Sử dụng bất đẳng thức AM - GM ta dễ thấy:

\(LHS=\sqrt{a-1+2\sqrt{a-2}}+\sqrt{a-1-2\sqrt{a-2}}\)

\(\ge2\sqrt{\left(a-1+2\sqrt{a-2}\right)\left(a-1-2\sqrt{a-2}\right)}\)

\(=2\sqrt{\left(a-1\right)^2-4\left(a-2\right)}=2\sqrt{a^2-6a+9}=2\sqrt{\left(a-3\right)^2}\ge2\)( vì a khác 3 ) 

Hoặc cách khác như thế này:

\(LHS=\sqrt{a-1+2\sqrt{a-2}}+\sqrt{a-1-2\sqrt{a-2}}\)

\(=\sqrt{\left[a-2+2\sqrt{a+2}+1\right]}+\sqrt{\left[a-2-2\sqrt{a-2}+1\right]}\)

\(=\sqrt{\left(\sqrt{a-2}+1\right)^2}+\sqrt{\left(\sqrt{a-2}-1\right)^2}\)

\(=\left|\sqrt{a-2}+1\right|+\left|\sqrt{a-2}-1\right|\)

\(=\left|\sqrt{a-2}+1\right|+\left|1-\sqrt{a-2}\right|\ge\left|\sqrt{a-2}+1+1-\sqrt{a-2}\right|=2\)

Đẳng thức tự tìm nha

bé hơn

= 4,65028154

(3\(x\) - 4)(\(x+1\))(2\(x-1\)) = 0

\(\left[\begin{array}{l}3x-4=0\\ x+1=0\\ 2x-1=0\end{array}\right.\)

\(\left[\begin{array}{l}3x=4\\ x=-1\\ 2x=1\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac43\\ x=-1\\ x=\frac12\end{array}\right.\)

Vậy \(x\) ∈ {-1; \(\frac12\); \(\frac43\)}

(3x - 4)(x + 1)(2x - 1) = 0

3x - 4 = 0 hoặc x + 1 = 0 hoặc 2x - 1 = 0

*) 3x - 4 = 0

3x = 4

*) x + 1 = 0

x = 0 - 1

x = -1

*) 2x - 1 = 0

2x = 1

Vậy:

\(A=\sqrt{5+\sqrt{3}}+\sqrt{5-\sqrt{3}}\)

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

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

=> \(A^2=10+2.22\)

=> \(A^2=54\)

=> \(A=\sqrt{54}=\sqrt{9.6}=3\sqrt{6}\)