11 D

12C

13B

14D

15A

B-A-B-B-C

1: \(\frac{1}{4+2\sqrt2}+\frac{1}{4-2\sqrt2}\)

\(=\frac{4-2\sqrt2+4+2\sqrt2}{\left(4-2\sqrt2\right)\left(4+2\sqrt2\right)}=\frac{8}{16-8}=1\)

2: \(\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}-\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}\)

\(=\frac{\left(\sqrt5+\sqrt3\right)^2-\left(\sqrt5-\sqrt3\right)^2}{\left(\sqrt5+\sqrt3\right)\left(\sqrt5-\sqrt3\right)}=\frac{8+2\sqrt{15}-\left(8-2\sqrt{15}\right)}{5-3}\)

\(=\frac{4\sqrt{15}}{2}=2\sqrt{15}\)

3: \(\left(\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}+1\right):\frac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}=\frac{\sqrt5-\sqrt3+\sqrt5+\sqrt3}{\sqrt5+\sqrt3}\cdot\frac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\)

\(=\frac{2\sqrt5}{\sqrt5-\sqrt3}=\frac{2\sqrt5\left(\sqrt5+\sqrt3\right)}{5-3}=\sqrt5\left(\sqrt5+\sqrt3\right)=5+\sqrt{15}\)

4: \(\sqrt{\left(1-\sqrt3\right)^2}+\frac{1}{\sqrt{\left(1+\sqrt3\right)^2}}\)

\(=\sqrt3-1+\frac{1}{\sqrt3+1}=\frac{\left(\sqrt3-1\right)\left(\sqrt3+1\right)+1}{\sqrt3+1}=\frac{3-1+1}{\sqrt3+1}\)

\(=\frac{3\left(\sqrt3-1\right)}{2}\)

5: \(\frac{3}{\sqrt7-2}-\frac{3}{\sqrt7+2}=\frac{3\left(\sqrt7+2\right)-3\left(\sqrt7-2\right)}{\left(\sqrt7-2\right)\left(\sqrt7+2\right)}\)

\(=\frac{3\sqrt7+6-3\sqrt7+6}{7-4}=\frac{12}{3}=4\)

6: \(\frac{5\sqrt2-2\sqrt5}{\sqrt5-\sqrt2}-\frac{9}{\sqrt{10}+1}\)

\(=\frac{\sqrt{10}\left(\sqrt5-\sqrt2\right)}{\sqrt5-\sqrt2}-\frac{9\left(\sqrt{10}-1\right)}{10-1}=\sqrt{10}-\left(\sqrt{10}-1\right)=1\)

7: \(\frac{2\sqrt3}{\sqrt3+\sqrt2}+\sqrt{24}\)

\(=\frac{2\sqrt3\left(\sqrt3-\sqrt2\right)}{\left(\sqrt3-\sqrt2\right)\left(\sqrt3+\sqrt2\right)}+2\sqrt6\)

\(=2\sqrt3\left(\sqrt3-\sqrt2\right)+2\sqrt6=6-2\sqrt6+2\sqrt6=6\)

8: \(\sqrt3\left(2\sqrt{27}-\sqrt{75}+\frac32\cdot\sqrt{12}\right)\)

\(=\sqrt3\left(2\cdot3\sqrt3-5\sqrt3+\frac32\cdot2\sqrt3\right)=\sqrt3\left(6\sqrt3-5\sqrt3+3\sqrt3\right)=\sqrt3\cdot4\sqrt3=12\)

9: \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt3-1}=\frac{\sqrt{\left(\sqrt6-\sqrt2\right)^2}}{\sqrt3-1}=\frac{\sqrt2\left(\sqrt3-1\right)}{\sqrt3-1}=\sqrt2\)

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

 3 bài 13, 14, 15??/

đâu ạ ^^?