a) \(\sqrt[]{x^2-4x+4}=x+3\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

a: \(3+\sqrt{2x-3}=x\)

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

=>x>=3 và 2x-3=(x-3)^2

=>x>=3 và x^2-6x+9=2x-3

=>x>=3 và x^2-8x+12=0

=>x>=3 và (x-2)(x-6)=0

=>x>=3 và \(x\in\left\{2;6\right\}\)

=>x=6

b: \(\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)-2x=-4\)

=>\(2x-3\sqrt{x}+2\sqrt{x}-3-2x=-4\)

=>\(-\sqrt{x}-3=-4\)

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

=>căn x=1

=>x=1(nhận)

c: \(\sqrt{2x+1}-x+1=0\)

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

=>x>=1 và (x-1)^2=2x+1

=>x>=1 và x^2-2x+1=2x+1

=>x>=1 và x^2-4x=0

=>x(x-4)=0 và x>=1

=>x=4

\(2,\\ a,\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(x\ge1\right)\\ \Leftrightarrow2\sqrt{x-1}+3\sqrt{x-1}-5\sqrt{x-1}=7\\ \Leftrightarrow0\sqrt{x-1}=7\Leftrightarrow x\in\varnothing\\ b,\sqrt{2x^2-3}=4\left(x\le-\dfrac{\sqrt{6}}{2};\dfrac{\sqrt{6}}{2}\le x\right)\\ \Leftrightarrow2x^2-3=16\\ \Leftrightarrow x^2=\dfrac{19}{2}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

\(1,\\ A=\sqrt{5+4x}+\sqrt{7-3x}\\ ĐKXĐ:\left\{{}\begin{matrix}5+4x\ge0\\7-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{5}{4}\\x\le\dfrac{7}{3}\end{matrix}\right.\)

Bài 2:

a) \(\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(đk:x\ge1\right)\)

\(\Leftrightarrow2\sqrt{x-1}+3\sqrt{x-2}-5\sqrt{x-1}=7\)

\(\Leftrightarrow0=7\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(\sqrt{2x^2-3}=4\left(đk:-\sqrt{\dfrac{3}{2}}\ge x\ge\sqrt{\dfrac{3}{2}}\right)\)

\(\Leftrightarrow2x^2-3=16\)

\(\Leftrightarrow2x^2=19\Leftrightarrow x^2=\dfrac{19}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

Bài 2

a . \(\sqrt{x-1}=3\Leftrightarrow x-1=9\Leftrightarrow x=10\)

b . \(\sqrt{x^2-6x+9}=1\Leftrightarrow\sqrt{\left(x-3\right)^2}=1\Leftrightarrow x-3=1\Leftrightarrow x=4\)

c . \(\sqrt{25x^2-10x+1}=5\Leftrightarrow\sqrt{\left(5x-1\right)^2}=5\Leftrightarrow5x-1=5\Leftrightarrow x=\frac{6}{5}\)

Bài 1

\(\sqrt{2}+\sqrt{3}+\sqrt{8}+\frac{\sqrt{16}}{\sqrt{2}}+\sqrt{3}+\sqrt{4}=\sqrt{2}+2\sqrt{3}+2\sqrt{2}+2\sqrt{2}+2\)

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

\(-\dfrac{4}{5}\sqrt{50-25x}-\dfrac{2}{3}\sqrt{18-9x}+6=0\left(x\le2\right)\\ \Leftrightarrow-\dfrac{4}{5}\cdot5\sqrt{2-x}-\dfrac{2}{3}\cdot3\sqrt{2-x}=-6\\ \Leftrightarrow-2\sqrt{2-x}=-6\\ \Leftrightarrow\sqrt{2-x}=3\Leftrightarrow2-x=9\\ \Leftrightarrow x=-7\left(tm\right)\)

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