a: \(x=\sqrt{28-10\sqrt3}+\sqrt[3]{-64}+\sqrt3\)
\(=\sqrt{\left(5-\sqrt3\right)^2}-4+\sqrt3\)
\(=5-\sqrt3-4+\sqrt3=5-4=1\)
Thay x=1 vào A, ta được:
\(A=\left(1^{2020}+1^{2021}-3\right)^{2020}=\left(1+1-3\right)^{2020}=\left(-1\right)^{2020}=1\)
b: \(\sqrt{x^2-10+\sqrt{x^2-4x+4}}=2\left(x-3\right)\)
=>\(\sqrt{x^2-10+\sqrt{\left(x-2\right)^2}}=2\left(x-3\right)\)
=>\(\sqrt{x^2-10+\left|x-2\right|}=2\left(x-3\right)\) (1)
TH1: x>=2
(1) sẽ trở thành: \(\sqrt{x^2-10+x-2}=2\left(x-3\right)\)
=>\(\sqrt{x^2+x-12}=2\left(x-3\right)\)
=>\(\begin{cases}2\left(x-3\right)\ge0\\ 4\left(x-3\right)^2=x^2+x-12=\left(x+4\right)\left(x-3\right)\end{cases}\)
=>\(\begin{cases}x-3\ge0\\ 4\left(x-3\right)^2-\left(x-3\right)\left(x+4\right)=0\end{cases}=>\begin{cases}x\ge3\\ \left(x-3\right)\left(4x-12-x-4\right)=0\end{cases}\)
=>\(\begin{cases}x\ge3\\ \left(x-3\right)\left(3x-16\right)=0\end{cases}\Rightarrow x\in\left\lbrace3;\frac{16}{3}\right\rbrace\)
TH2: x<2
=>x-3<0
=>\(\sqrt{x^2-10+\left|x-2\right|}=2\left(x-3\right)<0\)
=>Vô lý
=>Loại
