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\(1.P=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{2-\sqrt{a}}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}=\sqrt{a}+2+2+\sqrt{a}=4+2\sqrt{a}\left(a\ge0;a\ne4\right)\)
\(2.P=a+1\Leftrightarrow4+2\sqrt{a}=a+1\)
\(\Leftrightarrow a-2\sqrt{a}-3=0\)
\(\Leftrightarrow a+\sqrt{a}-3\sqrt{a}-3=0\)
\(\Leftrightarrow\left(\sqrt{a}-3\right)\left(\sqrt{a}+1\right)=0\)
\(\Leftrightarrow x=9\left(TM\right)\)
KL.............
a/ \(P=12\)
b/ \(Q=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c/ Ta có:
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Dấu = xảy ra khi x = 3 (thỏa tất cả các điều kiện )
a. Thay x = 3 vào biểu thức P ta được :
\(p=\frac{x+3}{\sqrt{x}-2}=\frac{9+3}{\sqrt{9}-2}=12\)
b, \(Q=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{x-4}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}+2}+\frac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x-3\sqrt{x}+2+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)
c, Ta có :
\(\frac{P}{Q}=\frac{\frac{x+3}{\sqrt{x}-2}}{\frac{\sqrt{x}}{\sqrt{x}-2}}=\frac{x+3}{\sqrt{x}}\ge\frac{2\sqrt{3x}}{\sqrt{x}}=2\sqrt{3}\)
Vậy GTNN \(\frac{P}{Q}=2\sqrt{3}\) khi và chỉ khi \(x=3\)
\(P=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{2-\sqrt{a}}=\sqrt{a}+2+2+\sqrt{a}\\ \)\(P=2\sqrt{a}+4\)
\(P=a+1\Leftrightarrow a-2\sqrt{a}-3=0\Rightarrow a=9\)
\(P=\frac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\frac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}\)
\(=\sqrt{a}+2+2+\sqrt{a}=2\sqrt{a}+4\)
\(P=a+1\Rightarrow2\sqrt{a}+4=a+1\)
\(\Leftrightarrow a-2\sqrt{a}-3=0\)
\(\Leftrightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{a}+1=0\left(vn\right)\\\sqrt{a}-3=0\end{matrix}\right.\) \(\Rightarrow\sqrt{a}=3\Rightarrow a=9\)
A=(\(\dfrac{\left(x+4\right)}{3\left(x+2\right)}-\dfrac{1}{\left(x+2\right)^2}\))(\(\dfrac{x+5+x-1}{x+5}\))
A=\(\dfrac{\left(x+4\right)\left(x+2\right)-3}{3\left(x+2\right)^2}\cdot\dfrac{2x+2}{x+5}\)
A=\(\dfrac{x^2+2x+4x+8-3}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{x^2+6x+5}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{x^2+6x+9-4}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{\left(x+3\right)^2-4}{3\left(x-2\right)}\cdot\dfrac{2}{x+5}\)
A=\(\dfrac{2\left(x+3-2\right)\left(x+3+2\right)}{3\left(x-2\right)\left(x+5\right)}\)
A=\(\dfrac{2\left(x+1\right)}{3\left(x-2\right)}\)
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 : \(0< a< 1\)
1: ĐKXĐ: a>=0; a<>16
\(A=\dfrac{-2a-\sqrt{a}+a-16+a+4\sqrt{a}+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}+4\right)}\)
\(=\dfrac{3\sqrt{a}-12}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-4\right)}=\dfrac{3}{\sqrt{a}+2}\)
2: \(C-\dfrac{3}{2}=\dfrac{3}{\sqrt{a}+2}-\dfrac{3}{2}=\dfrac{6-3\sqrt{a}-6}{2\left(\sqrt{a}+2\right)}=\dfrac{-3\sqrt{a}}{2\left(\sqrt{a}+2\right)}< =0\)
=>C<=3/2
=>0<=C<=3/2
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
a) \(P=\dfrac{a+4\sqrt{a}+4}{\sqrt{a}+2}+\dfrac{4-a}{2-\sqrt{a}}=\dfrac{\left(\sqrt{a}+2\right)^2}{\sqrt{a}+2}+\dfrac{\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)}{2-\sqrt{a}}\)
\(=\sqrt{a}+2+\sqrt{a}+2=2\sqrt{a}+4\)
b) \(P=a+1\Rightarrow2\sqrt{a}+4=a+1\Rightarrow a-2\sqrt{a}-3=0\)
\(\Rightarrow\left(\sqrt{a}+1\right)\left(\sqrt{a}-3\right)=0\) mà \(\sqrt{a}+1>0\Rightarrow\sqrt{a}=3\Rightarrow a=9\)