8d

9c

10d

11a

12d

8 disappointing

12 preservation

13 seriously

12:

\(\dfrac{x}{y}\sqrt{\dfrac{y}{x}}=\dfrac{\sqrt{y}}{y}\cdot\dfrac{x}{\sqrt{x}}=\dfrac{\sqrt{x}}{\sqrt{y}}\)

a) \(\dfrac{x}{y}\sqrt{\dfrac{y}{x}}=\sqrt{\dfrac{x^2}{y^2}\cdot\dfrac{y}{x}}\)

\(=\sqrt{\dfrac{x}{y}}\)

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

T

F(weak#strong)  

F(2#4)

F (march#january)

A  D

7 F (nation => nations)

8 F

9 F

10 F

B

11 A

12 D