\(x=\sqrt{40-x}.\sqrt{45-x}+\sqrt{45-x}.\sqrt{72-x}+\sqrt{72-x}.\sqrt{40-x}\)
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căn (40-x)=a , căn (45-x)=b,căn(72-x)=c (a,b,c >=0 )
đưa về hệ: ab+bc+ca=40-a^2 -> ab+bc+ca+a^2=40
ab+bc+ca=45-b^2......
ab+bc+ca=72-c^2.....
đến đó ok rồi
\(45+x=\sqrt{72}\)
\(\Rightarrow45+x=\sqrt{36\times2}\)
\(\Rightarrow45+x=\sqrt{36}\times\sqrt{2}\)
\(\Rightarrow45+x=6\sqrt{2}\)
\(\Rightarrow x=6\sqrt{2}-45\)
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a: Ta có: \(A=\left(\frac{\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\frac{3}{\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+2}{\sqrt{x}}-\frac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\frac{\sqrt{x}-4+3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}:\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4-x}=\frac{4\sqrt{x}-4}{-4}=-\sqrt{x}+1\)
b: Khi \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) thì \(A=-\sqrt{\left(\sqrt5-1\right)^2}+1\)
\(=-\left(\sqrt5-1\right)+1=-\sqrt5+1+1=-\sqrt5+2\)
\(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=20+4\sqrt{x-2}\)
\(\Leftrightarrow6\sqrt{x-2}-3\sqrt{x-2}-4\sqrt{x-2}=20\)
\(\Leftrightarrow-\sqrt{x-2}=20\)(vô lý)