Bạn chú ý lần sau ghi đầy đủ đề bài! Những bài cộc lốc thế này lần sau mình sẽ xóa không thương tiếc nhé.

Coi đây là bài toán rút gọn.

Lời giải:

$\sqrt{9-2\sqrt{18}}+\sqrt{9+2\sqrt{18}}$

$=\sqrt{3+6-2\sqrt{3}.\sqrt{6}}+\sqrt{3+6+2\sqrt{3}.\sqrt{6}}$

$=\sqrt{(\sqrt{3}-\sqrt{6})^2}+\sqrt{(\sqrt{3}+\sqrt{6})^2}$

$=|\sqrt{3}-\sqrt{6}|+|\sqrt{3}+\sqrt{6}|$

$=\sqrt{6}-\sqrt{3}+\sqrt{3}+\sqrt{6}=2\sqrt{6}$

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

=>\(\sqrt{x-2}=10\)

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

\(\sqrt{4\left(x-3\right)^2}=8\)

=>\(\sqrt{\left(2x-6\right)^2}=8\)

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

=>\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

`a)sqrt{5x-2}=3(x>=2/5)`

`<=>5x-2=9`

`<=>5x=11`

`<=>x=11/5(tm)`

`b)sqrt{x^2-4x+4}-5=0`

`<=>\sqrt{(x-2)^2}=5`

`<=>|x-2|=5`

`<=>` \(\left[ \begin{array}{l}x-2=5\\x-2=-5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=7\\x=-3\end{array} \right.\) 

`c)3sqrt{4x+8}-sqrt{9x+18}+9sqrt{(x+2)/9}=sqrt{72}(x>=-2)`

`<=>6sqrt{x+2}-3sqrt{x+2}+3sqrt{x+2}=sqrt{72}`

`<=>6sqrt{x+2}=6sqrt2`

`<=>sqrt{x+2}=sqrt2`

`<=>x+2=2`

`<=>x=0(tm)`

\(a,ĐK:x\ge\dfrac{2}{5}\)

\(\Leftrightarrow5x-2=9\)

\(\Leftrightarrow5x=11\)

\(\Leftrightarrow x=\dfrac{11}{5}\)

\(b,\)

\(\Leftrightarrow x^2-5x+4=25\)

\(\Leftrightarrow x^2-5x-21=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{109}}{2}\\x=\dfrac{5-\sqrt{109}}{2}\end{matrix}\right.\)

\(c,\)

\(\Leftrightarrow6\sqrt{x+2}-3\sqrt{x+2}+9\cdot\sqrt{\dfrac{x+2}{9}}=6\sqrt{2}\)

\(\Leftrightarrow2\sqrt{x+2}-\sqrt{x+2}+3\cdot\sqrt{\dfrac{x+2}{9}}=2\sqrt{2}\)

Đặt \(\sqrt{x+2}=a\) ta có (1)

\(2a-a+3\cdot\dfrac{a}{\sqrt{9}}=2\sqrt{2}\)

\(\Leftrightarrow a+3\cdot\dfrac{a}{3}=2\sqrt{2}\)

\(\Leftrightarrow2a=2\sqrt{2}\)

\(\Leftrightarrow a=\sqrt{2}\)

Thay \(a=\sqrt{2}\) vào (1) ta có

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

\(\Leftrightarrow x+2=2\)

\(\Leftrightarrow x=0\)

a: \(\sqrt{169}-\sqrt{225}\)

\(=\sqrt{13^2}-\sqrt{15^2}\)

=13-15

=-2

b: \(\dfrac{\sqrt{144}}{9}\)

\(=\dfrac{\sqrt{12^2}}{9}\)

\(=\dfrac{12}{9}=\dfrac{4}{3}\)

c: \(\sqrt{18}:\sqrt{2}=\sqrt{\dfrac{18}{2}}=\sqrt{9}=3\)

\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)

=\(\sqrt{21-2\sqrt{54}}+\sqrt{6+2\sqrt{18}+3}-\sqrt{4\cdot\left(6+3\sqrt{3}\right)}\)

=\(\sqrt{18-2\sqrt{54}+3}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+12\sqrt{3}}\)

=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{24+2\sqrt{108}}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\sqrt{\sqrt{18}+2\sqrt{108}+\sqrt{6}}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\left|\sqrt{18}-\sqrt{6}\right|\)

=\(\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}\)

= 0

Hic câu dưới bị giải nhầm nha bạn :<

\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)

=\(\sqrt{21-2\sqrt{54}}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+12\sqrt{3}}\)

=\(\sqrt{18-2\sqrt{54}+3}+\sqrt{6+2\sqrt{18}+3}-\sqrt{24+2\sqrt{108}}\)

=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{18+2\sqrt{108}+6}\)

=\(\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}\)

=\(\left|\sqrt{18}-\sqrt{3}\right|+\left|\sqrt{6}+\sqrt{3}\right|-\left|\sqrt{18}+\sqrt{6}\right|\)

=\(\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}\)

=0

0 nhé bạn, thực ra thì tui bấm máy tính, chớ tui ms hc lớp 7 hà,

tích vs nhé

nếu bấm máy thì chắc chị giải đc rồi em :> 

Bấm mode 53 chưa bạn??

\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)

\(=3\sqrt{2}-\sqrt{3}+\sqrt{3}+\sqrt{6}-3\sqrt{2}-\sqrt{6}=0\)