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

\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\\ =2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\\ =3-\sqrt{7}\)

\(\sqrt{\left(x-3\right)^2}\\ =\left|x-3\right|\\ =x-3\left(vì.x>3\right)\)

\(\sqrt{\left(1-x\right)^2}\\ =\left|1-x\right|\\ =x-1\left(vì.x>1\right)\)

\(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}\\ =\left|3a^2\right|\\ =3a^2\)

\(\sqrt{100a^2}\\ =\sqrt{\left(10a\right)^2}\\ =\left|10a\right|\\ =-10a\left(vì.a< 0\right)\)

Lời giải:

a. $=|2-\sqrt{5}|+|2\sqrt{2}-\sqrt{5}|$

$=(\sqrt{5}-2)+(2\sqrt{2}-\sqrt{5})=-2+2\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|+|3-2\sqrt{2}|=2\sqrt{2}-\sqrt{7}+(3-2\sqrt{2})$

$=3-\sqrt{7}$

c.

$=|x-3|=x-3$
d.

$=|1-x|=x-1$

$=\sqrt{(3a^2)^2}=|3a^2|=3a^2$
e.

$=\sqrt{(10a)^2}=|10a|=-10a$

a) Ta có: \(\left(\dfrac{1}{2-\sqrt{3}}-\dfrac{3}{\sqrt{7}-2}\right):\dfrac{2}{\sqrt{7}+\sqrt{3}}\)

\(=\left(2+\sqrt{3}-\sqrt{7}-2\right):\dfrac{\left(\sqrt{7}-\sqrt{3}\right)}{2}\)

\(=\dfrac{-\left(\sqrt{7}-\sqrt{3}\right)}{1}\cdot\dfrac{2}{\sqrt{7}-\sqrt{3}}\)

=-2

b) Ta có: \(\left(\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-1\right):\left(\sqrt{x}-x\right)+\dfrac{1}{x}\)

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

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

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

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

\(a,\cdot\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}\\ =\left[\left(8:2,4\right)\cdot\left(5,25:7\right)\right]:\left[\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)\right]\\ =\left(\dfrac{10}{3}\cdot\dfrac{3}{4}\right):\left(3:\dfrac{9}{2}\right)\\ =\dfrac{5}{2}:\dfrac{2}{3}\\ =\dfrac{15}{4}\)

a: \(\dfrac{\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}^2\right)\right]\right\}}{\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}}\)

\(=\dfrac{\dfrac{8}{2,4}\cdot\dfrac{5,25}{7}}{\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)}\)

\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{3}{4}}{3:\left(4\cdot\dfrac{9}{8}\right)}\)

\(=\dfrac{\dfrac{10}{4}}{3:\left(\dfrac{9}{2}\right)}=\dfrac{5}{2}:\left(3\cdot\dfrac{2}{9}\right)=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{15}{4}\)

b: \(\sqrt{\left(x-\sqrt{2}\right)^2}=\left|x-\sqrt{2}\right|>=0\forall x\)

\(\sqrt{\left(y+\sqrt{2}\right)^2}=\left|y+\sqrt{2}\right|>=0\forall y\)

\(\left|x+y+z\right|>=0\forall x,y,z\)

Do đó: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{matrix}\right.\)

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

1) \(\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)

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

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

= \(\frac{7+2\sqrt{35}+5+7-2\sqrt{35}+5}{7-5}\) = \(\frac{24}{2}=12\)

2) \(x+2y-\sqrt{\left(x^2-4xy+4y^2\right)^2}\left(x\ge2y\right)\)

= \(x+2y-\sqrt{\left(x-2y\right)^4}\) = \(x+2y-|x-2y|\)

= \(x+2y-\left(x-2y\right)\) = \(x+2y-x+2y=4y\)

3)\(4x+\sqrt{\left(x-12\right)^2}\left(x\ge2\right)\)

= \(4x+x-12=5x-12\)

Yêu cầu?

đặt \(\sqrt[3]{2-x}=a;\sqrt[3]{7+x}=b\rightarrow a^3+b^3=9\)

thay vào pt ta đc

\(a^2+b^2-ab=\dfrac{\left(a^3+b^3\right)}{3}\)

\(a^2+b^2-ab=\dfrac{\left(a+b\right)\left(a^2+b^2-ab\right)}{3}\)

do \(a^2+b^2-ab>0\)nên

a+b=3

\(\rightarrow\sqrt[3]{2-x}+\sqrt[3]{7+x}=3\)

\(\left(\sqrt[3]{2-x}+\sqrt[3]{7+x}\right)^3=27\)

\(2=\sqrt[3]{\left(7+x\right)\left(2-x\right)}\)

0=6-5x-x^2 đến đấy khá đơn giản rồi nhỉ

(x-1)(x+6)=0

vậy pt có nghiệm x=1;x=-6

Lời giải:

Đặt \(\sqrt[3]{2-x}=a; \sqrt[3]{7+x}=b(*)\). Ta có hệ phương trình:

\(\left\{\begin{matrix} a^3+b^3=9\\ a^2+b^2-ab=3\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} (a+b)(a^2-ab+b^2)=9\\ a^2+b^2-ab=3\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2-ab=3\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} a+b=3\\ (a+b)^2-3ab=3\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a+b=3\\ ab=2\end{matrix}\right.\)

Theo định lý Viete đảo thì $a,b$ là nghiệm của pt:

\(x^2-3x+2=0\), do đó \((a,b)=(1,2)\) hoặc \((a,b)=(2,1)\)

Thay vào $(*)$ suy ra $x=1$ hoặc $x=-6$