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1,\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
2, (tương tự ý 1 cũng tách thành hằng đẳng thức \(\sqrt{46-6\sqrt{5}}=\sqrt{\left(3\sqrt{5}-1\right)^2}\)và \(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}\)
3,\(\left(\sqrt{2}-\sqrt{9}\right)\left(\sqrt{\left(3+\sqrt{2}\right)^2}\right)=\left(\sqrt{2}-3\right)\left(\sqrt{2}+3\right)=2-9=-7\)
4, tương tự ý 3
mình ghi nhầm pn ơi.. bài 2 là \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{6}}\)
Do mình xem trong bộ đề kiểm tra đang làm thử xem cách giải là 12-18 nên thấy kì kì
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\(a.\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3=11-4\sqrt{6}\)
\(b.\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=\left(1+\sqrt{3}\right)^2-2=4+2\sqrt{3}-2=2+2\sqrt{3}\) \(c.\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{9-5}=6+4=10\) \(d.\left(\sqrt{\sqrt{11}+\sqrt{7}}-\sqrt{\sqrt{11}-\sqrt{7}}\right)^2=\sqrt{11}+\sqrt{7}+\sqrt{11}-\sqrt{7}-2\sqrt{11-7}=2\sqrt{11}-4\) \(e.\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\sqrt{2}\) \(f.\sqrt{21-12\sqrt{3}}-\sqrt{3}=\sqrt{12-2.2\sqrt{3}.3+9}-\sqrt{3}=2\sqrt{3}-3-\sqrt{3}=\sqrt{3}-3\)
\(g.\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{3+2\sqrt{3}+1}=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)=2\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)=2\left(3-4\right)=-2\)
\(h.\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{16-2.4\sqrt{2}+2}}}=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)
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
Nhìn nó có dạng \(\sqrt{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}=\sqrt{a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}\)
Thấy \(2\left(\sqrt{6}+\sqrt{12}+\sqrt{18}\right)=2\left(\sqrt{2.3}+\sqrt{2.6}+\sqrt{3.6}\right)\)
nên:
\(\sqrt{11+2\left(\sqrt{6}+\sqrt{12}+\sqrt{18}\right)}=\sqrt{\sqrt{2}^2+\sqrt{6}^2+\sqrt{3}^2+2\left(\sqrt{2.3}+\sqrt{2.6}+\sqrt{3.6}\right)}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{6}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{6}\)
Nguyễn Việt Lâm Anh ơi! Anh làm cho em câu này ạ!