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

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

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

\(\left\{{}\begin{matrix}2>0\left(luondung\right)\\x-5=4\end{matrix}\right.\)\(\Rightarrow x=9\left(tm\right)\)

ĐK: \(x\ge0\)\(4\sqrt{x}-2\sqrt{9x}+16\sqrt{x}=5\)  5  (=) \(\sqrt{x}\left(4-2\sqrt{9}+16\right)=5\) (=) \(\sqrt{x}.14=5\)(=) x=\(\frac{25}{196}\)

ĐK: \(x\ge-5\)PT(=) \(\sqrt{5+x}\left(\sqrt{4}-3+\frac{4}{3}.3\right)=6\) (=) \(\sqrt{5+x}.3=6\) (=)\(\sqrt{5+x}=2\)(=) X = -1 (nhận)

\(a,\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)\(ĐKXĐ:x\ge-\frac{5}{7}\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow9x-7x=5+7\)

\(\Leftrightarrow2x=12\)

\(\Leftrightarrow x=6\)

\(b,\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3.\frac{\sqrt{x-5}}{\sqrt{9}}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\)

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

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

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

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

\(\Leftrightarrow x-5=4\)

\(\Leftrightarrow x=9\)

a) \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (1)

\(\Leftrightarrow9x-7=\sqrt{\left(7x+5\right)\left(7x+5\right)}\)

\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)\left(7x+5\right)}=7\)

\(\Leftrightarrow9x-\sqrt{\left(7x+5\right)^2}=7\)

\(\Leftrightarrow9x-\left|7x+5\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}9x-\left(7x+5\right)=7\left(đk:7x+5\ge0\right)\\9x-\left[-\left(7x+5\right)\right]=7\left(đk:7x+5< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\left(đk:x\ge-\dfrac{5}{7}\right)\\x=\dfrac{1}{8}\left(đk:x< -\dfrac{5}{7}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x\in\varnothing\end{matrix}\right.\)

\(\Leftrightarrow x=6\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{6\right\}\)

b) \(\sqrt{4x-20}+3\sqrt{\dfrac{x+5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\) (2)

\(\Leftrightarrow\sqrt{4\left(x-5\right)}+3\cdot\dfrac{\sqrt{x+5}}{3}-\dfrac{1}{3}\cdot\sqrt{9\left(x-5\right)}=4\)

\(\Leftrightarrow\sqrt{4}\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot\sqrt{9}\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

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

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

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

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

\(\Leftrightarrow x-5=16-8\sqrt{x+5}+x+5\)

\(\Leftrightarrow-5=16-8\sqrt{x+5}+5\)

\(\Leftrightarrow-5=21-8\sqrt{x+5}\)

\(\Leftrightarrow8\sqrt{x+5}=21+5\)

\(\Leftrightarrow8\sqrt{x+5}=26\)

\(\Leftrightarrow\sqrt{x+5}=\dfrac{13}{4}\)

\(\Leftrightarrow x+5=\dfrac{169}{16}\)

\(\Leftrightarrow x=\dfrac{169}{16}-5\)

\(\Leftrightarrow x=\dfrac{89}{16}\)

Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{89}{16}\right\}\)

Giải:

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

\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}\) + 3\(\frac{\sqrt{x-5}}{\sqrt{9}}\)-\(\frac{1}{3}\)\(\sqrt{9\left(x-5\right)}\)=4

\(\Leftrightarrow\)\(\sqrt{4}\)\(\sqrt{x-5}\)+ 3\(\frac{\sqrt{x-5}}{3}\)-\(\frac{1}{3}\)\(\sqrt{9}\)\(\sqrt{x-5}\)= 4

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

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

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

\(\Leftrightarrow\)\(\sqrt{x-5}\)= 2 . Đk : x \(\ge\)5

\(\Rightarrow\)x - 5 = 4

\(\Leftrightarrow\)x = 9 ( thỏa mãn )

Vậy phương trình đã cho có tập nghiệm S = \(\left\{9\right\}\)

a) \(\sqrt{\left(2x-1\right)^2}=3\)

⇔ \(\left|2x-1\right|=3\)

⇔ \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

⇔ \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

b) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)

ĐKXĐ : \(x\ge0\)

⇔ \(3\sqrt{x}-2\sqrt{3^2x}+\sqrt{4^2x}=5\)

⇔ \(3\sqrt{x}-2\cdot3\sqrt{x}+4\sqrt{x}=5\)

⇔ \(7\sqrt{x}-6\sqrt{x}=5\)

⇔ \(\sqrt{x}=5\)

⇔ \(x=25\)( tm )

c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)

ĐKXĐ : \(x\ge-5\)

⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)

⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot3\sqrt{x+5}=6\)

⇔ \(-\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)

⇔ \(\frac{5}{4}\sqrt{x+5}=6\)

⇔ \(\sqrt{x+5}=\frac{24}{5}\)

⇔ \(x+5=\frac{576}{25}\)

⇔ \(x=\frac{451}{25}\left(tm\right)\)

\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)