\(\left(2-\sqrt{3}\right)^x+\left(7-4\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)
Ta có: \(2-\sqrt{3}=\frac{1}{2+\sqrt{3}}\)
\(7-4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\left(2-\sqrt{3}\right)^x+\left(7-4\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)
<=> \(\frac{1}{\left(2+\sqrt{3}\right)^x}+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\)
<=> \(1+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^x\left(2+\sqrt{3}\right)^x=4\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x\)
<=> \(1+\left(2-\sqrt{3}\right)^2\left(2+\sqrt{3}\right)^{2x}=4\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x\)
Đặt: \(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=t\)
Ta có pt ẩn t: \(1+t^2=4t\)
<=> \(t^2-4t+1=0\Leftrightarrow\orbr{\begin{cases}t=2-\sqrt{3}\\t=2+\sqrt{3}\end{cases}}\)
+) Với \(t=2+\sqrt{3}\), ta có:
\(\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^x=2+\sqrt{3}\)
<=> \(\left(2+\sqrt{3}\right)^x=\frac{2+\sqrt{3}}{2-\sqrt{3}}=\left(2+\sqrt{3}\right)^2\)
<=> x=2
Trường hợp còn lại em làm tương tự
