\(A=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\\ \Rightarrow A^2=2-\sqrt{3}+2\sqrt{2-\sqrt{3}}\sqrt{2+\sqrt{3}}+2+\sqrt{3}\\ \Rightarrow A^2=4+2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\\ \Rightarrow A^2=4+2\sqrt{2^2-\sqrt{3^2}}\\ \Rightarrow A^2=4+2\sqrt{1}\\ \Rightarrow A^2=6\\ \Rightarrow A=\pm\sqrt{6}\)

Mà \(A>0\Rightarrow A=\sqrt{6}\)

Lời giải:

\(B\sqrt{2}=\sqrt{8+2\sqrt{15}}-\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}.\sqrt{3}+3}-\sqrt{5-2\sqrt{5}+1}\)

\(=\sqrt{(\sqrt{5}+\sqrt{3})^2}-\sqrt{(\sqrt{5}-1)^2}=|\sqrt{5}+\sqrt{3}|-|\sqrt{5}-1|\)

\(=\sqrt{5}+\sqrt{3}-(\sqrt{5}-1)=\sqrt{3}+1\)

$\Rightarrow B=\frac{\sqrt{3}+1}{\sqrt{2}}$

---------------

\(C\sqrt{2}=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}-2\)

\(=\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2}-2=|\sqrt{5}+1|-|\sqrt{5}-1|-2\)

\(=(\sqrt{5}+1)-(\sqrt{5}-1)-2=0\Rightarrow C=0\)

------------------------------

\(D\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+\sqrt{14}\)

\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}+\sqrt{14}=|\sqrt{7}-1|-|\sqrt{7}+1|+\sqrt{14}\)

\(=\sqrt{7}-1-(\sqrt{7}+1)+\sqrt{14}=-2+\sqrt{14}\)

\(\Rightarrow D=-\sqrt{2}+\sqrt{7}\)

a: Xét tứ giác AKCI có 

AK//CI

AK=CI

Do đó: AKCI là hình bình hành

hơi mờ và còn nhiều ,mik nghĩ bạn nên cắt ra đi

1 not going on holiday with you

2 such easy questions that all the students got them right

3 we had had time, we would have visited the museum

4 to Daisy for breaking her vase

5 studying ENglish 5 years ago

6 are thought to be the most popular dance in Brazil

7 sooner had she received the exam result than she phoned her mom

8 It was the absence of leadership that caused most of the problems on the committee

9 you have any complants about the product, return it to the shop

10 The more fondness for the game increased , the more proficiency he has

11 been a dramatical rise in house prices this year

12 believed to have escaped in a stolen car

13 gone out with him for 2 years

14 he was going to meet his sister in front of the station

đk x khác 1 ; y khác -2 

\(\left\{{}\begin{matrix}\dfrac{8}{x-1}+\dfrac{15}{y+2}=1\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x-1}+\dfrac{15}{y+2}=1\\\dfrac{8}{x-1}+\dfrac{8}{y+2}=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y+2}=-7\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=-1\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\\dfrac{1}{x-1}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\1=2x-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=\dfrac{3}{2}\end{matrix}\right.\)

Đặt ẩn: a=1/x-1 ;b=1/y+2

⇔\(\left[{}\begin{matrix}8a+15b=1\\a+b=1\end{matrix}\right.\) 

sau đó bạn giải hệ PT rồi kết luận :

thay a=....=1/x-1

       b=.....=1/y+2

Bài 1:

1: =>x(2x-5)(2x+5)=0

hay \(x\in\left\{0;\dfrac{5}{2};-\dfrac{5}{2}\right\}\)

2: \(\Leftrightarrow2x^2+8x-x-4=0\)

=>(x+4)(2x-1)=0

hay \(x\in\left\{-4;\dfrac{1}{2}\right\}\)

3: \(\Leftrightarrow x^3-x-6x-6=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\)

=>(x+1)(x-3)(x+2)=0

hay \(x\in\left\{-1;3;-2\right\}\)

4: \(\Leftrightarrow3x^2-12x-5x-3x^2=-34\)

=>-17x=-34

hay x=2

5: \(\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\)

=>(4x-3)(7x-3)=0

hay \(x\in\left\{\dfrac{3}{4};\dfrac{3}{7}\right\}\)

Thôi cảm ơn mình tự làm đc r :)