NT
0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
AH
Akai Haruma
Giáo viên
3 tháng 7 2019
1.
Đặt \(\sqrt{a^2+x^2}=m,\sqrt{a^2-x^2}=n\Rightarrow x^2=\frac{m^2-n^2}{2}\)
\(\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{a^4}{x^4}-1}=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\sqrt{\frac{(a^2+x^2)(a^2-x^2)}{x^4}}\)
\(=\frac{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}{\sqrt{a^2+x^2}-\sqrt{a^2-x^2}}-\frac{\sqrt{(a^2+x^2)(a^2-x^2)}}{x^2}\)
\(=\frac{m+n}{m-n}-\frac{mn}{\frac{m^2-n^2}{2}}=\frac{(m+n)^2}{m^2-n^2}-\frac{2mn}{m^2-n^2}=\frac{m^2+n^2}{m^2-n^2}\)
\(=\frac{2a^2}{2x^2}=\frac{a^2}{x^2}\)
2.
\(=\left[\frac{(1-\sqrt{a})(1+\sqrt{a}+a)}{1-\sqrt{a}}+\sqrt{a}\right].\left[\frac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}\right]\)
\(=(1+\sqrt{a}+a+\sqrt{a})(1-\sqrt{a}+a-\sqrt{a})\)
\(=(a+2\sqrt{a}+1)(a-2\sqrt{a}+1)=(\sqrt{a}+1)^2(\sqrt{a}-1)^2\)
\(=(a-1)^2\)
3.
\(=\frac{3(1-x)}{\sqrt{1+x}.\sqrt{1-x}}:\frac{3+\sqrt{1-x^2}}{\sqrt{1-x^2}}=\frac{3(1-x)}{\sqrt{1-x^2}}.\frac{\sqrt{1-x^2}}{3+\sqrt{1-x^2}}=\frac{3(1-x)}{3+\sqrt{1-x^2}}\)
AH
Akai Haruma
Giáo viên
3 tháng 7 2019
4. Bạn xem lại đề xem đã đúng chưa?
5.
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{\sqrt{b}(a+\sqrt{ab})+\sqrt{b}(a-\sqrt{ab})}{(a-\sqrt{ab})(a+\sqrt{ab})}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\frac{2a\sqrt{b}}{a^2-ab}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}}.\frac{1}{a-b}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}\)
\(=\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{1}{a+\sqrt{ab}}=\frac{\sqrt{a}+\sqrt{b}}{a+\sqrt{ab}}=\frac{1}{\sqrt{a}}\)
M
11 tháng 8 2020
A= \(\left(\frac{\sqrt{b}}{a-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-b}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{\sqrt{b}}{\sqrt{a}.\sqrt{a}-\sqrt{ab}}-\frac{\sqrt{a}}{\sqrt{ab}-\sqrt{b}.\sqrt{b}}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{b}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)
A = \(\left(\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(\sqrt{a}.\sqrt{a}.\sqrt{b}-\sqrt{b}.\sqrt{b}\sqrt{a}\right)\)
A = \(\left(\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\left(\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\right)\)
A = b-a
B = \(\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}}{a-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-1}\)
B = \(\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\frac{\sqrt{a}\left(a+\sqrt{a}\right)}{a^2-a}\right).\frac{a-1}{\sqrt{a}+1}\)
B = \(\left(\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{a-1}-\frac{\sqrt{a}.\sqrt{a}\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
\(B=\left(\frac{a\sqrt{a}\left(\sqrt{a}+1\right)}{a\left(a-1\right)}-\frac{a\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
B= \(\left(\frac{a\sqrt{a}\left(\sqrt{a}+1\right)-a\left(\sqrt{a}+1\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
B= \(\left(\frac{\left(\sqrt{a}+1\right)\left(a\sqrt{a}-a\right)}{a\left(a-1\right)}\right).\frac{a-1}{\sqrt{a}+1}\)
B = \(\frac{\left(\sqrt{a}+1\right)a\left(\sqrt{a}-1\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
\(B=\frac{a\left(\sqrt{a}^2-1^2\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
\(B=\frac{a\left(a-1\right)}{a\left(a-1\right)}.\frac{a-1}{\sqrt{a}+1}\)
B = \(\frac{a-1}{\sqrt{a}+1}\)
15 tháng 7 2020
Bài 1:
b) Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(=\frac{\sqrt{2\left(4+\sqrt{7}\right)}}{\sqrt{2}}-\frac{\sqrt{2\left(4-\sqrt{7}\right)}}{\sqrt{2}}\)
\(=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}-\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}-\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}-\frac{\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}+1\right|}{\sqrt{2}}-\frac{\left|\sqrt{7}-1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Bài 2:
a) Ta có: \(\frac{a^2-\sqrt{a}}{a+\sqrt{a}+1}-\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{a+\sqrt{a}+1}-\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}\)
\(=\sqrt{a}\left(\sqrt{a}-1\right)-\sqrt{a}\left(\sqrt{a}+1\right)\)
\(=a-\sqrt{a}-a-\sqrt{a}\)
\(=-2\sqrt{a}\)
b) Ta có: \(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\frac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
\(=\sqrt{ab}-\sqrt{ab}=0\)
d) Ta có: \(\frac{a+b+2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)}\)
\(=\sqrt{a}+\sqrt{b}-\left(\sqrt{a}+\sqrt{b}\right)\)
=0
Bài 3:
a) ĐKXĐ: x≥0
Ta có: \(\frac{\sqrt{27x}}{\sqrt{3}}=6\)
\(\Leftrightarrow\frac{\sqrt{27}\cdot\sqrt{x}}{\sqrt{3}}=6\)
\(\Leftrightarrow3\cdot\sqrt{x}=6\)
\(\Leftrightarrow\sqrt{x}=\frac{6}{3}=2\)
hay \(x=4\)(thỏa mãn)
Vậy: S={4}
b) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x+1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-1\end{matrix}\right.\Leftrightarrow x\ge0\)
Ta có: \(\sqrt{x+1}=3-\sqrt{x}\)
\(\Leftrightarrow\left(\sqrt{x+1}\right)^2=\left(3-\sqrt{x}\right)^2\)
\(\Leftrightarrow x+1=9-6\sqrt{x}+x\)
\(\Leftrightarrow x+1-9+6\sqrt{x}-x=0\)
\(\Leftrightarrow-8+6\sqrt{x}=0\)
\(\Leftrightarrow6\sqrt{x}=8\)
\(\Leftrightarrow\sqrt{x}=\frac{8}{6}=\frac{4}{3}\)
hay \(x=\frac{16}{9}\)(thỏa mãn)
Vậy: \(S=\left\{\frac{16}{9}\right\}\)
20 tháng 7 2021
\(a,\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{4}{2+a+b}\)( BĐT cô-si dạng engel)
\(\frac{4}{2+a+b}\le\frac{4}{2+2\sqrt{ab}}=\frac{2}{1+\sqrt{ab}}=VP\)(bđt tương đương)
vậy cả hai bđt dấu "=" xảy ra đồng thời
\(\hept{\begin{cases}\frac{1}{1+a}=\frac{1}{1+b}\\a=b=1\end{cases}}\)
vậy \(\frac{1}{1+a}+\frac{1}{1+b}=\frac{2}{1+\sqrt{ab}}\)khi \(a=b=1\)
\(b,\)\(\frac{1}{1+a}+\frac{1}{1+b}>\frac{2}{1+\sqrt{ab}}\)khi và chỉ khi bđt cô -si không xảy ra dấu bằng
và bđt tương đương xảy ra dấu bằng
\(\hept{\begin{cases}\frac{1}{1+a}+\frac{1}{1+b}>\frac{4}{2+a+b}\\\frac{4}{2+a+b}=\frac{2}{1+\sqrt{ab}}\end{cases}}\)
\(\hept{\begin{cases}\frac{2+a+b}{1+a+b+ab}>\frac{4}{2+a+b}\\4+4\sqrt{ab}=4+2a+2b\end{cases}}\)
\(\hept{\begin{cases}4+a^2+b^2+4a+4b+2ab>4+4a+4a+4ab\\2\sqrt{ab}=a+b\end{cases}}\)
\(\hept{\begin{cases}a^2+b^2>2ab\\a^2+b^2=0\end{cases}}\)
\(0>2ab\)
\(ab< 0\)
rồi chia ra từng TH
ra đc \(TH1:\hept{\begin{cases}a< 0\\b>0\end{cases}}\)
\(TH2:\hept{\begin{cases}a>0\\b< 0\end{cases}}\)
\(c,\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{2}{1+\sqrt{ab}}\)khi và chỉ khi
bđt cô- si dạng engel lớn hơn hoặc bằng còn bđt tương đương thì dấu bằng xảy ra
\(\hept{\begin{cases}\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{4}{2+a+b}\\\frac{4}{2+a+b}=\frac{2}{1+\sqrt{ab}}\end{cases}}\)
\(\hept{\begin{cases}a^2+b^2\ge2ab\\a^2+b^2=0\end{cases}}\)
\(< =>0\ge2ab\)
vì đề bài cho \(a,b>0\)lên dấu bằng không xảy ra
vậy không có giá trị a,b nào thỏa mãn \(\frac{1}{1+a}+\frac{1}{1+b}\ge\frac{2}{1+\sqrt{ab}}\)
câu d lập luận như các câu trên cậu làm nốt nha
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 3 2019
2/
a/ \(\sqrt{a}+\frac{1}{\sqrt{a}}\ge2\sqrt{\sqrt{a}.\frac{1}{\sqrt{a}}}=2\), dấu "=" khi \(a=1\)
b/ \(a+b+\frac{1}{2}=a+\frac{1}{4}+b+\frac{1}{4}\ge2\sqrt{a.\frac{1}{4}}+2\sqrt{b.\frac{1}{4}}=\sqrt{a}+\sqrt{b}\)
Dấu "=" khi \(a=b=\frac{1}{4}\)
c/ Có lẽ bạn viết đề nhầm, nếu đề đúng thế này thì mình ko biết làm
Còn đề như vậy: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\) thì làm như sau:
\(\frac{1}{x}+\frac{1}{y}\ge\frac{2}{\sqrt{xy}}\) ; \(\frac{1}{y}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\); \(\frac{1}{x}+\frac{1}{z}\ge\frac{2}{\sqrt{yz}}\)
Cộng vế với vế ta được:
\(\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\ge\frac{2}{\sqrt{xy}}+\frac{2}{\sqrt{yz}}+\frac{2}{\sqrt{xz}}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\)
Dấu "=" khi \(x=y=z\)
d/ \(\frac{\sqrt{3}+2}{\sqrt{3}-2}-\frac{\sqrt{3}-2}{\sqrt{3}+2}=\frac{\left(\sqrt{3}+2\right)\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\frac{\left(\sqrt{3}-2\right)\left(\sqrt{3}-2\right)}{\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)}\)
\(=\frac{7+4\sqrt{3}}{3-4}-\frac{7-4\sqrt{3}}{3-4}=-7-4\sqrt{3}+7-4\sqrt{3}=-8\sqrt{3}\)
e/ \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}:\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}=\frac{\left(a-b\right)\left(a+b-\sqrt{ab}\right)}{\sqrt{ab}}\)
\(=\frac{a^2-b^2}{\sqrt{ab}}-\left(a-b\right)\) (bạn chép đề sai)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 6 2020
\(\frac{1}{2}\left(a+b\right)^2\le a^2+b^2=ab\left(a+b\right)+ab\le ab\left(a+b\right)+\frac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\frac{1}{2}\left(a+b\right)^2\le ab\left(a+b\right)+\frac{1}{4}\left(a+b\right)^2\)
\(\Rightarrow\frac{1}{4}\left(a+b\right)^2\le ab\left(a+b\right)\Rightarrow a+b\le4ab\)
\(\Rightarrow\frac{a+b}{ab}\le4\)
\(P=\frac{\sqrt{b\left(a+b\right)}}{ab}+\frac{\sqrt{a\left(a+b\right)}}{ab}=\frac{1}{2\sqrt{2}}\left(\frac{2\sqrt{2b\left(a+b\right)}+2\sqrt{2a\left(a+b\right)}}{ab}\right)\)
\(P\le\frac{1}{2\sqrt{2}}\left(\frac{2b+a+b+2a+a+b}{ab}\right)=\sqrt{2}\left(\frac{a+b}{ab}\right)\le4\sqrt{2}\)
\(P_{max}=4\sqrt{2}\) khi \(a=b=\frac{1}{2}\)
30 tháng 10 2020
a) Ta có: \(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
b)Sửa đề: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
Ta có: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}\)
\(=-2\sqrt{b}\)
c) Ta có: \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
d) Ta có: \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-2\sqrt{ab}+b\right)\cdot\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
e) Ta có: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)