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1. \(=\sqrt{\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{3}{2}}\right)^2}+\sqrt{\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{3}{2}}\right)^2}-2\sqrt{4\sqrt{7}}=\frac{7}{2}+\frac{3}{2}+\frac{7}{2}-\frac{3}{2}-2\sqrt{4\sqrt{7}}\)
\(=7-2\sqrt{4\sqrt{7}}\)
cho hỏi tại sao có số \(\frac{7}{2};\frac{3}{2}\)zậy chỉ với
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
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
a, Nghe đề sai sai là lạ
b, Ta có : \(B=\left(\sqrt{2}-\sqrt{3+\sqrt{5}}\right)\sqrt{2}+2\sqrt{5}\)
\(=\sqrt{4}-\sqrt{6+2\sqrt{5}}+2\sqrt{5}=2+2\sqrt{5}-\sqrt{5+2\sqrt{5}+1}\)
\(=2+2\sqrt{5}-\sqrt{5}-1=\sqrt{5}+1\)
c, Ta có : \(C=\left(\sqrt{14}-\sqrt{10}\right)\left(\sqrt{6}+\sqrt{35}\right)\)
\(=\sqrt{84}-\sqrt{60}+\sqrt{490}-\sqrt{350}=2\sqrt{21}-2\sqrt{15}+7\sqrt{10}-5\sqrt{14}\)
d, Ta có : \(D=\sqrt{11-4\sqrt{7}}-\sqrt{2}\sqrt{8+3\sqrt{7}}\)
\(=\sqrt{4-4\sqrt{7}+7}-\sqrt{9+6\sqrt{7}+7}\)
\(=\sqrt{7}-2-3-\sqrt{7}=-5\)
1) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
Xin lỗi xin lỗi :v
1)\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
= \(\sqrt{7}.\left(3\sqrt{7}-2\sqrt{14}\right)+14\sqrt{2}\)
= 21 - \(14\sqrt{2}+14\sqrt{2}\)
= 21
2) \(\left(\sqrt{8}-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{18}-\sqrt{8}+\sqrt{5}\right)\)
= \(\left(2\sqrt{2}-\sqrt{2}-\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{5}-2\sqrt{2}\right)\)
= \(\left(\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}+\sqrt{5}\right)\)
=\(\left(\sqrt{2}\right)^2-\left(\sqrt{5}\right)^2\)
= -3
b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)
\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)
c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)
\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)
\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)
d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)