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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)\(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+7\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)
b)\(\left(7+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(7+\sqrt{14}\right)\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=\sqrt{7}\left(7-2\right)=5\sqrt{7}\)
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 : \(D=\sqrt{11-4\sqrt{7}}-\sqrt{2}\sqrt{8+3\sqrt{7}}\)
\(=\sqrt{4-4\sqrt{7}+7}-\sqrt{9+6\sqrt{7}+7}\)
\(=\sqrt{7}-2-3-\sqrt{7}=-5\)
a)\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
b) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}=20\sqrt{3}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}=4\sqrt{3}\)
c)\(\sqrt{12}+2\sqrt{75}-3\sqrt{48}-\frac{2}{7}\sqrt{147}=2\sqrt{3}+10\sqrt{3}-12\sqrt{3}-2\sqrt{3}=-2\sqrt{3}\)
d) \(\sqrt{\left(3+\sqrt{5}\right)^2}-\sqrt{9-4\sqrt{5}}\)
\(=\left|3+\sqrt{5}\right|-\sqrt{\left(\sqrt{5}-2\right)^2}=3+\sqrt{5}-\left|\sqrt{5}-2\right|=3+\sqrt{5}-\sqrt{5}+2=5\)
e) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{\sqrt{5}+\sqrt{2}}{3}\)
\(=\left[\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right]\cdot\frac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\cdot\frac{3}{\sqrt{5}+\sqrt{2}}=-3\)
Nản k lm nữa ^^
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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\(B=\sqrt{2}-\sqrt{3-\sqrt{5}}=\dfrac{2-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{2-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{2-\sqrt{5}+1}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}\)
--
\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
còn lại lúc nx mk lm nốt nhé, h bận
1. Đặt A =\(\sqrt{\frac{129}{16}+\sqrt{2}}\)
\(\sqrt{16}\)A = \(\sqrt{129+16\sqrt{2}}\)
4A = \(\sqrt{\left(8\sqrt{2}+1\right)^2}\)
4A = \(8\sqrt{2}+1\)
⇒ A = \(\frac{\text{}8\sqrt{2}+1}{4}\)= \(2\sqrt{2}\) + \(\frac{1}{4}\)
2. Đặt B = \(\sqrt{\frac{289+4\sqrt{72}}{16}}\)
\(\sqrt{16}\)B = \(\sqrt{289+24\sqrt{2}}\)
4B = \(\sqrt{\left(12\sqrt{2}+1\right)^2}\)
4B = \(12\sqrt{2}+1\)
⇒ B = \(\frac{12\sqrt{2}+1}{4}\)= \(3\sqrt{2}+\frac{1}{4}\)
3. \(\sqrt{2-\sqrt{3}}\). \(\left(\sqrt{6}+\sqrt{2}\right)\)
= \(\sqrt{2-\sqrt{3}}\). \(\sqrt{2}.\left(\sqrt{3}+1\right)\)
= \(\sqrt{4-2\sqrt{3}}\) . \(\left(\sqrt{3}+1\right)\)
= \(\sqrt{\left(\sqrt{3}-1\right)^2}\) . \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}-1\right)\). \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}\right)^2\) - 12
= 3 - 1
= 2
4. \(\left(\sqrt{21}+7\right)\). \(\sqrt{10-2\sqrt{21}}\)
= \(\left(\sqrt{21}+7\right)\) . \(\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
= \(\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\) . \(\left(\sqrt{7}-\sqrt{3}\right)\)
= \(\sqrt{7}\) \(\left[\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2\right]\)
= \(\sqrt{7}\) . (7 - 3)
= 4\(\sqrt{7}\)
5. \(2.\left(\sqrt{10}-\sqrt{2}\right)\). \(\sqrt{4+\sqrt{6-2\sqrt{5}}}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{4+\sqrt{5}-1}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{3+\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{12+4\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\left(\sqrt{10}+\sqrt{2}\right)\)
= \(\left(\sqrt{10}\right)^2-\left(\sqrt{2}\right)^2\)
= 10 - 2
= 8
6. \(\left(4\sqrt{2}+\sqrt{30}\right)\). \(\left(\sqrt{5}-\sqrt{3}\right)\). \(\sqrt{4-\sqrt{15}}\)
= \(\sqrt{2}\)\(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{4-\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{8-2\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)^2\)
= \(\left(4+\sqrt{15}\right)\). \(\left(8-2\sqrt{15}\right)\)
= 32 - \(8\sqrt{15}\) + \(8\sqrt{15}\) - 30
= 2
7. \(\left(7-\sqrt{14}\right)\) . \(\sqrt{9-2\sqrt{14}}\)
= \(\sqrt{7}\) \(\left(\sqrt{7}-\sqrt{2}\right)\). \(\left(\sqrt{7}-\sqrt{2}\right)\)
= \(\sqrt{7}\). \(\left(\sqrt{7}-\sqrt{2}\right)^2\)
= \(\sqrt{7}\) . \(\left(9-2\sqrt{14}\right)\)
= 9\(\sqrt{7}\) - 14\(\sqrt{2}\)
TICK MÌNH NHA!
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.
\(E=\sqrt{4+2\sqrt{2}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2^2-(2+\sqrt{2})}\)
\(=\sqrt{4+2\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}.\sqrt{2^2-(\sqrt{2})^2}=\sqrt{2}.\sqrt{2}=2\)
\(F=(\sqrt{2}-\sqrt{3+\sqrt{5}})\sqrt{2}+2\sqrt{5}\)
\(=2-\sqrt{6+2\sqrt{5}}+2\sqrt{5}\)
\(=2-\sqrt{5+1-2\sqrt{5}}+2\sqrt{5}\)
\(=2-\sqrt{(\sqrt{5}-1)^2}+2\sqrt{5}\)
\(=2-(\sqrt{5}-1)+2\sqrt{5}=3+\sqrt{5}\)
\(G=(\sqrt{14}-\sqrt{10}).\sqrt{6+\sqrt{35}}=\sqrt{2}(\sqrt{7}-\sqrt{5})\sqrt{6+\sqrt{35}}\)
\(=(\sqrt{7}-\sqrt{5})\sqrt{12+2\sqrt{35}}=(\sqrt{7}-\sqrt{5}).\sqrt{7+5+2\sqrt{7.5}}\)
\(=(\sqrt{7}-\sqrt{5}).\sqrt{(\sqrt{7}+\sqrt{5})^2}=(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})=7-5=2\)
\(H=\sqrt{11-4\sqrt{7}}-\sqrt{2}.\sqrt{8+3\sqrt{7}}\)
\(=\sqrt{2^2+7-2.2.\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{(2-\sqrt{7})^2}-\sqrt{3^2+7+2.3\sqrt{7}}=\sqrt{(2-\sqrt{7})^2}-\sqrt{(3+\sqrt{7})^2}\)
\(=|2-\sqrt{7}|-|3+\sqrt{7}|=\sqrt{7}-2-(3+\sqrt{7})=-5\)
1: \(\left(\sqrt{10}-\sqrt{14}\right)\cdot\sqrt{6+\sqrt{35}}\)
\(=\left(\sqrt{5}-\sqrt{7}\right)\cdot\sqrt{12+2\sqrt{35}}\)
\(=\left(\sqrt{5}-\sqrt{7}\right)\left(\sqrt{5}+\sqrt{7}\right)\)
=5-7=-2
2: Sửa đề: \(\sqrt{4+\sqrt{8}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{2^2-\left(2+\sqrt{2}\right)}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}=\sqrt{2}\)