a) \(\sqrt{\left(x-3\right)^2}=2\Leftrightarrow\left|x-3\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\left(đk:x\ge-2\right)\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\Leftrightarrow\sqrt{x+2}=3\Leftrightarrow x+2=9\Leftrightarrow x=7\)

a: \(\sqrt{\left(x-3\right)^2}=2\)

\(\Leftrightarrow\left|x-3\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

a) Ta có: \(\sqrt{\left(x-3\right)^2}=2\)

\(\Leftrightarrow\left|x-3\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b) ĐKXĐ: \(x\ge-2\)

Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+\dfrac{4}{5}\cdot5\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7(thỏa ĐK)

a) \(\Leftrightarrow\left|x-3\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

Vậy:.....

b) ĐKXĐ: x ≥ -2

 \(\Leftrightarrow\sqrt{9}.\sqrt{x+2}-5.\sqrt{x+2}+\dfrac{4}{5}.\sqrt{25}.\sqrt{x+2}=6\)

<=> \(\sqrt{x+2}.\left(3-5+\dfrac{4}{5}.5\right)=6\)

\(\Leftrightarrow2.\sqrt{x+2}=6\)

\(\Leftrightarrow\sqrt{x+2}=3\)

<=> x + 2 = 9

<=> x = 7

Tham khảo ạ:

a) Ta có: 

⇔[x−3=2x−3=−2⇔[x=5x=1⇔[x−3=2x−3=−2⇔[x=5x=1

b) ĐKXĐ: 

Ta có: ⇔3√x+2−5√x+2+45⋅5√x+2=6⇔3x+2−5x+2+45⋅5x+2=6

hay x=7(thỏa ĐK)

a) \(\sqrt{\left(x-3\right)^2}=2\Rightarrow\left|x-3\right|=2\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)

b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Rightarrow\sqrt{9\left(x+2\right)}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25\left(x+2\right)}=6\)

\(\Rightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Rightarrow2\sqrt{x+2}=6\Rightarrow\sqrt{x+2}=3\Rightarrow x+2=9\Rightarrow x=7\)

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

Ta có: \(x-2\sqrt{x}+3=x-2\sqrt{x}+1+2=\left(\sqrt{x}-1\right)^2+2\ge2\)

\(\Rightarrow\dfrac{1}{x-2\sqrt{x}+3}\le2\Rightarrow Q_{max}=2\) khi \(x=1\)

a) Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2\sqrt{x-2}+\sqrt{9x-18}\)

\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}=2\sqrt{x-2}+3\sqrt{x-2}\)

\(\Leftrightarrow\sqrt{25x+75}=\sqrt{4x-8}\)

\(\Leftrightarrow25x-4x=-8-75\)

\(\Leftrightarrow21x=-83\)

hay \(x=-\dfrac{83}{21}\)

b) Ta có: \(\sqrt{\left(2x-1\right)^2}=4\)

\(\Leftrightarrow\left|2x-1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

c) Ta có: \(\sqrt{\left(2x+1\right)^2}=3x-5\)

\(\Leftrightarrow\left|2x+1\right|=3x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x-5\left(x\ge-\dfrac{1}{2}\right)\\2x+1=5-3x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3x=-5-1\\2x+3x=5-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(nhận\right)\\x=\dfrac{4}{5}\left(loại\right)\end{matrix}\right.\)

d) Ta có: \(\sqrt{4x-12}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-3}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

\(\Leftrightarrow2\sqrt{x-3}-5\sqrt{x-2}=8\)

\(\Leftrightarrow4\left(x-3\right)+25\left(x-2\right)-20\sqrt{x^2-5x+6}=8\)

\(\Leftrightarrow4x-12+25x-50-8=20\sqrt{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow20\sqrt{\left(x-2\right)\left(x-3\right)}=29x-70\)

\(\Leftrightarrow x^2-5x+6=\dfrac{\left(29x-70\right)^2}{400}\)

\(\Leftrightarrow x^2-5x+6=\dfrac{841}{400}x^2-\dfrac{203}{20}x+\dfrac{49}{4}\)

\(\Leftrightarrow\dfrac{-441}{400}x^2+\dfrac{103}{20}x-\dfrac{25}{4}=0\)

\(\Delta=\left(\dfrac{103}{20}\right)^2-4\cdot\dfrac{-441}{400}\cdot\dfrac{-25}{4}=-\dfrac{26}{25}\)(Vô lý)

vậy: Phương trình vô nghiệm

Bài 1:

Ta có: \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\)

\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)

\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)

\(\Leftrightarrow6\sqrt{x-2}=18\)

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

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

\(\Leftrightarrow x-2=9\)

\(\Leftrightarrow x=11\)

Vậy tập nghiệm của PT \(S=\left\{11\right\}\)

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x+3\right)^2}=12\)

=>\(\left|x+3\right|=12\)

=>\(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)

b: ĐKXĐ: x>=1

\(\sqrt{25x-25}-\sqrt{9x-9}=10\)

=>\(5\sqrt{x-1}-3\sqrt{x-1}=10\)

=>\(2\sqrt{x-1}=10\)

=>x-1=25

=>x=26(nhận)

a) \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

<=> \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)

<=>\(\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

<=>\(\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+\frac{6}{2}\right)=-17\)

<=>\(\sqrt{x-1}=-17\)

<=>x-1=17

<=>x=18

Vậy pt có nghiệm là x=18

\(a.ĐK:x-1\ge0\Leftrightarrow x\ge1\)

\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{27}{2}\sqrt{x-1}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{27}{2}+24\sqrt{\frac{1}{64}}\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}.\left(-10\right)=-17\)

\(\Leftrightarrow\sqrt{x-1}=\frac{-17}{-10}=\frac{17}{10}\)

\(\Leftrightarrow x-1=\left(\frac{17}{10}\right)^2\)

\(\Leftrightarrow x=\frac{289}{100}+1=3,89\left(TM\right)\)

Vậy \(S=\left\{3,89\right\}\)

\(b.ĐK:x^2+2\ge0\)

\(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)

\(\Leftrightarrow9\sqrt{x^2+2}+2\sqrt{x^2+2}-25\sqrt{x^2+2}=-3\)

\(\Leftrightarrow\sqrt{x^2+2}\left(9+2-25\right)=-3\)

\(\Leftrightarrow\sqrt{x^2+2}=\frac{-3}{-14}=\frac{3}{14}\)

\(\Leftrightarrow x^2+2=\left(\frac{3}{14}\right)^2\)

\(\Leftrightarrow x=\sqrt{\frac{9}{196}-2}=\sqrt{-\frac{383}{196}}\left(vl\right)\)

Vậy \(S=\varnothing\)

Mấy câu kia làm tương tự

b)\(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)

<=>\(\sqrt{9\left(x^2+2\right)}+2\sqrt{x^2+2}-\sqrt{25\left(x^2+2\right)}+3=0\)

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

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

<=>0x=-3

Vậy pt vô nghiệm

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

a: \(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a, \(\sqrt{\left(2x-3\right)^2}=7\\ \Rightarrow\left|2x-3\right|=7\\ \Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

c, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\sqrt{x+3}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)