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\(=\frac{-40\sqrt{3}+30\sqrt{2}}{-4\sqrt{3}+3\sqrt{2}}=10\)
=\(\sqrt{3-\sqrt{5}}\)\(\sqrt{2}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{6-2\sqrt{5}}\)(\(\sqrt{5}-1\)) (\(3+\sqrt{5}\))
=\(\sqrt{\left(\sqrt{5}+1\right)^2}\)(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=(\(\sqrt{5}+1\))(\(\sqrt{5}-1\))(\(3+\sqrt{5}\))
=4(\(3+\sqrt{5}\))
=12+4\(\sqrt{5}\)
Mk sửa lại đề nha
\(A=\left(\frac{x-5\sqrt{x}}{x-25}-1\right):\left(\frac{25-x}{x+2\sqrt{x}-15}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\left(ĐKXĐ:x\ne25\right)\)
\(A=\left(\frac{x-5\sqrt{x}-x+25}{x-25}\right):\left(\frac{25-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}+\frac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(A=\left(\frac{25-5\sqrt{x}}{x-25}\right):\left(\frac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(A=\left(\frac{5.\left(5-\sqrt{x}\right)}{x-25}\right):\left(\frac{9-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(1a.A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\) ( x ≥ 0 ; x # 9 )
\(b.A>\dfrac{1}{3}\) ⇔ \(\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\text{⇔}\dfrac{3-\sqrt{x}}{3\left(\sqrt{x}+3\right)}>0\)
⇔ \(3-\sqrt{x}>0\)
⇔ \(x< 9\)
Kết hợp ĐKXĐ , ta có : \(0\text{≤}x< 9\)
\(c.\) Tìm GTLN chứ ?
\(A=\dfrac{2}{\sqrt{x}+3}\text{≤}\dfrac{2}{3}\)
⇒ \(A_{MAX}=\dfrac{2}{3}."="x=0\left(TM\right)\)
\(a.VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9=VP\)Vậy , đẳng thức được chứng minh .
\(b.VT=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}+\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}=VP\)Vậy , đẳng thức được chứng minh .
\(c.VT=\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{\sqrt{5}+2}=\dfrac{2\left(\sqrt{5}+2\right)-2\left(\sqrt{5}-2\right)}{5-4}=8=VP\)Vậy , đẳng thức được chứng minh .
\(A=\dfrac{10.\sqrt{18}+5\sqrt{3}-15\sqrt{27}}{\sqrt{3}.\left(\sqrt{6}-4\right)}\)
\(A=\dfrac{10.\sqrt{3.6}+5\sqrt{3}-15.\sqrt{3.3^2}}{\sqrt{3}.\left(\sqrt{6}-4\right)}\)
\(A=\dfrac{10.\sqrt{3}.\sqrt{6}+5\sqrt{3}-15.\sqrt{3}.3}{\sqrt{3}\left(\sqrt{6}-4\right)}\)
\(A=\dfrac{\sqrt{3}.\left(10.\sqrt{6}+5-15.3\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}\)
\(A=\dfrac{10.\sqrt{6}+5-45}{\sqrt{6}-4}=\dfrac{10.\sqrt{6}-40}{\sqrt{6}-4}\)
\(A=\dfrac{10.\left(\sqrt{6}-4\right)}{\sqrt{6}-4}=10\)
Vậy \(A=10\)
Chúc bạn học tốt!!!
cảm ơn