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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: \(B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}=\dfrac{3}{\sqrt{x}+3}\)
Cau 1:
a: ĐKXĐ: x-2<>0
=>x<>2
b: ĐKXĐ: 1-x>=0
=>x<=1
c: ĐKXĐ: \(x\in R\)
d: ĐKXĐ: 4-3x>=0 và x<>0
=>x<=3/4 và x<>0
3) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\)
\(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\)
\(\Leftrightarrow\sqrt{4x-20}=4\)
\(\Leftrightarrow4x-20=16\)
\(\Leftrightarrow4x=36\)
\(\Leftrightarrow x=9\)
vậy ...
1)
\(A=\dfrac{\sqrt{x}-2}{x-4}=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}\right)^2-2^2}\\ A=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{1}{\sqrt{x}+2}\)
\(B=\dfrac{x^2-2x\sqrt{2}+2}{x^2-2}=\dfrac{x^2-2x\sqrt{2}+\left(\sqrt{2}\right)^2}{x^2-\sqrt{2}}\\ B=\dfrac{\left(x-\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}=\dfrac{\left(x-\sqrt{2}\right)}{\left(x+\sqrt{2}\right)}\)
\(C=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+5}=\dfrac{x+\sqrt{5}}{x^2+2x\sqrt{5}+\left(\sqrt{5}\right)^2}\\ C=\dfrac{x+\sqrt{5}}{\left(x+\sqrt{5}\right)^2}=\dfrac{1}{x+\sqrt{5}}\)
\(D=\dfrac{\sqrt{a}-2a}{2\sqrt{a}-1}=\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)}{2\sqrt{a}-1}=\sqrt{a}\)
\(E=\dfrac{x^2-2}{x-\sqrt{2}}=\dfrac{x^2-\left(\sqrt{2}\right)^2}{x-\sqrt{2}}\\ E=\dfrac{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}{x-\sqrt{2}}=x+\sqrt{2}\)
\(F=\dfrac{\sqrt{x}-3}{x-9}=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}\right)^2-3^2}\\ F=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ F=\dfrac{1}{\sqrt{x}+3}\)