Chương II - Hàm số bậc nhất

\(PT\Leftrightarrow\left(\sqrt{x^2-4x+29}-5\right)+\left(x^2-4x+4\right)=0\\ \Leftrightarrow\dfrac{x^2-4x+4}{\sqrt{x^2-4x+29}+5}+\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x^2-4x+4\right)\left(\dfrac{1}{\sqrt{x^2-4x+29}+5}+1\right)=0\\ \Leftrightarrow\left(x-2\right)^2\left(\dfrac{1}{\sqrt{x^2-4x+29}+5}+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\\dfrac{1}{\sqrt{x^2-4x+29}+5}+1=0\left(1\right)\end{matrix}\right.\)

Dễ thấy \(\left(1\right)>0\)

Thử lại ta thấy x=2 thỏa mãn

Vậy PT có nghiệm x=2

\(PT\Leftrightarrow\sqrt{2x-1}=1-2x\)

\(ĐK:\left\{{}\begin{matrix}1-2x\ge0\\2x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\ge\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2}\)

Vậy PT có nghiệm \(x=\dfrac{1}{2}\)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(2x=1-\sqrt{2x-1}\\ \Leftrightarrow1-2x=\sqrt{2x-1}\\ \Leftrightarrow1-4x+4x^2=2x-1\left(x\le\dfrac{1}{2}\right)\\ \Leftrightarrow4x^2-6x+2=0\\ \Leftrightarrow2x^2-3x+1=0\\ \Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\\ \Leftrightarrow2x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(\dfrac{\sqrt{x}+3}{\sqrt{x}-1}=\sqrt{x}\\ \Leftrightarrow\sqrt{x}+3=\sqrt{x}\left(\sqrt{x}-1\right)\\ \Leftrightarrow\sqrt{x}+3=x-\sqrt{x}\\ \Leftrightarrow x-\sqrt{x}-\sqrt{x}+3=0\\ \Leftrightarrow x-2\sqrt{x}-3=0\\ \Leftrightarrow\left(x-2\sqrt{x}+1\right)-4=0\\ \Leftrightarrow\left(\sqrt{x}-1\right)^2-2^2=0\\ \Leftrightarrow\left(\sqrt{x}-1-2\right)\left(\sqrt{x}-1+2\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=-1\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\)

ĐKXĐ:\(x\ge3\)

\(2\sqrt{x-3}+\dfrac{5}{3}\sqrt{9x-27}-2=0\\ \Leftrightarrow2\sqrt{x-3}+\dfrac{5}{3}\sqrt{9\left(x-3\right)}-2=0\\ \Leftrightarrow2\sqrt{x-3}+\dfrac{5}{3}.3\sqrt{x-3}-2=0\\ \Leftrightarrow2\sqrt{x-3}+5\sqrt{x-3}=2\\ \Leftrightarrow7\sqrt{x-3}=2\\ \Leftrightarrow\sqrt{x-3}=\dfrac{2}{7}\\ \Leftrightarrow x-3=\dfrac{4}{49}\\ \Leftrightarrow x=\dfrac{151}{49}\left(tm\right)\)

ĐK: x-5 >= 0 <=> x>= 5

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

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

\(a,ĐK:x\ge0\\ PT\Leftrightarrow\sqrt{x}+2=3\sqrt{x}-3\\ \Leftrightarrow2\sqrt{x}=5\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\\ b,\Leftrightarrow\sqrt{\left(x+3\right)^2}=2\\ \Leftrightarrow\left|x+3\right|=2\Leftrightarrow\left[{}\begin{matrix}x+3=2\\x+3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)

\(F=\sqrt{\dfrac{10-2\sqrt{21}}{10+2\sqrt{21}}}+\sqrt{\dfrac{10+2\sqrt{21}}{10-2\sqrt{21}}}\\ F=\sqrt{\dfrac{\left(\sqrt{7}-\sqrt{3}\right)^2}{\left(\sqrt{7}+\sqrt{3}\right)^2}}+\sqrt{\dfrac{\left(\sqrt{7}+\sqrt{3}\right)^2}{\left(\sqrt{7}-\sqrt{3}\right)^2}}\\ F=\dfrac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}}+\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}\\ F=\dfrac{10-2\sqrt{21}+10+2\sqrt{21}}{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)}\\ F=\dfrac{20}{4}=5\)

\(E=\left[\dfrac{4\left(\sqrt{5}+2\right)}{1}-\dfrac{11\left(2\sqrt{5}+3\right)}{11}\right]\left(2\sqrt{5}-5\right)\\ E=\sqrt{5}\left(2-\sqrt{5}\right)\left(4\sqrt{5}+8-2\sqrt{5}-3\right)\\ E=\left(2\sqrt{5}-5\right)\left(2\sqrt{5}+5\right)\\ E=20-25=-5\)

\(D=\dfrac{3\sqrt{2}+9-3+\sqrt{2}}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}\left(3-2\sqrt{2}\right)\\ D=\dfrac{2\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}{7}=\dfrac{2\left(9-1\right)}{7}=\dfrac{16}{7}\)