Đáp án:
1: a) =2$\sqrt[]{4.3}$ - 3$\sqrt[]{16.3}$ + 4$\sqrt[]{25.3}$ - 2$\sqrt[]{9.3}$
= 4$\sqrt[]{3}$-12$\sqrt[]{3}$ + 20$\sqrt[]{3}$ - 6$\sqrt[]{3}$
= (4-12+20-6)$\sqrt[]{3}$ = 6$\sqrt[]{3}$
b) =$\frac{2+\sqrt[]{3}}{(2-\sqrt[]{3})(2+\sqrt[]{3})}$ + $\frac{2-\sqrt[]{3}}{(2+\sqrt[]{3})(2-\sqrt[]{3})}$
= $2-\sqrt[]{3}$ + $2+\sqrt[]{3}$ = 4
c) = $\sqrt[]{(\sqrt[]{3})^{2}+2.2.\sqrt[]{3}+2^{2}}$ - ($2-\sqrt[]{3}$)
=$\sqrt[]{(2+\sqrt[]{3})^{2}}$ +2^{2}}$ - ($2-\sqrt[]{3}$)
= $2+\sqrt[]{3}$ - ($2-\sqrt[]{3}$)
= 2$\sqrt[]{3}$
2.
a) B= 2$\sqrt[]{2x-1}$ - 3$\sqrt[]{2x-1}$ + 10$\sqrt[]{2x-1}$
= 9$\sqrt[]{2x-1}$
b) B=18 <=> 9$\sqrt[]{2x-1}$ = 18
$\sqrt[]{2x-1}$= 2 => x= 5/2
3.
a) A = x^2 + x + 2 A=x^2+x+2
= x ^2 + 2.1/2 x + 1/4 + 7/4
=x^2+2.1/2x+1/4+7/4
= ( x + 1/2 )^ 2 + 7/4 ≥ 7/4
=(x+1/2)^2+7/4≥7/4 Vậy Amin = 7/4 <=> x = -1/2
