\(\sqrt{2-\sqrt{3}}\)=\(\sqrt{\dfrac{2\left(2-\sqrt{3}\right)}{2}}\)=\(\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)=\(\sqrt{\dfrac{\sqrt{3}^2-2\sqrt{3}+1}{2}}\)=\(\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}\)=\(\dfrac{\sqrt{3}-1}{\sqrt{2}}\)

\(1\)

bài 1

Xét tam giác vuông ABC có :

AB²=BC²+AC² (pytago)

<=> AC=\(\sqrt{AB^2-BC^2}\)=\(\sqrt{12^2-7^2}\)=\(\sqrt{95}\)

sinA=\(\dfrac{BC}{AB}\)=\(\dfrac{7}{12}\)

=> Â\(\approx\)35⁰

ta có Â+ góc B=90⁰ (vì tam giác ABC là tam giác vuông)

=> góc B = 90⁰- 35⁰ =55⁰