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1: \(=\sqrt{36}=6\)
2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)
3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)
4: \(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)
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 = \(\sqrt{2}\left(\sqrt{8}-\sqrt{32}-2\sqrt{18}\right)=\sqrt{16}-\sqrt{64}-2\sqrt{36}=4-8-2\cdot6=-4-12=-16\)
--
\(B=\sqrt{2}-\sqrt{3-\sqrt{5}}=\dfrac{2-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{2-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{2-\sqrt{5}+1}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}\)
--
\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
còn lại lúc nx mk lm nốt nhé, h bận
1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)
\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)
\(=6-8=-2\)
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
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
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.=\(\left(\sqrt{8-\sqrt{35}}\right)^2\)+\(\left(\sqrt{8+\sqrt{35^{ }}}\right)^2\)
= 8 -\(\sqrt{35}\)+8+\(\sqrt{35}\)
= 16
câu còn lại làm tương tự chỉ cần phân tách thôi