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\(A=43+24\sqrt{3}-8\sqrt{20+2\sqrt{\left(3\sqrt{3}+4\right)^2}}\)
\(=43+24\sqrt{3}-8\sqrt{20+2\left(3\sqrt{3}+4\right)}\)
\(=43+24\sqrt{3}-8\sqrt{28+6\sqrt{3}}\)
\(=43+24\sqrt{3}-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)
\(=43+24\sqrt{3}-8\left(3\sqrt{3}+1\right)\)
\(=43-8=35\)
\(=\frac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=\frac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)
\(=\frac{15}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)
\(=\frac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{\sqrt{3}\left(2+\sqrt{6}\right)+\sqrt{3}\left(2-\sqrt{6}\right)}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}-\frac{1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(-2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{1}{\sqrt{2}}-\frac{2-\sqrt{6}}{\left(2-\sqrt{6}\right)\left(2+\sqrt{6}\right)}+\frac{\left(\sqrt{2}-1\right)\left(-2\sqrt{6}+6\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\frac{1}{\sqrt{2}}\)
\(=\frac{2-\sqrt{6}}{2}-4\sqrt{3}+6\sqrt{2}+2\sqrt{6}-6\)
\(=6\sqrt{2}-4\sqrt{3}+\frac{3\sqrt{6}}{2}-5\)
Kết quả xấu quá, chắc bạn ghi nhầm đề
Đã kiểm tra đáp án bằng casio
Ta có:
\(P=\sqrt{\frac{15}{2}}\cdot\sqrt{\frac{10\left(a-1\right)^2}{3}}\\ =\sqrt{\frac{15}{2}\cdot\frac{10\left(a-1\right)^2}{3}}\\ =\sqrt{25\left(a-1\right)^2}\\ =5\left|a-1\right|\\ =\left[{}\begin{matrix}5\left(a-1\right)\left(a=1\right)\\5\left(1-a\right)\left(a< 1\right)\end{matrix}\right.\\ =\left[{}\begin{matrix}5a-5\\5-5a\end{matrix}\right.\)
P.s: Ko chắc lắm nha :v
