Question

tìm x : ( \((\sqrt{3} - \sqrt{2} )^x +( \sqrt{3} + \sqrt{2})^x =10\)

Answer 1

Với x < 2

=> \(\left(\sqrt{3}-\sqrt{2}\right)^x+\left(\sqrt{3}+\sqrt{2}\right)^x< \left(\sqrt{3}-\sqrt{2}\right)^2+\left(\sqrt{3}+\sqrt{2}\right)^2=5-2\sqrt{6}+5+2\sqrt{6}=10\)

=> \(\left(\sqrt{3}-\sqrt{2}\right)^x+\left(\sqrt{3}+\sqrt{2}\right)^x< 10\) (không thỏa mãn)

Với x > 2

=> \(\left(\sqrt{3}-\sqrt{2}\right)^x+\left(\sqrt{3}+\sqrt{2}\right)^x>\left(\sqrt{3}-\sqrt{2}\right)^2+\left(\sqrt{3}+\sqrt{2}\right)^2=5-2\sqrt{6}+5+2\sqrt{6}=10\)

=> \(\left(\sqrt{3}-\sqrt{2}\right)^x+\left(\sqrt{3}+\sqrt{2}\right)^x>10\) (không thỏa mãn)

=> x = 2

........Kaito Kid.......

Answer 2

(\(\sqrt{3}-\sqrt{2}\))^x + (\(\sqrt{3}+\sqrt{2}\))^x = 10

\(\Leftrightarrow\dfrac{1}{\left(\sqrt{3}+\sqrt{2}\right)^x}\) + \(\left(\sqrt{3}+\sqrt{2}\right)^x\)= 10

Đặt \(\left(\sqrt{3}+\sqrt{2}\right)^x\) = a

\(\Rightarrow\dfrac{1}{a}\) + a = 10

Giờ thì đơn giản rồi nhé