Giúp em Bài 4 và bài 5 ạ Question Giúp em Bài 4 và bài 5 ạ in progress 0 Môn Toán Latifah 4 years 2020-10-29T07:07:09+00:00 2020-10-29T07:07:09+00:00 1 Answers 71 views 0
Answers ( )
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
x + \dfrac{1}{x} = 3\\
a,\\
A = {x^2} + \dfrac{1}{{{x^2}}} = \left( {{x^2} + 2.x.\dfrac{1}{x} + \dfrac{1}{{{x^2}}}} \right) – 2.x.\dfrac{1}{x}\\
= {\left( {x + \dfrac{1}{x}} \right)^2} – 2\\
= {3^2} – 2\\
= 7\\
b,\\
B = {x^3} + \dfrac{1}{{{x^3}}}\\
= \left( {{x^3} + 3.{x^2}.\dfrac{1}{x} + 3.x.\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}} \right) – \left( {3.{x^2}.\dfrac{1}{x} + 3.x.\dfrac{1}{{{x^2}}}} \right)\\
= {\left( {x + \dfrac{1}{x}} \right)^3} – 3.x.\dfrac{1}{x}.\left( {x + \dfrac{1}{x}} \right)\\
= {3^3} – 3.1.3\\
= 18\\
5,\\
{x^2} + {y^2} – 4x – 6y + 13 = 0\\
\Leftrightarrow \left( {{x^2} – 4x + 4} \right) + \left( {{y^2} – 6y + 9} \right) = 0\\
\Leftrightarrow {\left( {x – 2} \right)^2} + {\left( {y – 3} \right)^2} = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x – 2} \right)^2} = 0\\
{\left( {y – 3} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 3
\end{array} \right.\\
S = {\left( {x – y} \right)^{2020}} = {\left( {2 – 3} \right)^{2020}} = {\left( { – 1} \right)^{2020}} = 1
\end{array}\)