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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
ĐKXĐ:\(x>0;x\ne4\)
B=\(\left(\dfrac{1}{x-4}-\dfrac{1}{x-4\sqrt{x}+4}\right)\cdot\dfrac{x+2\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}-2-\sqrt{x}-2}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}}=\dfrac{-4}{\left(\sqrt{x}-2\right)^2\left(\sqrt{x}+2\right)}\cdot\left(\sqrt{x}+2\right)=-\dfrac{4}{\left(\sqrt{x}-2\right)^2}\)
Vậy...
có phải/....
1) \(A=\dfrac{x+3}{\sqrt{x}-2}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-2}{x-4}\) hay \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{5\left(\sqrt{x}-2\right)}{x-4}\)
2) \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
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 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
\(Q=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+5\sqrt{x}-8}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(P=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-3x}{x-4}\)
\(=\dfrac{-\sqrt{x}}{x-4}\)
E = \(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\) = \(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
E = \(\sqrt{x}+1+\sqrt{x}\) = \(2\sqrt{x}+1\)
F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}+1}{3-\sqrt{x}}-\dfrac{3-11\sqrt{x}}{x-9}\)
F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
F = \(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
F = \(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
F = \(\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) = \(\dfrac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\) = \(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{4\sqrt{x}-4}{4-x}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{4\sqrt{x}-4}{x-4}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)-\left(4\sqrt{x}-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{x+2\sqrt{x}+3\sqrt{x}+6-\left(x-2\sqrt{x}-\sqrt{x}+2\right)-\left(4\sqrt{x}-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{x+5\sqrt{x}+6-x+2\sqrt{x}+\sqrt{x}-2-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
G = \(\dfrac{4\sqrt{x}+8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{4\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\) = \(\dfrac{4}{\sqrt{x}-2}\)
Bài 2:
a: \(A=\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}-\dfrac{3\sqrt{5}\left(3-\sqrt{5}\right)}{4}\)
\(=-5+3\sqrt{5}+\dfrac{5+\sqrt{5}-9\sqrt{5}+15}{4}\)
\(=-5+3\sqrt{5}+5-2\sqrt{5}=\sqrt{5}\)
b: \(B=\left(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right):\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+3\sqrt{x}+6-2\sqrt{x}-6}=1\)
\(ĐK:x>0;x\ne4\\ B=\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ B=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ B=\dfrac{x+2\sqrt{x}+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ B=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+3\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}}{x-4}\)