right. \\ \Rightarrow x^2+y^2=10\left(x+y\right) \end{gathered}$$

Khi đó:

$$\begin{gathered} x^3+y^3=a{.10}^{3z}+b{.10}^{2z} \\ \iff \left(x+y\right)\left(x^2-xy+y^2\right)=a.{\left({10}^z\right)}^3+b.{\left({10}^z\right)}^2 \\ \iff \left(x+y\right)\left(x^2-xy+y^2\right)=a.{\left(x+y\right)}^3+b.{\left(x+y\right)}^2 \\ \iff x^2-xy+y^2=a.{\left(x+y\right)}^2+b.\left(x+y\right) \\ \iff x^2-xy+y^2=a.\left(x^2+2xy+y^2\right)+\frac{b}{10}.\left(x^2+y^2\right) \\ \iff x^2+y^2-xy=\left(a+\frac{b}{10}\right).\left(x^2+y^2\right)+2a.xy \end{gathered}$$

Đồng nhất hệ số, ta được:

$\left\{\begin{array}{l}a+\frac{b}{10}=1\\ 2a=-1\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=-\frac{1}{2}\\ b=15\end{array}\right..$

Vậy $a+b=\frac{29}{2}.$