\(\dfrac{4}{\sqrt{5}-3}-\dfrac{4}{\sqrt{5}+3}\\ =\dfrac{4\left(\sqrt{5}+3\right)}{5-9}-\dfrac{4\left(\sqrt{5}-3\right)}{5-9}\\ =\dfrac{4\left(\sqrt{5}+3\right)}{-4}-\dfrac{4\left(\sqrt{5}-3\right)}{-4}\\ =-\left(\sqrt{5}+3\right)+\sqrt{5}-3\\ =-\sqrt{5}-3+\sqrt{5}-3\\ =-6\)

ĐK: \(x\ge5;x\le1\)

PT trở thành:

\(\sqrt{4}.\sqrt{x-5}-\dfrac{3\sqrt{x-5}}{3}=\sqrt{1-x}\\ \Leftrightarrow2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\\ \Leftrightarrow\sqrt{x-5}=\sqrt{1-x}\\ \Leftrightarrow x-5=1-x\\ \Leftrightarrow x-5-1+x=0\\ \Leftrightarrow2x-6=0\\ \Leftrightarrow x=3\left(loại\right)\)

Vậy PT vô nghiệm.

`HaNa♬D`

a: \(=\dfrac{4\left(\sqrt{5}+3\right)-4\left(\sqrt{5}-3\right)}{5-9}=\dfrac{4\left(\sqrt{5}+3-\sqrt{5}+3\right)}{-4}=-6\)

b: ĐKXĐ: x-5>=0 và 1-x<=0

=>x>=5 và x<=1

=>Không có x thỏa mãn ĐKXĐ

=>PT vô nghiệm

a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

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

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

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

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

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

=>\(\dfrac{3}{2}\sqrt{x-5}=3\)

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

=>x-5=4

=>x=9(nhận)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

\(1,\sqrt{x+2+4\sqrt{x-2}}=5\left(x\ge2\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-2}+4\right)^2}=5\\ \Leftrightarrow\sqrt{x-2}+4=5\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\Leftrightarrow x=3\\ 2,\sqrt{x+3+4\sqrt{x-1}}=2\left(x\ge1\right)\\ \Leftrightarrow\sqrt{\left(\sqrt{x-1}+4\right)^2}=2\\ \Leftrightarrow\sqrt{x-1}+4=2\\ \Leftrightarrow\sqrt{x-1}=-2\\ \Leftrightarrow x\in\varnothing\left(\sqrt{x-1}\ge0\right)\)

\(3,\sqrt{x+\sqrt{2x-1}}=\sqrt{2}\left(x\ge\dfrac{1}{2};x\ne1\right)\\ \Leftrightarrow x+\sqrt{2x-1}=2\\ \Leftrightarrow x-2=-\sqrt{2x-1}\\ \Leftrightarrow x^2-4x+4=2x-1\\ \Leftrightarrow x^2-6x+5=0\\ \Leftrightarrow\left(x-5\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(loại\right)\end{matrix}\right.\)

\(4,\sqrt{x-2+\sqrt{2x-5}}=3\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}=6\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}=6\\ \Leftrightarrow\sqrt{2x-5}+1=6\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(TM\right)\)

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

a) ĐKXĐ: \(x\ge0\)

Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)

\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)

\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)

a, ĐK: \(x\ge11\)

\(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)

\(\Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{x^2-x+11}=16\)

\(\Leftrightarrow2x+2\sqrt{x^2-x+11}=16\)

\(\Leftrightarrow x+\sqrt{x^2-x+11}=8\)

Ta thấy \(x+\sqrt{x^2-x+11}>11>\text{​​}8\)

\(\Rightarrow\) phương trình vô nghiệm.

\(a,\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\left(x\ge11\right)\\ \Leftrightarrow x+\sqrt{x-11}+x-\sqrt{x-11}+2\sqrt{\left(x+\sqrt{x-11}\right)\left(x-\sqrt{x-11}\right)}=16\\ \Leftrightarrow2x+2\sqrt{x^2-x+11}=16\\ \Leftrightarrow x+\sqrt{x^2-x+11}=8\\ \Leftrightarrow\sqrt{x^2-x+11}=8-x\\ \Leftrightarrow x^2-x+11=x^2-16x+64\\ \Leftrightarrow15x=53\\ \Leftrightarrow x=\dfrac{53}{15}\left(ktm\right)\)

\(b,\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\left(x\ge\dfrac{5}{2}\right)\\ \Leftrightarrow\sqrt{2x-5+6\sqrt{2x-5}+9}+\sqrt{2x-5-2\sqrt{2x-5}+1}=4\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\\ \Leftrightarrow\sqrt{2x-5}+3+\left|\sqrt{2x-5}-1\right|=4\\ \Leftrightarrow\left|\sqrt{2x-5}-1\right|=1-\sqrt{2x-5}\\ \Leftrightarrow\sqrt{2x-5}-1\le0\\ \Leftrightarrow\sqrt{2x-5}\le1\\ \Leftrightarrow2x-5\le1\Leftrightarrow x\le\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{2}\)

b, ĐK: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)

\(\Leftrightarrow\sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-\sqrt{2x-5}}=4\)

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

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

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\):

\(\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}-1\right|\)

\(=\left|\sqrt{2x-5}+3\right|+\left|1-\sqrt{2x-5}\right|\)

\(\ge\left|\sqrt{2x-5}+3+1-\sqrt{2x-5}\right|\)

\(=4\)

Đẳng thức xảy ra khi: 

\(\left(\sqrt{2x-5}+3\right)\left(1-\sqrt{2x-5}\right)\ge0\)

\(\Leftrightarrow1-\sqrt{2x-5}\ge0\)

\(\Leftrightarrow\sqrt{2x-5}\le1\)

\(\Leftrightarrow0\le2x-5\le1\)

\(\Leftrightarrow\dfrac{5}{2}\le x\le3\)

a) đk: \(1\le x\le5\)

 \(\sqrt[4]{5-x}+\sqrt[4]{x-1}=\sqrt{2}\)

<=> \(\left(\sqrt[4]{5-x}+\sqrt[4]{x-1}\right)^4=\sqrt{2}^4\)

<=> \(5-x+x-1+4\sqrt[4]{5-x}^3.\sqrt[4]{x-1}+6\sqrt[4]{5-x}^2.\sqrt[4]{x-1}^2+4\sqrt[4]{5-x}.\sqrt[4]{x-1}^3=4\)

<=> \(\sqrt[4]{\left(5-x\right)\left(x-1\right)}.\left(2\sqrt[4]{5-x}^2+3\sqrt[4]{5-x}.\sqrt[4]{x-1}+2\sqrt[4]{x-1}^2\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt[4]{\left(5-x\right)\left(x-1\right)}=0\left(2\right)\\2\sqrt[4]{5-x}^2+3\sqrt[4]{\left(5-x\right)\left(x-1\right)}+2\sqrt[4]{x-1}^2=0\left(1\right)\end{matrix}\right.\)

Giải (2) <=> \(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\left(tm\right)\)

Giải (1) : Đặt \(\sqrt[4]{5-x}=a;\sqrt[4]{x-1}=b\)(đk : a, b \(\ge\)0)

Khi đó, ta có: \(2a^2+3ab+2b^2=0\)

<=> 2(a² + 3/2ab + 9/16b²) + \(\dfrac{7}{8}b^2=0\)

<=> \(2\left(a+\dfrac{3}{4}b\right)^2+\dfrac{7}{8}b^2=0\)

<=> \(\left\{{}\begin{matrix}a+\dfrac{3}{4}b=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\sqrt[4]{x-1}=0\\\sqrt[4]{5-x}=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)(vô lí)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x-3>=0\\5-x>=0\end{matrix}\right.\)

=>3<=x<=5

\(\sqrt{x-3}+\sqrt{5-x}=2\)

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

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

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

=>x-4=0

=>x=4