Đáp án:
$a)\dfrac{512}{161}\\
b)
x=\pm 6\\
c)A=2-\dfrac{1}{2^{2020}}$
Giải thích các bước giải:
$a)\left ( \dfrac{2}{3}-1-4+\dfrac{5}{11} \right ):\left ( \dfrac{5}{12}-1-\dfrac{7}{11} \right )\\
=\left ( \dfrac{2.11}{33}-\dfrac{5.33}{33}+\dfrac{15}{33} \right ):\left ( \dfrac{5.11}{12.11}-\dfrac{11.12}{11.12}-\dfrac{7.12}{12.11} \right )\\
=\dfrac{-128}{3.11}:\dfrac{-161}{11.12}\\
=\dfrac{128}{3.11}.\dfrac{11.12}{161}\\
=\dfrac{128.4}{161}\\
=\dfrac{512}{161}\\
b)
2.x^2-72=0\\
\Leftrightarrow 2.x^2=72\\
\Leftrightarrow x^2=36\\
\Leftrightarrow x=\pm \sqrt{36}=\pm 6\\
c)A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}\\
\Rightarrow 2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\\
A=2A-A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}-\left ( 1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}} \right )\\
=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}- 1-\dfrac{1}{2}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^{2020}} \\
=2-\dfrac{1}{2^{2020}}$
