bạn nhập pt vào máy tính rồi nhấn shift slove = ,sẽ ra nghiệm là 0,5 .lấy 0,5 thể vào căn thức rồi nhân liên hợp là ok

đặt x+5=a\(\left(a\ge0\right)\) khi đó phương trình trở thành:

\(a^2-4+\sqrt{a}+\sqrt{16-a}=0\)

lại có \(\sqrt{a}+\sqrt{16-a}\ge\sqrt{a+16-a}=4\)

nên ta có:

\(a^2-4+\sqrt{a}+\sqrt{16-a}\ge a^2\)

Suy ra \(0\ge a^2\)

\(\Rightarrow a=0\)hay x+5=0

\(\Leftrightarrow x=-5\)

Liên hợp đi

Đặt \(\left\{{}\begin{matrix}2x+3=a\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)

Pt trở thành:

\(a^2+2b^2-3ab=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

a: ĐKXĐ: x>=5

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

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

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

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

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x>=1/2

\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)

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

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

=>\(\sqrt{2x-1}=5\)

=>2x-1=25

=>2x=26

=>x=13(nhận)

c: \(\sqrt{x^2-10x+25}=2\)

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

=>\(\left|x-5\right|=2\)

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

d: \(\sqrt{x^2-14x+49}-5=0\)

=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)

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

=>|x-7|=5

=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)

\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)

\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a)ĐKXĐ:x\ge5\\ \sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=\dfrac{4}{2}\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2=2^2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=4+5\\ \Leftrightarrow x=9\left(tmđk\right)\)

Vậy \(S=\left\{9\right\}\)

\(b)ĐKXĐ:x\ge2\\ \sqrt{2x-1}-\sqrt{8x-4}+5=0\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{8x-4}=0-5\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow-\left(\sqrt{2x-1}\right)=\left(-5\right)^2\\ \Leftrightarrow-2x+1=-25\\ \Leftrightarrow-2x=\left(-25\right)-1\\ \Leftrightarrow-2x=-26\\ \Leftrightarrow x=\dfrac{-26}{-2}\\ \Leftrightarrow x=13\left(tmđk\right)\)

Vậy \(S=\left\{13\right\}\)

\(c)\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+5\\x=\left(-2\right)+5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{7;3\right\}\)

\(d)\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{x^2-14x+49}=0+5\\ \Leftrightarrow\sqrt{x^2-14x+49}=5\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5+7\\x=\left(-5\right)+7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{12;2\right\}.\)