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\(\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(\sqrt{3}+2\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{\sqrt{5}+2}+\sqrt{x}}\\ =\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+1-\sqrt{x}=1\)
\(K=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[3]{2+\sqrt{3}}-x}{\sqrt{\sqrt{5}-2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{1-x}{1+\sqrt{x}}=\sqrt{x}+1-\sqrt{x}=1\)
Vậy K không phụ thuộc vào x
\(P=\sqrt{x}+\dfrac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{\left(2+\sqrt{3}\right)^2}-x}{\sqrt[4]{\left(\sqrt{5}-2\right)^2}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{\sqrt[3]{\left(2-\sqrt{3}\right).\left(2+\sqrt{3}\right)}-x}{\sqrt{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\sqrt{x}}\)
\(=\sqrt{x}+\dfrac{1-x}{1+\sqrt{x}}=\sqrt{x}+1-\sqrt{x}=1\)
Ta có \(\sqrt[3]{2-\sqrt{3}}\times\sqrt[6]{7+4\sqrt{3}}\)
= \(\sqrt[3]{2-\sqrt{3}}\times\sqrt[3]{2+\sqrt{3}}\)
= 1
Và \(\sqrt[4]{9-4\sqrt{5}}\times\sqrt{2+\sqrt{5}}\)
= \(\sqrt{\sqrt{5}-2}\times\sqrt{2+\sqrt{5}}\)
= 1
Vậy A = \(\frac{\sqrt{x}+1-x}{1+\sqrt{x}}\)\(=1-\frac{x}{1+\sqrt{x}}\)
Vậy A phải phụ thuộc vào x. Có thể đề sai
Bài 2:
Ta có: \(B=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)
\(=\frac{\sqrt{\sqrt{5}-1}\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)}{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\frac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}}{2\sqrt{2}}-\left(\sqrt{2}-1\right)\)
\(=\frac{\sqrt{5}+1+3-\sqrt{5}}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\frac{4}{2\sqrt{2}}-\sqrt{2}+1\)
\(=\sqrt{2}-\sqrt{2}+1\)
=1
câu 3: C = \(\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)}{\left(\text{4+\sqrt{15}}\right)\left(\sqrt{10-\sqrt{6}}\right)\sqrt{4-\sqrt{15}}}\)
\(=\frac{\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}}{\sqrt{4+\sqrt{15}}.\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}}\)
=\(\frac{\sqrt{9-\left(\sqrt{5}\right)^2}\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}}{\sqrt{16-\left(\sqrt{15}\right)^2}.\left(\sqrt{10}-\sqrt{6}\right).\sqrt{4+\sqrt{15}}}\)
\(=\frac{2\left(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\right)}{\sqrt{40+10\sqrt{15}}-\sqrt{24-6\sqrt{15}}}\)
\(=2.\frac{\left(\sqrt{5}+5\right)-\left(\sqrt{5}+1\right)}{\left(\sqrt{15}+5\right)-\left(\sqrt{15}+3\right)}\)
= 4
Câu 2b đề là tìm x chứ nhỉ???
b) \(\sqrt{x^2-4}+\sqrt{x-2}=0\)
Ta có: \(\left\{{}\begin{matrix}\sqrt{x^2-4}\ge0\\\sqrt{x-2}\ge0\end{matrix}\right.\)
=> Dấu = xảy ra <=> \(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x-2}=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=\pm2\\x=2\end{matrix}\right.\) <=> x = 2
Vậy x = 2
bài 2 câu b) đề sai rồi bạn
còn bài 1 câu b) mình cảm thấy sai sai