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\(F=\left(\dfrac{1}{3-\sqrt{5}}+\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{6}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}:\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\dfrac{3}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{3}{2\sqrt{5}}\)
\(G=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}=\dfrac{\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3.\sqrt{5}+5}-2}{\sqrt{2}}=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(H=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{x-2-2\sqrt{2}.\sqrt{x-2}+2}=\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\left(x\ge2\right)\)
1) \(\sqrt{\text{x^2− 20x + 100 }}=10\)
<=> \(\sqrt{\left(x-10\right)^2}=10\)
<=> \(\left|x-10\right|=10\)
=> \(\left[{}\begin{matrix}x-10=10\\x-10=-10\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=10+10\\x=\left(-10\right)+10\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=20\\x=0\end{matrix}\right.\)
Vậy S = \(\left\{20;0\right\}\)
2) \(\sqrt{x +2\sqrt{x}+1}=6\)
<=> \(\sqrt{\left(\sqrt{x^2}+2.\sqrt{x}.1+1^2\right)}=6\)
<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}=6\)
<=> \(\left|\sqrt{x}+1\right|=6\)
=> \(\left[{}\begin{matrix}\sqrt{x}+1=6\\\sqrt{x}+1=-6\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{x}=6-1=5\\\sqrt{x}=\left(-6\right)-1=-7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=25\\x=-49\left(loai\right)\end{matrix}\right.\)
Vậy S = \(\left\{25\right\}\)
3) \(\sqrt{x^2-6x+9}=\sqrt{4+2\sqrt{3}}\)
<=> \(\sqrt{\left(x-3\right)^2}=\sqrt{\sqrt{3^2}+2.\sqrt{3}.1+1^2}\)
<=> \(\left|x-3\right|=\sqrt{\left(\sqrt{3}+1\right)^2}\)
<=> \(\left|x-3\right|=\sqrt{3}+1\)
=> \(\left[{}\begin{matrix}x-3=\sqrt{3}+1\\x-3=-\left(\sqrt{3}+1\right)\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\sqrt{3}+4\\x=-\sqrt{3}+2\end{matrix}\right.\)
Vậy S = \(\left\{\sqrt{3}+4;-\sqrt{3}+2\right\}\)
4) \(\sqrt{3x+2\sqrt{3x}+1}=5\)
<=> \(\sqrt{\sqrt{3x}^2+2.\sqrt{3x}.1+1^2}=5\)
<=> \(\sqrt{\left(\sqrt{3x}+1\right)^2}=5\)
<=> \(\left|\sqrt{3x}+1\right|=5\)
=> \(\left[{}\begin{matrix}\sqrt{3x}+1=5\\\sqrt{3x}+1=-5\end{matrix}\right.\)=> \(\left[{}\begin{matrix}\sqrt{3x}=5-1=4\\\sqrt{3x}=\left(-5\right)-1=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3x=16\\3x=-6\left(loai\right)\end{matrix}\right.\)=> x = \(\dfrac{16}{3}\) Vậy S = \(\left\{\dfrac{16}{3}\right\}\)
5) \(\sqrt{x^2+2x\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}\)
<=> \(\sqrt{\left(x-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
<=> \(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)
<=> \(\left[{}\begin{matrix}x-\sqrt{3}=\sqrt{3}-1\\x-\sqrt{3}=-\left(\sqrt{3}-1\right)\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=-1\\x=-2\sqrt{3}+1\end{matrix}\right.\)
Vậy S = \(\left\{-1;-2\sqrt{3}+1\right\}\)
6) \(\sqrt{6x+4\sqrt{6x}+4}=7\)
<=> \(\sqrt{\sqrt{6x}^2+2.\sqrt{6x}.2+2^2}=7\)
<=> \(\sqrt{\left(\sqrt{6}+2\right)^2}=7\)
<=> \(\left|\sqrt{6x}+2\right|=7\)
=> \(\left[{}\begin{matrix}\sqrt{6x}+2=7\\\sqrt{6x}+2=-7\end{matrix}\right.\)=>\(\left[{}\begin{matrix}\sqrt{6x}=7-2=5\\\sqrt{6x}=\left(-7\right)-2=-9\left(loai\right)\end{matrix}\right.\)
=> \(\sqrt{6x}=5=>6x=25=>x=\dfrac{25}{6}\)
Điều kiện: \(x\ge2\)
\( x - 5\sqrt {x - 2} = - 2\\ \Leftrightarrow - 5\sqrt {x - 2} = - 2 - x\\ \Leftrightarrow 25\left( {x - 2} \right) = 4 + 4x + {x^2}\\ \Leftrightarrow 25x - 50 - 4 - 4x - {x^2} = 0\\ \Leftrightarrow - {x^2} + 21x - 54 = 0\\ \Leftrightarrow {x^2} - 21x + 54 = 0\\ \Delta = {b^2} - 4ac = {\left( { - 21} \right)^2} - 4.1.54 = 225 > 0 \)
Phương trình có hai nghiệm phân biệt:
\( {x_1} = \dfrac{{ - b + \sqrt \Delta }}{{2a}} = \dfrac{{ - \left( { - 21} \right) + \sqrt {225} }}{{2.1}} = 18 (TM)\\ {x_2} = \dfrac{{ - b - \sqrt \Delta }}{{2a}} = \dfrac{{ - \left( { - 21} \right) - \sqrt {225} }}{{2.1}} = 3(TM) \)
\(\sqrt {x - 2 + \sqrt {2x- 5} } + \sqrt {x + 2 + 3\sqrt {2x - 5} } = 7\sqrt 2 \)
Điều kiện: \(x \ge \dfrac{5}{2}\)
Nhân cả hai vế của phương trình với \(\sqrt{2}\), rồi biến đổi về dạng:
\(\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\\ \Leftrightarrow\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\\ \Leftrightarrow\sqrt{2x-5}=10\\ \Leftrightarrow2x-5=100\\ \Leftrightarrow2x=150\\ \Leftrightarrow x=15\left(TM\right)\)
ĐKXĐ: \(x\ge\frac{5}{2}\)
Ta có:
\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\\ \Leftrightarrow\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)
(nhân cả hai vế với \(\sqrt{2}\))
\(\Leftrightarrow\sqrt{2x-5+2\cdot\sqrt{2x-5}\cdot1+1}+\sqrt{2x-5+2\cdot2\sqrt{2x-5}\cdot3+9}=14\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\\ \Leftrightarrow\left|\sqrt{2x-5}+1\right|+\left|\sqrt{2x-5}+3\right|=14\\ \Leftrightarrow\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\\ \Leftrightarrow2\sqrt{2x-5}=10\\ \Leftrightarrow\sqrt{2x-5}=5\\ \Leftrightarrow2x-5=25\Leftrightarrow x=15\left(tm\right)\)
Vậy x=15
P/s: Chỗ GTTĐ | | phá được vì cả hai biểu thức đều > 0