g: \(\sqrt3<\sqrt4\)

=>\(\sqrt3<2\)

=>\(\frac{\sqrt3}{2}<\frac22=1\)

h: \(\frac{\sqrt{10}}{2}=\sqrt{\frac{10}{4}}=\sqrt{2,5}<\sqrt{20}=2\sqrt5\)

=>\(-\frac{\sqrt{10}}{2}>-2\sqrt5\)

j: \(2\sqrt5-5\sqrt2=\sqrt{20}-\sqrt{50}<0\)

mà 0<1

nên \(2\sqrt5-5\sqrt2<1\)

k: \(\frac{\sqrt8}{3}=\frac{2\sqrt2}{3}=\frac{2\sqrt2\cdot4}{3\cdot4}=\frac{8\sqrt2}{12}\)

\(\frac34=\frac{3\cdot3}{4\cdot3}=\frac{9}{12}\)

mà \(8\sqrt2=\sqrt{128}>\sqrt{81}=9\)

nên \(\frac{\sqrt8}{3}>\frac34\)

l: \(6\sqrt{\frac14}=\sqrt{6^2\cdot\frac14}=\sqrt9;4\sqrt{\frac12}=\sqrt{4^2\cdot\frac12}=\sqrt8\)

\(2\sqrt3=\sqrt{2^2\cdot3}=\sqrt{12};\sqrt{\frac{15}{5}}=\sqrt3\)

mà 12>9>8>3

nên \(2\sqrt3>6\sqrt{\frac14}>4\sqrt{\frac12}>\sqrt{\frac{15}{5}}>0\)

mà \(0>-\sqrt{132}\)

nên \(2\sqrt3>6\sqrt{\frac14}>4\sqrt{\frac12}>\sqrt{\frac{15}{5}}>-\sqrt{132}\)

m: \(-2\sqrt6=-\sqrt{2^2\cdot6}=-\sqrt{24}<-\sqrt{23}\)

n: \(2\sqrt6=\sqrt{24}<\sqrt{25}=5\)

=>\(2\sqrt6-2<5-2\)

=>\(2\sqrt6-2<3\)
o: \(28\sqrt2=\sqrt{2\cdot28^2}=\sqrt{1568}\)

\(\sqrt{14}=\sqrt{14}\)

\(2\sqrt{147}=\sqrt{147\cdot4}=\sqrt{588}\)

\(36\sqrt4=\sqrt{4\cdot36^2}=\sqrt{5184}\)

mà 14<588<1568<5184

nên \(\sqrt{14}<2\sqrt{147}<28\sqrt2<36\sqrt4\)

q: \(\sqrt9=3;\sqrt{25}-\sqrt{16}=5-4=1\)

mà 3>1

nên \(\sqrt9>\sqrt{25}-\sqrt{16}\)

r: \(\sqrt{111}<\sqrt{121}=11\)

=>\(\sqrt{111}-7<11-7\)

=>\(\sqrt{111}-7<4\)
p: \(4\sqrt3=\sqrt{3\cdot4^2}=\sqrt{48};16\sqrt5=\sqrt{16^2\cdot5}=\sqrt{1280}\)

\(21\sqrt2=\sqrt{2\cdot21^2}=\sqrt{882}\)

mà 1280>882>48

nên \(16\sqrt5>21\sqrt2>4\sqrt3>0\)

mà 0>-27

nên \(16\sqrt5>21\sqrt2>4\sqrt3>-27\)

\(\sqrt{7-2\sqrt{10}}=\sqrt{5-2\sqrt{10}+2}\\ =\sqrt{\sqrt{5}^2-2.\sqrt{5}.\sqrt{2}+\sqrt{2}^2}\\ =\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\left|\sqrt{5}-\sqrt{2}\right|\\ =\sqrt{5}-\sqrt{2}\)

\(\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{2}.\sqrt{5}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\left(vì\sqrt{5}-\sqrt{2}>0\right)\)

Lời giải:

$\sqrt{11-2\sqrt{10}}=\sqrt{10-2\sqrt{10}+1}=\sqrt{(\sqrt{10}-1)^2}$

$=|\sqrt{10}-1|=\sqrt{10}-1$

P/s: Lần sau bạn lưu ý đăng đầy đủ yêu cầu đề bài.

\(\sqrt{11-2\sqrt{10}}=\sqrt{10-2\sqrt{10}+1}\\ =\sqrt{\sqrt{10}^2-2.\sqrt{10}.1+1^2}\\ =\sqrt{\left(\sqrt{10}-1\right)^2}\\ =\left|\sqrt{10}-1\right|=\sqrt{10}-1\)

`a)`\(\sqrt{3x+1}-\sqrt{x-1}=2\)

\(ĐK:x\ge1\)

\(\Leftrightarrow3x+1+x-1-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)

\(\Leftrightarrow4x-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)

\(\Leftrightarrow\sqrt{\left(3x+1\right)\left(x-1\right)}=2x-2\)

\(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=4x^2-8x+4\)

\(\Leftrightarrow3x^2-3x+x-1=4x^2-8x+4\)

\(\Leftrightarrow x^2-6x+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\) `(tm)`

Vậy \(S=\left\{5;1\right\}\)

`b)`\(\dfrac{\sqrt{x+27}+\sqrt{27-x}}{\sqrt{27+x}-\sqrt{27-x}}=\dfrac{27}{x}\)

\(ĐK:-27\le x\le27;x\ne0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}+\sqrt{27-x}\right)\left(\sqrt{x+27}-\sqrt{27-x}\right)}{\left(\sqrt{27+x}-\sqrt{27-x}\right)\left(\sqrt{27+x}-\sqrt{27-x}\right)}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}\right)^2-\left(\sqrt{27-x}\right)^2}{\left(\sqrt{27+x}-\sqrt{27-x}\right)^2}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{2x}{54-2\sqrt{\left(27+x\right)\left(27-x\right)}}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{x}{27-\sqrt{27^2-x^2}}=\dfrac{27}{x}\)
\(\Leftrightarrow x^2=27^2-\sqrt{27^2-x^2}\)

\(\Leftrightarrow27^2-x^2-\sqrt{27^2-x^2}=0\)

\(\Leftrightarrow\sqrt{27^2-x^2}\left(\sqrt{27^2-x^2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}27^2-x^2=0\\\sqrt{27^2-x^2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm27\\x=\pm\sqrt{27^2-1}\end{matrix}\right.\)

Thế vào pt ta được \(x=\pm27\) là thỏa mãn

Vậy \(S=\left\{\pm27\right\}\)

a) ĐKXĐ: \(x\ge1\)

Ta có: \(\sqrt{3x+1}-\sqrt{x-1}=2\)

\(\Leftrightarrow\sqrt{3x+1}-4+2-\sqrt{x-1}=0\)

\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{4-x+1}{2+\sqrt{x-1}}=0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{5-x}{2+\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}\right)=0\)

+ Nếu \(x-5=0\Leftrightarrow x=5\left(tmđkxđ\right)\)

+ Nếu \(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}=0\Leftrightarrow\dfrac{3}{\sqrt{3x+1}+4}=\dfrac{1}{2+\sqrt{x-1}}\)

\(\Leftrightarrow3\sqrt{x-1}+2=\sqrt{3x+1}\Rightarrow6x-6+12\sqrt{x-1}=0\)

\(\sqrt{x-1}\left(6\sqrt{x-1}+12\right)=0\Leftrightarrow\sqrt{x-1}=0\)

\(\Leftrightarrow x=1\left(tmđkxđ\right)\)

Vậy ....

ĐKXĐ: \(a>0;a\ne1\)

Ta có: \(P=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right):\left(\dfrac{a^2+a\sqrt{a}}{\sqrt{a}+1}\right)\)

\(=\left(\dfrac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\left(\dfrac{\sqrt{a}+1}{a\sqrt{a}\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\dfrac{4a\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right).\left(\dfrac{1}{a\sqrt{a}}\right)\)

\(=\dfrac{4}{a-1}\)

Để \(P>2\) thì \(\dfrac{4}{a-1}>2\Leftrightarrow\dfrac{4}{a-1}-2>2\)

\(\Leftrightarrow\dfrac{2\left(a-3\right)}{1-a}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a-3>0\\1-a>0\end{matrix}\right.\\\left\{{}\begin{matrix}a-3< 0\\1-a< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a>3\\a< 1\end{matrix}\right.\\\left\{{}\begin{matrix}a< 3\\a>1\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow1< a< 3\)

Vậy khi \(1< a< 3\) thì \(P>2\)

VI. Rewrite each of the following sentences in such a way that it means exacly the same as the provided sentences

1. I am sorry I can't finish the report now

-> I wish..I could finish the report now....

2. We started building the bridge six months ago

->We have..built the bridge for 6 months.

3. You must hand in your report before the end of this month

->Your report.must be handed in before the end of this month...

4.Be careful or you will cut yourself

->If..you aren't careful, you will cut yourself...

5. As far as I know, there is no reason for Linda to be so unhappy

->To the..best of my knowledge, there is no reason for Linda to be so unhappy..

VII.Use the follwing sets od words or phrase to write complete sentences.

1. His uncle/teach/English/our school/ five years/ago.

->.His uncle taught English at our school five years ago..

2. It/be/difficult/Lan/speak/English/fluently

-> It is difficult for Lan to speak English fluently.

1. I wish I had finished the report now.

2. We have started building the bridge for six month.

3. Your repotrt must be handed in before the end of this month.

4. 

5. 

VII.

1. His uncle taught English at our school five years ago.

2. It is difficult for Lan to speak English fluently.

\(R=\dfrac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(x>=0;x\ne1\right)\\ =\dfrac{3x+3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{3x+3\sqrt{x}-3-\left(x-1\right)-\left(x+4\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

\(R=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1-2}{\sqrt{x}-1}=1-\dfrac{2}{\sqrt{x}-1}\\ R\in Z=>\dfrac{2}{\sqrt{x}-1}\in Z\\ =>2⋮\left(\sqrt{x}-1\right)=>\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\\ =>\sqrt{x}\in\left\{2;3;0;-1\right\}\\ =>x\in\left\{4;9;0\right\}\left(TMDK\right)\)