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\(B=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
B=2/3A
=>3căn x/căn x+2=2/3*3=2
=>3căn x=2căn x+4
=>x=16
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
c: \(=\sqrt{\dfrac{4}{16-6\sqrt{7}}}+\sqrt{7}\)
\(=\dfrac{2}{3-\sqrt{7}}+\sqrt{7}\)
\(=3+2\sqrt{7}\)
d: \(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}\)
\(=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
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)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
\(< =>\sqrt[3]{x+5}=-2\)
<=> \(\left(\sqrt[3]{x+5}\right)^3=-8\)
<=> \(x+5=-8\)
<=> x=-13
b) \(B=\dfrac{x-\sqrt{x}}{1-\sqrt{x}}-\dfrac{x\sqrt{x}}{\sqrt{x}}=\dfrac{\sqrt{x}\left(x-\sqrt{x}\right)-x\sqrt{x}\left(1-\sqrt{x}\right)}{\sqrt{x}\left(1-\sqrt{x}\right)}\) = \(\dfrac{x\sqrt{x}-x-x\sqrt{x}+x^2}{\sqrt{x}-x}=\dfrac{x^2-x}{\sqrt{x}-x}\)
c) \(C=\dfrac{x+2\sqrt{x}}{\sqrt{x}-x}-\dfrac{x\sqrt{x}}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)-x\sqrt{x}\left(\sqrt{x}-x\right)}{\left(\sqrt{x}-x\right)\left(\sqrt{x}+1\right)}=x+2\sqrt{x}-x\sqrt{x}\)
\(d,D=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\) \(\dfrac{\left(x+2\sqrt{x}\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+7\sqrt{x}-2}{\sqrt{x}+2}\)
e) \(E=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\) = \(\dfrac{2\sqrt{x}-24}{\sqrt{x}+3}\)
F) F = \(\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{23-2\sqrt{x}}{\sqrt{x}+5}\)
\(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\left(dkxd:x\ge0,x\ne4\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+2}{x-4}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}-2\right)-\left(5\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x-2\sqrt{x}+\sqrt{x}-2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3x-10\sqrt{x}-4}{x-4}\)
\(\sqrt{x}\) - 2 đâu