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a) A có nghĩa\(\Leftrightarrow x-y\ne0\Leftrightarrow x\ne y\)
b) \(A=\frac{x+y-2\sqrt{xy}}{x-y}=\frac{\left(\sqrt{x-\sqrt{y}}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
a.
\(B=\left(\frac{x+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\left(\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\\ =\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
b. Ta có :
\(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\\ =\sqrt{25+2\cdot5\cdot\sqrt{2}+2}-\sqrt{16+2\cdot4\cdot\sqrt{2}+2}\\ =\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\\ =5+\sqrt{2}-4-\sqrt{2}=1\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}+3}=\frac{1+1}{1+3}=\frac{2}{4}=\frac{1}{2}\)
c. Giả sử B>\(\frac{1}{3}\), ta có
\(B=\frac{\sqrt{x}+1}{\sqrt{x}+3}>\frac{1}{3}\\ \Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}+3}-\frac{1}{3}>0\\ \Leftrightarrow\\\frac{3\left(\sqrt{x}+1\right)-\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow\frac{2\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\left(luondungvoix>0\right)\)
Vậy.........
1)
a)
\(\sqrt{11-6\sqrt{2}}=\sqrt{2-2.3.\sqrt{2}+9}=\left|\sqrt{2}-3\right|=3-\sqrt{2}\)
\(A=3-\sqrt{2}+3+\sqrt{2}=6\)
b)
\(B^2=24+2\sqrt{12^2-4.11}=24+2\sqrt{100}=24+20=44\)
\(B=\sqrt{44}=2\sqrt{11}\)
a)
\(M=2+\sqrt{\left(2x\right)^2-2.2x.3+3^2}\)
\(\Rightarrow M=2+\sqrt{\left(2x-3\right)^2}\)
\(\Rightarrow M=2+2x-3\)
\(\Rightarrow M=2x-1\)
b)
(+) x=5/2
=> \(M=2.\frac{5}{2}-1=5-1=4\)
(+) x= - 1/5
=> \(M=2.\frac{\left(-1\right)}{5}-1=-\frac{2}{5}-1=-\frac{7}{5}\)
a/ Giải rồi
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)
Pt trở thành:
\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)
\(\Leftrightarrow...\)
e/ ĐKXD: \(x>0\)
\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)
\(\Rightarrow t^2=x+\frac{1}{4x}+1\)
Pt trở thành:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)
\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)
\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)
Dấu '=' xảy ra khi x=0
2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)
Dấu '=' xảy ra khi x=0
3: \(A=-2x-3\sqrt{x}+2< =2\)
Dấu '=' xảy ra khi x=0
5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)
Dấu '=' xảy ra khi x=1