11) \(\dfrac{2\sqrt{3}-\sqrt{15}}{\sqrt{3}}\)

\(=\dfrac{\left(2\sqrt{3}-\sqrt{15}\right)\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}\)

\(=\dfrac{2\cdot\sqrt{3}\cdot\sqrt{3}-\sqrt{15}\cdot\sqrt{3}}{3}\)

\(=\dfrac{6-3\sqrt{5}}{3}\)

\(=2-\sqrt{5}\)

12) \(\dfrac{2\sqrt{2}+\sqrt{2}}{5\sqrt{2}}\)

\(=\dfrac{3\sqrt{2}}{5\sqrt{2}}\)

\(=\dfrac{3}{5}\)

11: \(=\dfrac{\sqrt{3}\left(2-\sqrt{5}\right)}{\sqrt{3}}=2-\sqrt{5}\)

12: \(=\dfrac{3\sqrt{2}}{5\sqrt{2}}=\dfrac{3}{5}\)

8d

9c

10d

11a

12d

12. have been

10. built

11. was talking

12. have been

13. were eating

8 disappointing

12 preservation

13 seriously

Bài 13:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\x\ne3\end{matrix}\right.\)

b: Ta có: \(P=\dfrac{x-3}{\sqrt{x-1}-\sqrt{2}}\)

\(=\dfrac{\left(\sqrt{x-1}-\sqrt{2}\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}\)

\(=\sqrt{x-1}+\sqrt{2}\)

c: Thay \(x=12-6\sqrt{2}\) vào P, ta được:

\(P=\sqrt{11-6\sqrt{2}}+\sqrt{2}=3-\sqrt{2}+\sqrt{2}=3\)

d: Ta có: \(\sqrt{x-1}\ge0\forall x\) thỏa mãn ĐKXĐ

\(\Leftrightarrow\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=1

a: Xét ΔABE và ΔADC có

AB=AD

\(\widehat{BAE}=\widehat{DAC}\)

AE=AC

Do đó: ΔABE=ΔADC

12 I consider that to manage to know what other people are thinking is quite impossible

they had to work in a quiet office

Bạn đọc lại câu hỏi giùm mình, đừng trả lời bừa

I wish they didn't have to work in a noisy office.

Câu này thêm chủ ngữ I vào đắng trước và làm theo cấu trúc câu ước là được em nhé!