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9 T I C H sai buồn
\(A=\frac{\sqrt{x^3}}{\sqrt{xy}-2y}-\frac{2x}{x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}}.\frac{1-x}{1-\sqrt{x}}..\)
nhờ vào năng lực rinegan tối hậu của ta , ta có thể dễ dàng nhìn thấy mẫu chung
\(x+\sqrt{x}-2\sqrt{xy}-2\sqrt{y}=\sqrt{x}\left(\sqrt{x}-2\sqrt{xy}\right)+\left(\sqrt{x}-2\sqrt{y}\right)=\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+1\right)\)
\(A=\frac{\sqrt{x^3}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}-\frac{2x\left(x-1\right)}{\left(\sqrt{x}-2\sqrt{y}\right)\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}.\)
\(\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
\(A=\frac{\sqrt{x^3}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\sqrt{x}-2x\sqrt{y}}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x\left(\sqrt{x}-2\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}-2\sqrt{y}\right)}=\frac{x}{\sqrt{y}}\)
b) thay y=625 vào ta được
\(\frac{x}{\sqrt{625}}=\frac{x}{25}< 0.2\Leftrightarrow x< 5\)
vậy \(0< x< 5\)
Ta có \(P=\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{8}-2}-\frac{\sqrt{15}-\sqrt{3}}{2-2\sqrt{5}}\right):\frac{1}{\sqrt{7}-\sqrt{3}}\)
\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{2\left(1-\sqrt{5}\right)}\right).\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\left(\frac{\sqrt{7}}{2}+\frac{\sqrt{3}}{2}\right).\left(\sqrt{7}-\sqrt{3}\right)=\frac{\sqrt{7}+\sqrt{3}}{2}.\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\frac{7-3}{2}=2\)
Vậy \(P=2\)
\(=\sqrt{8-2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}\)
\(\sqrt{2}.\sqrt{4-\sqrt{7}}\\ =\dfrac{1}{2}.\sqrt{2}.\sqrt{8-2\sqrt{7}}\\ =\dfrac{1}{2}.\sqrt{2}.\sqrt{7-2\sqrt{7}.1+1}\\ =\dfrac{1}{2}.\sqrt{2}.\sqrt{\left(\sqrt{7}-1\right)^2}\\ =\dfrac{1}{2}.\sqrt{2}.\left(\sqrt{7}-1\right)\\ =\dfrac{1}{2}.\sqrt{2}.\sqrt{7}-\dfrac{1}{2}.\sqrt{2}\\ =\dfrac{1}{2}.\sqrt{14}-\dfrac{1}{2}.\sqrt{2}\\ =\dfrac{1}{2}.\left(\sqrt{14}-\sqrt{2}\right)\)
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}-\sqrt{2}\)
\(=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7-2\sqrt{7}.1+1}}{\sqrt{2}}-\dfrac{\sqrt{7+2\sqrt{7}.1+1}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=-\dfrac{2}{\sqrt{2}}\)
\(=-\sqrt{2}\)
\(P=\dfrac{1-\sqrt{x-1}}{\sqrt{x-2\sqrt{x-1}}}\)
\(=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1-2\sqrt{x-1}\cdot1+1}}\)
\(=\dfrac{1-\sqrt{x-1}}{\sqrt{x-1}-1}\)
=-1
ĐKXĐ: \(x>0\)
Áp dụng BĐT Cauchy cho 2 số dương:
\(\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x}\right)^2=1\Leftrightarrow x=1\left(tm\right)\)
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