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\(\frac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)=\(\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}\left(\sqrt{3}+\sqrt{5}\right)}\)=\(\frac{\sqrt{2}}{\sqrt{7}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Viết lại đề cho bạn nè
\(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
= \(\sqrt{9-6\sqrt{3}+3}-\sqrt{12-12\sqrt{3}+9}\)
= \(\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}\right)^2-2.3.\sqrt{12}+3^2}\)
= \(\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}-3\right)^2}\)
= |\(3-\sqrt{3}\)| - |\(\sqrt{12}-3\)|
= \(3-\sqrt{3}-\sqrt{12}+3\)
= \(6-\sqrt{3}-2\sqrt{3}\)
= \(6-3\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{9}+2\sqrt{21}\)
=\(\left(\sqrt{4}\sqrt{7}-\sqrt{7}-\sqrt{12}\right).3+2\sqrt{21}\)
=\(\left(2\sqrt{7}-\sqrt{7}-\sqrt{4}\sqrt{3}\right).3+2\sqrt{21}\)
=\(\left(\sqrt{7}-2\sqrt{3}\right).3+2\sqrt{21}\)
=\(3\sqrt{7}-6\sqrt{3}+2\sqrt{21}\)
đề có sai ko nhưng kết quả ra thế
\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{9}+2\sqrt{21}=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right).3+2\sqrt{21}=\left(\sqrt{7}-2\sqrt{3}\right).3+2\sqrt{21}=3\sqrt{7}-6\sqrt{3}+2\sqrt{21}\)
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\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\frac{16a^4b^6}{128a^6b^6}}=\sqrt{\frac{1}{8a^2}}=\frac{\sqrt{1}}{\sqrt{8a^2}}=\frac{1}{\sqrt{2}\sqrt{4}\sqrt{a}}\)
=\(\frac{1}{2\sqrt{2}a}\)
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tui hướng dẫn thui nha,,,,\(\sqrt{33-12\sqrt{6}}=\sqrt{24-2.2\sqrt{6}.3+9}=\sqrt{\left(2\sqrt{6}-3\right)^2}=2\sqrt{6}-3\)
ấy dễ ko???,,,bạn lm tương tự típ nhá,,,,
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đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
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-Nếu \(x\le2\) thì A=-81-25(2-x)+40=-81-50+25x+40=25x-91
-Nếu \(x\ge2\) thì A=-81-25(x-2)+40=-81-25x+50+40=9-25x