x^2-(m-6)x+m+2=0
Tim m để pt có 2 nghiệm x1;x2 thỏa x1^8*x2+x1*x2^8=0
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\(M=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(M=\frac{x^3}{\left(x-2\right)\left(x+2\right)}-\frac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(M=\frac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(M=\frac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(M=\frac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(M=\frac{\left(x-1\right)\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(M=x-1\)
\(M=\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3}{\left(x-2\right)\left(x+2\right)}-\frac{x}{x-2}-\frac{2}{x+2}\)
\(=\frac{x^3-x\left(x+2\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^3-x^2-4x-4}{\left(x-2\right)\left(x+2\right)}\)
\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
?? :v
\(\sqrt{\frac{3\sqrt{5}+1}{2\sqrt{5}-3}}\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{\frac{\left(3\sqrt{5}+1\right)\left(2\sqrt{5}+3\right)}{\left(2\sqrt{5}-3\right)\left(2\sqrt{5}+3\right)}}.\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{3+\sqrt{5}}.\sqrt{2}.\left(\sqrt{5}-1\right)\)
\(=\sqrt{6+2\sqrt{5}}.\left(\sqrt{5}-1\right)\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=4\)