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\(x=\sqrt{x^2-2x+5}=\sqrt{x^2-2x+1+4}\\ =\sqrt{\left(x-1\right)^2+4}\ge\sqrt{4}=2\)
dấu "=" xảy ra khi x=1
vậy min x=2 khi x=1
\(y=\sqrt{\dfrac{x^2}{4}-\dfrac{x}{6}+1}=\sqrt{\left(\dfrac{x}{2}\right)^2-2.\dfrac{x}{2}.\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{35}{36}}\\ =\sqrt{\left(\dfrac{x}{2}-\dfrac{1}{6}\right)^2+\dfrac{35}{36}}\ge\sqrt{\dfrac{35}{36}}\)
dấu "=" xảy ra khi \(\dfrac{x}{2}-\dfrac{1}{6}=0\Rightarrow x=\dfrac{1}{3}\)
vậy min y =\(\sqrt{\dfrac{35}{36}}\) tại \(x=\dfrac{1}{3}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
Bài 3:
a: \(A=\dfrac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{x-25}\)
\(=\dfrac{x-10\sqrt{x}+25}{x-25}=\dfrac{\sqrt{x}-5}{\sqrt{x}+5}\)
b: \(B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}=\dfrac{3}{\sqrt{x}+3}\)
1/ Rút gọn: \(a)3\sqrt{2a}-\sqrt{18a^3}+4\sqrt{\dfrac{a}{2}}-\dfrac{1}{4}\sqrt{128a}\left(a\ge0\right)=3\sqrt{2a}-3a\sqrt{2a}+2\sqrt{2a}-2\sqrt{2a}=3\sqrt{2a}\left(1-a\right)\)b)\(\dfrac{\sqrt{2}-1}{\sqrt{2}+2}-\dfrac{2}{2+\sqrt{2}}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-1-2}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3}{\sqrt{2}+2}+\dfrac{\sqrt{2}+1}{\sqrt{2}}=\dfrac{\sqrt{2}-3+2+1+2\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3\sqrt{2}}{\sqrt{2}\left(1+\sqrt{2}\right)}=\dfrac{3}{1+\sqrt{2}}\)c)\(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{\left(\sqrt{2}+\sqrt{3+\sqrt{5}}\right)\sqrt{2}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{6-2\sqrt{5}}}=\dfrac{2\sqrt{2}+\sqrt{10}}{2+\sqrt{\left(\sqrt{5}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{10}}{2-\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{2+\sqrt{5}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{2-\sqrt{5}+1}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)}{3+\sqrt{5}}+\dfrac{\sqrt{2}\left(2-\sqrt{5}\right)}{3-\sqrt{5}}=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{\sqrt{2}\left(6-2\sqrt{5}+3\sqrt{5}-5+6+2\sqrt{5}-3\sqrt{5}-5\right)}{9-5}=\dfrac{2\sqrt{2}}{4}=\dfrac{1}{\sqrt{2}}\)
Làm nốt nè :3
\(2.a.P=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}}{x-2\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x}=\dfrac{x-1}{x}\left(x>0;x\ne1\right)\)\(b.P>\dfrac{1}{2}\Leftrightarrow\dfrac{x-1}{x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x-2}{2x}>0\)
\(\Leftrightarrow x-2>0\left(do:x>0\right)\)
\(\Leftrightarrow x>2\)
\(3.a.A=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a-1}=\dfrac{\sqrt{a}-1}{\sqrt{a}-1}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}+1}=\sqrt{a}-1\left(a>0;a\ne1\right)\)
\(b.Để:A< 0\Leftrightarrow\sqrt{a}-1< 0\Leftrightarrow a< 1\)
Kết hợp với DKXĐ : \(0< a< 1\)
Bài 1:
a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)
= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)
= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)
=\(\sqrt{xy}\)
b.ĐK: x ≠ 1
Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)
*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)
⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)
⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)
c.Ta có:
1. \(\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{\sqrt{a}-2}\)
\(=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}-\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\sqrt{a}-2}\)
\(=\sqrt{a}+2-\sqrt{a}-2\)
= 0
2: \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\dfrac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}\)
\(=\sqrt{x}-\sqrt{y}+\sqrt{y}-\sqrt{x}=0\)
4: \(=\left(1+\sqrt{a}+\sqrt{a}+a\right)\cdot\dfrac{1}{1+\sqrt{a}}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}+1}=\sqrt{a}+1\)
a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{x-9}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}+3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
b: Khi \(x=4-2\sqrt{3}\) vào A, ta được:
\(A=\dfrac{-3\left(\sqrt{3}-1\right)+3}{\left(\sqrt{3}-1+3\right)\left(\sqrt{3}-1+1\right)}\)
\(=\dfrac{-3\sqrt{3}+6}{\sqrt{3}\cdot\left(\sqrt{3}+2\right)}=\dfrac{-3+2\sqrt{3}}{2+\sqrt{3}}\)
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