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1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)
Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)
\(x=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}}}\)
\(\Leftrightarrow x=\sqrt{5+\sqrt{13+x}}\) (\(x\ge0\))
\(\Leftrightarrow x^2=5+\sqrt{13+x}\)
\(\Leftrightarrow x^2-9=\sqrt{13+x}-4\)
\(\Leftrightarrow\left(x-3\right).\left(x+3\right)=\dfrac{x-3}{\sqrt{13+x}+4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=\dfrac{1}{\sqrt{x+13}+4}\left(∗\right)\end{matrix}\right.\)
Xét (*) ta có VT \(\ge3\) (1)
mà \(VP=\dfrac{1}{\sqrt{x+13}+4}\le\dfrac{1}{4}\) (2)
Từ (1) và (2) dễ thấy (*) vô nghiệm
Hay x = 3
Ta có A > 0
Từ đó \(A^2=2+\sqrt{2+\sqrt{2+...}}\Leftrightarrow A^2=2+A\Leftrightarrow A^2-A-2=0\)
\(\Leftrightarrow\left(A+1\right)\left(A-2\right)=0\Leftrightarrow\orbr{\begin{cases}A+1=0\\A-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}A=-1\\A=2\end{cases}}\)
Do A > 0 nên A= 2
b, tương tự
c,\(C>2\)
Xét \(C^2=5+\sqrt{13+\sqrt{5+\sqrt{13...}}}\)
\(\left(C^2-5\right)^2=13+C\Leftrightarrow C^4-10C^2-C+12=0\Leftrightarrow\left(C^4-9C^2\right)-\left(C^2-9\right)-\left(C-3\right)=0\)
\(\Leftrightarrow\left(C-3\right)\left[\left(C+3\right)\left(C-1\right)\left(C+1\right)-1\right]=0\)
VÌ C> 2 => C-3 = 0 => C=3
chú ý\(x=\sqrt{x}^2\) tương tự với y , và các số tự nhiên dương
\(A=\frac{\sqrt{x}^2+2\sqrt{x}-3}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)}=\sqrt{x}+3\)
\(B=\frac{\left(2\sqrt{y}\right)^2+3\sqrt{y}-7}{4\sqrt{y}+7}=\frac{\left(\sqrt{y}-1\right)\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}=\sqrt{y}-1\)
\(C=\frac{\sqrt{x}^2\sqrt{y}-\sqrt{y}^2\sqrt{x}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{xy}\)
\(D=\frac{\sqrt{x}^2-3\sqrt{x}-4}{\sqrt{x}^2-\sqrt{x}-12}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)}\)
\(E=\sqrt{1+2\sqrt{5}+5}+\sqrt{\sqrt{5}-2\sqrt{5}+1}=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
=>\(E=1+\sqrt{5}+\sqrt{5}-1=2\sqrt{5}\)
CÂU CUỐI chưa làm đc
ý cuối cùng này :
\(D=\sqrt{13-4\sqrt{10}}+\sqrt{13+4\sqrt{10}}\)lấy bình phương 2 vế ta có
\(D^2=13-4\sqrt{10}+13+4\sqrt{10}+2\sqrt{13-4\sqrt{10}}\sqrt{13+4\sqrt{10}}\)
\(D^2=26+2\sqrt{13^2-16\sqrt{10}^2}\Leftrightarrow D^2=26+2\sqrt{9}\)
\(D^2=32\Leftrightarrow D=\sqrt{32}=4\sqrt{2}\)
\(a.\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}+\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)}=\sqrt{x}-\sqrt{y}+\sqrt{x}+\sqrt{y}=2\sqrt{x}\)
\(b.\sqrt{\left(\sqrt{5}-1\right)\sqrt{13-\sqrt{49-2.7.2\sqrt{5}+20}}}=\sqrt{\left(\sqrt{5}-1\right)\sqrt{5+2\sqrt{5}+1}}=\sqrt{\left(\sqrt{5}-1\right)\left(\sqrt{5+1}\right)}=\sqrt{5}-1\)
\(c.\dfrac{\sqrt{3+\sqrt{5}}\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{10}+\sqrt{2}\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}=\dfrac{\sqrt{2}.\sqrt{5+2\sqrt{5}+1}\left(\sqrt{3}+1\right)\left(\sqrt{5}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}=\dfrac{\sqrt{2}\left(\sqrt{5}+1\right)^2\left(\sqrt{3}+1\right)\left(3-\sqrt{5}\right)}{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}=\dfrac{2\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{\sqrt{3+2\sqrt{3}+1}}=2\left(9-5\right)=2.4=8\)
Câu a
\(\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}+\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\sqrt{x}+\sqrt{y}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+\sqrt{x^2y}+\sqrt{xy^2}}{\sqrt{xy}}\\ =\dfrac{x\sqrt{y}-y\sqrt{x}+x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\\ =\dfrac{2x\sqrt{y}}{\sqrt{xy}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
+) ta có : \(A=\sqrt{13+4\sqrt{10}}-\sqrt{13-4\sqrt{10}}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}-2\sqrt{2}+\sqrt{5}=2\sqrt{5}\) (sữa đề)
+) ta có : \(B=\sqrt{\dfrac{3-2\sqrt{2}}{17-12\sqrt{2}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(3-2\sqrt{2}\right)^2}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)^2}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{1}{\sqrt{2}-1}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\sqrt{\dfrac{\left(2-\sqrt{3}\right)\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}\)
\(=\sqrt{2}+1+2-\sqrt{3}=3-\sqrt{3}+\sqrt{2}\) (sữa đề )
+) đk : \(x\ne-3\)
ta có : \(C=\dfrac{\sqrt{x^2+6x+9}}{x+3}=\dfrac{\sqrt{\left(x+3\right)^2}}{x+3}=\dfrac{\left|x+3\right|}{x+3}\)
\(\left[{}\begin{matrix}C=1\left(x>-3\right)\\C=-1\left(x< -3\right)\end{matrix}\right.\)
+) \(m\ge\dfrac{5}{2}\)
ta có : \(D=\sqrt{2m+4+6\sqrt{2m-5}}-\sqrt{2m-5}\)
\(=\sqrt{\left(\sqrt{2m-5}+3\right)^2}-\sqrt{2m-5}=\left|\sqrt{2m-5}+3\right|-\sqrt{2m-5}\)
\(\Leftrightarrow\left[{}\begin{matrix}C=3\left(m\ge7\right)\\C=-3-2\sqrt{2m-5}\left(\dfrac{5}{2}\le m\le7\right)\end{matrix}\right.\)
13) x+ \(\sqrt{x}\) - 6
14) x - 2\(\sqrt{x}\) - 5
15) x + 2\(\sqrt{x}\) -15
16) 2x - 5\(\sqrt{x}\) + 3
13: \(x+\sqrt{x}-6=\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)\)
15: \(x+2\sqrt{x}-15=\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)\)
16: \(2x-5\sqrt{x}+3=\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)\)
Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
Dễ dàng nhận thấy \(x>0\)
a/ \(x^2=6+\sqrt{6+\sqrt{6+...}}\)
\(\Leftrightarrow x^2=6+x\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow x=3\)
b/ \(x^2=5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}\)
\(\Leftrightarrow x^2=5+\sqrt{13+x}\)
\(\Leftrightarrow x^2-5=\sqrt{x+13}\) (\(x\ge\sqrt{5}\))
\(\Leftrightarrow\left(x^2-5\right)^2=x+13\)
\(\Leftrightarrow x^4-10x^2-x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2-x-4\right)=0\)
Do \(x\ge\sqrt{5}\Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x^3-x=x^2\left(x-1\right)>0\\x^2\ge5\Rightarrow3x^2-4>0\end{matrix}\right.\)
\(\Rightarrow x^3+3x^2-x-4>0\)
\(\Rightarrow x=3\)