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a)\(\sqrt{13-4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{12-2.2\sqrt{3}.1+1}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2\sqrt{3}-1\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|2\sqrt{3}-1\right|+\left|2-\sqrt{3}\right|\)
\(=2\sqrt{3}-1+2-\sqrt{3}=\sqrt{3}+1\)
b)\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5+2\sqrt{5}.1+1}+\sqrt{5-2\sqrt{5}.1+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\left(\sqrt{5}+1\right)+\left(\sqrt{5}-1\right)=2\sqrt{5}\)
c)\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)=2\)
d)\(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(2+\sqrt{3}\right)+\left(2-\sqrt{3}\right)=4\)
e)\(\sqrt{9+4\sqrt{5}}=\sqrt{5+2.\sqrt{5}.2+4}=\sqrt{\left(\sqrt{5}+2\right)^2}=\sqrt{5}+2\)
f)\(\sqrt{23+8\sqrt{7}}=\sqrt{16+2.4.\sqrt{7}+7}=\sqrt{\left(4+\sqrt{7}\right)^2}=4+\sqrt{7}\)
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)
\(=3^2-\left(\sqrt{5}\right)^2\)
\(=9-5=4\)
3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)
=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn
A = \(\sqrt{2}\left(\sqrt{8}-\sqrt{32}-2\sqrt{18}\right)=\sqrt{16}-\sqrt{64}-2\sqrt{36}=4-8-2\cdot6=-4-12=-16\)
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\(B=\sqrt{2}-\sqrt{3-\sqrt{5}}=\dfrac{2-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{2-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{2-\sqrt{5}+1}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}\)
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\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
còn lại lúc nx mk lm nốt nhé, h bận
\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)
\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}=6\sqrt{3}-12\sqrt{3}+20\sqrt{3}=14\sqrt{3}\)
câu tiếp tương tự câu thứ 2 nha
b) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
= \(\sqrt{3.4-3\sqrt{7}}-\sqrt{3.4+3\sqrt{7}}\)
= \(\sqrt{3.\left(4-\sqrt{7}\right)}-\sqrt{3.\left(4+\sqrt{7}\right)}\)
= \(\sqrt{3}.\sqrt{4-\sqrt{7}}-\sqrt{3}.\sqrt{4+\sqrt{7}}\)
= \(\sqrt{3}.\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)\)
\(\)≈ \(-2,449\)
\(\sqrt{\dfrac{13}{4}+\sqrt{3}}-\sqrt{\dfrac{7}{4}-\sqrt{3}}\)
= \(\sqrt{\dfrac{13}{4}+\dfrac{4\sqrt{3}}{4}}-\sqrt{\dfrac{7}{4}-\dfrac{4\sqrt{3}}{4}}\)
= \(\sqrt{\dfrac{13+4\sqrt{3}}{4}}-\sqrt{\dfrac{7-4\sqrt{3}}{4}}\)
= \(\dfrac{\sqrt{13+4\sqrt{3}}}{\sqrt{4}}-\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)
= \(\dfrac{\sqrt{13+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}}{\sqrt{4}}\)
≈ \(2,098\)