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......................?
mik ko biết
mong bn thông cảm
nha ................
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\(C=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[6]{\left(7-4\sqrt{3}\right).\left(7+4\sqrt{3}\right)}-x}{\sqrt[4]{\left(9+4\sqrt{5}\right).\left(9-4\sqrt{5}\right)}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1+\sqrt{x}\right).\left(1-\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{12-3}.\sqrt{7}-\sqrt{12+3}.\sqrt{7}\)
\(=\sqrt{7}.\sqrt{12^2-3^2}\)
\(=\sqrt{7}.\sqrt{135}\)
\(=\sqrt{945}\)
\(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
Ta có :
\(\left(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\right)^2\)
= \(12-3\sqrt{7}+12+3\sqrt{7}-2\sqrt{12-3\sqrt{7}}.\sqrt{12+3\sqrt{7}}\)
= \(24-2.\sqrt{12^2-\left(3\sqrt{7}\right)^2}\)
= \(24-2.\sqrt{144-63}\)
= \(24-18=6\)
Mặt khác ta dễ thấy : \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}< 0\)
\(\Rightarrow\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}=-\sqrt{6}\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{12-6\sqrt{3}}-\)\(\sqrt{7-4\sqrt{3}}=\)\(\sqrt{3^2-2.3.\sqrt{3}+3}-\)\(\sqrt{2^2-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(3-\sqrt{3}\right)^2}-\)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=3-\sqrt{3}-\)\(\left(2-\sqrt{3}\right)\)
\(=3-\sqrt{3}-2+\sqrt{3}\)\(=1\)
các bạn ơi giúp mình vs
Rút gon P=\((-3x +5\sqrt(x)-2)/(x+4\sqrt(x)-5) \)