11 D

12C

13B

14D

15A

B-A-B-B-C

b) \(B=\sqrt{12+2\sqrt{35}}=\sqrt{12+2.\sqrt{7}.\sqrt{5}}=\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}=\left|\sqrt{7}+\sqrt{5}\right|\)

Vì \(\sqrt{7}>\sqrt{5}\) nên \(\left|\sqrt{7}+\sqrt{5}\right|=\sqrt{7}+\sqrt{5}\)

d) \(D=\sqrt{6-\sqrt{35}}+\sqrt{10}=\sqrt{2}.\sqrt{6-\sqrt{35}}+\sqrt{5}=\sqrt{12-\sqrt{35}}+\sqrt{5}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}+\sqrt{5}=\left|\sqrt{7}-\sqrt{5}\right|+\sqrt{5}=\sqrt{7}-\sqrt{5}+\sqrt{5}=\sqrt{7}\)

II

1 doesn't snow

2 will be heated

3 playing

4 was waiting

5 were

Câu 3: 

a: Ta có: \(\left(1-4x\right)\left(x-1\right)+\left(2x+1\right)\left(2x+3\right)=38\)

\(\Leftrightarrow x-1-4x^2+4x+4x^2+6x+2x+3=38\)

\(\Leftrightarrow13x=36\)

hay \(x=\dfrac{36}{13}\)

b: Ta có: \(\left(2x+3\right)\left(x+2\right)-\left(x-4\right)\left(2x-1\right)=75\)

\(\Leftrightarrow2x^2+4x+3x+6-2x^2+x+8x-4=75\)

\(\Leftrightarrow15x=73\)

hay \(x=\dfrac{73}{15}\)

?

a: góc xOt=góc yOt=100/2=50 độ

b: góc xOt'=180 độ-góc xOt=130 độ

A, B thuộc đường tròn nên \(IA=IB=R=4\left(cm\right)\)

Chu vi tam giác: \(IA+IB+AB=4+4+3=11\left(cm\right)\)

13 B

14 B

15 C