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\(x=\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\)
\(\Leftrightarrow x^3=4\left(\sqrt{5}+1\right)-4\left(\sqrt{5}-1\right)-3.\sqrt[3]{4\left(\sqrt{5}+1\right).4\left(\sqrt{5}-1\right)}x\)
\(\Leftrightarrow x^3=8-3.\sqrt[3]{4^2.\left(5-1\right)}x\)
\(\Leftrightarrow x^3=8-3.4x=8-12x\)
\(\Rightarrow M=\left(x^3+12x-9\right)^{2014}=\left(8-12x+12x-9\right)^{2014}=\left(-1\right)^{2014}=1\)
ĐKXĐ: ...
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x\left(\sqrt{x}+1\right)}{\left(x+2\sqrt{x}\right)}=\frac{x}{\sqrt{x}-1}\)
\(x=\frac{2}{2-\sqrt{3}}=\frac{4}{4-2\sqrt{3}}=\left(\frac{2}{\sqrt{3}-1}\right)^2\)
\(\Rightarrow P=\frac{\frac{2}{2-\sqrt{3}}}{\frac{2}{\sqrt{3}-1}-1}=\frac{\frac{2}{2-\sqrt{3}}}{\frac{3-\sqrt{3}}{\sqrt{3}-1}}=\frac{2}{2\sqrt{3}-3}\)
\(\sqrt{P}\) xác định khi \(x>1\)
Khi đó: \(\sqrt{P}=\sqrt{\frac{x}{\sqrt{x}-1}}=\sqrt{\frac{x}{\sqrt{x}-1}-4+4}=\sqrt{\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge2\)
\(\sqrt{P}_{min}=2\) khi \(x=4\)
http://olm.vn/hoi-dap/question/104313.html
coi hỉu j ko tui đang mò
- \(A=\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}-1\right)^2}=\sqrt{10}-1\)
- \(B=\left(\sqrt{28}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}=\left(2\sqrt{7}-2\sqrt{4}+\sqrt{7}\right).\sqrt{7}+7\sqrt{7}\)
\(=\left(3\sqrt{7}-4\right).\sqrt{7}+7\sqrt{7}=3\sqrt{7}+3\sqrt{7}=6\sqrt{7}\)
- \(C=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\frac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
- \(D=0,2.\sqrt{10^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=2\sqrt{3}+2\left(\sqrt{3}-\sqrt{5}\right)=4\sqrt{3}-2\sqrt{5}\)
`Ta có : \(x=\sqrt[3]{4\sqrt{5}+4}-\sqrt[3]{4\sqrt{5}-4}\)
\(\Rightarrow x^3=8-3\sqrt[3]{\left(4\sqrt{5}\right)^2-4^2}.x\Leftrightarrow x^3+12x-8=0\Rightarrow x^3-12x-9=-1\)
Từ đó tính được P = (-1)2016 = 1
\(x+\sqrt{xy}=3\sqrt{xy}+15y\Leftrightarrow x-2\sqrt{xy}+y=16y\Leftrightarrow\sqrt{x}=\sqrt{y}+4\sqrt{y}=5\sqrt{y}\Leftrightarrow x=25y\)
\(E=\frac{50y+5y+3y}{25y+5y-y}=\frac{58}{29}=2\)
Đề có sai không vậy bạn?
Phải là \(4\left(\sqrt{5}+1\right)\) chứ
k có sai đâu bn giải giúp mk vs