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câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
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
a)\(\left(4\sqrt{2}+\sqrt{30}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)
\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(=\sqrt{10\left(4-\sqrt{15}\right)}+\sqrt{6\left(4-\sqrt{15}\right)}\)
\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)
\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(3-\sqrt{15}\right)^2}\)
\(=5-\sqrt{15}+\sqrt{15}-3\)
\(=2\)
b) \(2\left(\sqrt{10}-\sqrt{2}\right)\left(4+\sqrt{6-2\sqrt{5}}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{\left(1-\sqrt{5}\right)^2}\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(4+\sqrt{5}-1\right)\)
\(=\left(2\sqrt{10}-2\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(=6\sqrt{10}+2\sqrt{50}-6\sqrt{2}-2\sqrt{10}\)
\(=6\sqrt{10}+10\sqrt{2}-6\sqrt{2}-2\sqrt{10}\)
\(=4\sqrt{10}+4\sqrt{2}\)
c) \(\left(\sqrt{7}+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\sqrt{\left(\sqrt{2}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{7}+\sqrt{14}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=7\sqrt{7}-7\sqrt{2}+\sqrt{98}-\sqrt{28}\)
\(=7\sqrt{7}-7\sqrt{2}+7\sqrt{2}-2\sqrt{7}\)
\(=5\sqrt{7}\)
d) \(\sqrt{\dfrac{289+4\sqrt{72}}{16}}\)
\(=\sqrt{\dfrac{289+42\sqrt{2}}{16}}\)
\(=\dfrac{\sqrt{289+42\sqrt{2}}}{\sqrt{4^2}}\)
\(=\dfrac{\sqrt{\left(1+12\sqrt{2}\right)^2}}{4}\)
\(=\dfrac{1+12\sqrt{2}}{4}\)
e) \(\left(\sqrt{21}+7\right)\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\sqrt{\left(\sqrt{3}-\sqrt{7}\right)^2}\)
\(=\left(\sqrt{21}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{147}-\sqrt{63}+7-\sqrt{21}\)
\(=7\sqrt{3}-\sqrt{63}+7-\sqrt{21}\)
f) bạn xem đề lại nhé
Lời giải:
\(B=(\sqrt{2}-\sqrt{3-\sqrt{5}})\sqrt{2}=2-\sqrt{6-2\sqrt{5}}\)
\(=2-\sqrt{5+1-2\sqrt{5}}=2-\sqrt{(\sqrt{5}-1)^2}=2-(\sqrt{5}-1)=3-\sqrt{5}\)
\(C=\sqrt{4-\sqrt{7}}-\sqrt{4}+\sqrt{7}=\sqrt{\frac{8-2\sqrt{7}}{2}}-2+\sqrt{7}\)
\(=\sqrt{\frac{7+1-2\sqrt{7}}{2}}-2+\sqrt{7}\)
\(=\sqrt{\frac{(\sqrt{7}-1)^2}{2}}-2+\sqrt{7}\)
\(=\frac{|\sqrt{7}-1|}{\sqrt{2}}-2+\sqrt{7}=\frac{\sqrt{7}-1}{\sqrt{2}}-2+\sqrt{7}\)
$D$: bạn xem lại đề, mình thấy biểu thức không rút gọn được nữa.
\(E=\sqrt{4+2\sqrt{2}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-(2+\sqrt{2})}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{2^2-(\sqrt{2})^2}=\sqrt{2}.\sqrt{2}=2\)
\(F=(\sqrt{2}-\sqrt{3+\sqrt{5}})\sqrt{2}+2\sqrt{5}\)
\(=2-\sqrt{6+2\sqrt{5}}+2\sqrt{5}\)
\(=2-\sqrt{5+1-2\sqrt{5}}+2\sqrt{5}\)
\(=2-\sqrt{(\sqrt{5}-1)^2}+2\sqrt{5}\)
\(=2-(\sqrt{5}-1)+2\sqrt{5}=3+\sqrt{5}\)
\(G=(\sqrt{14}-\sqrt{10}).\sqrt{6+\sqrt{35}}=\sqrt{2}(\sqrt{7}-\sqrt{5})\sqrt{6+\sqrt{35}}\)
\(=(\sqrt{7}-\sqrt{5})\sqrt{12+2\sqrt{35}}=(\sqrt{7}-\sqrt{5}).\sqrt{7+5+2\sqrt{7.5}}\)
\(=(\sqrt{7}-\sqrt{5}).\sqrt{(\sqrt{7}+\sqrt{5})^2}=(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})=7-5=2\)
\(H=\sqrt{11-4\sqrt{7}}-\sqrt{2}.\sqrt{8+3\sqrt{7}}\)
\(=\sqrt{2^2+7-2.2.\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{(2-\sqrt{7})^2}-\sqrt{3^2+7+2.3\sqrt{7}}=\sqrt{(2-\sqrt{7})^2}-\sqrt{(3+\sqrt{7})^2}\)
\(=|2-\sqrt{7}|-|3+\sqrt{7}|=\sqrt{7}-2-(3+\sqrt{7})=-5\)
a: \(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
c: \(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-\left(2+\sqrt{2}\right)}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2\left(\sqrt{2}+1\right)\cdot\sqrt{2}\cdot\left(\sqrt{2}-1\right)}=\sqrt{2\sqrt{2}}\)
d: \(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{12+2\sqrt{35}}\right)\)
\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)\)
=7-5=2
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, 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 : \(D=\sqrt{11-4\sqrt{7}}-\sqrt{2}\sqrt{8+3\sqrt{7}}\)
\(=\sqrt{4-4\sqrt{7}+7}-\sqrt{9+6\sqrt{7}+7}\)
\(=\sqrt{7}-2-3-\sqrt{7}=-5\)