Đáp án:
Bài 3: $=\dfrac{125}{27}$
Bài 4: $(x;y)=\left(-\dfrac{13}{14};\dfrac{8}{7}\right)$
Giải thích các bước giải:
Bài 3:
$\dfrac{10^3+5.10^2+5^3}{6^3+3.6^2+3^3}$
$=\dfrac{(2.5)^3+5.(2.5)^2+5^3}{(2.3)^3+3.(2.3)^2+3^3}$
$=\dfrac{2^3.5^3+2^2.5.5^2+5^3}{2^3.3^3+2^2.3.3^2+3^3}$
$=\dfrac{8.5^3+4.5^3+5^3}{8.3^3+4.3^3+3^3}$
$=\dfrac{13.5^3}{13.3^3}$
$=\dfrac{5^3}{3^3}$
$=\dfrac{125}{27}$
Bài 4:
$\left|x+\dfrac{13}{14}\right|+\left|y-\dfrac{8}{7}\right|=0$
Ta có: $\begin{cases}\left|x+\dfrac{13}{14}\right|\ge 0\\\left|y-\dfrac{8}{7}\right|\ge 0\end{cases}$
$⇒\left|x+\dfrac{13}{14}\right|+\left|y-\dfrac{8}{7}\right|=0$
$⇒\begin{cases}\left|x+\dfrac{13}{14}\right|= 0\\\left|y-\dfrac{8}{7}\right|= 0\end{cases}⇒\begin{cases}x+\dfrac{13}{14}= 0\\y-\dfrac{8}{7}= 0\end{cases}⇒\begin{cases}x=-\dfrac{13}{14}\\y=\dfrac{8}{7}\end{cases}$
Vậy $(x;y)=\left(-\dfrac{13}{14};\dfrac{8}{7}\right)$.
