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23 tháng 10 2016
a/ Đặt \(\hept{\begin{cases}\sqrt{3+\sqrt{5}}=a\\\sqrt{3-\sqrt{5}}=b\end{cases}}\)
Khi đó ta có a2 + b2 = 6; ab = 2; a + b = \(\sqrt{10}\) ; a - b = \(\sqrt{2}\); a2 - b2 = \(2\sqrt{5}\)
Ta có cái ban đầu
\(=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}\)=
\(\frac{\sqrt{10}a^2+a^2b-\sqrt{10}b^2-ab^2}{10+\sqrt{10}a+\sqrt{10}b+ab}\)
\(=\frac{10\sqrt{2}+2\sqrt{2}}{10+10+2}=\frac{6\sqrt{2}}{11}\)
8 tháng 5 2019
a, \(\sqrt{2}A=\sqrt{10-2\sqrt{3.7}}+\sqrt{10+2\sqrt{3.7}}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{3}+\sqrt{7}=2\sqrt{7}\)
\(\Rightarrow A=\sqrt{14}\)
b, \(B=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{3\sqrt{5}}{2}\)
c, \(C=\left(1-\sqrt{11}\right)\left(\sqrt{11}+1\right)=1-11=-10\)
d, \(D=\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}{2-3}-\frac{\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}{2-3}\)
\(=-2-\sqrt{6}+2-\sqrt{6}=-2\sqrt{6}\)
PN
29 tháng 7 2020
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
9 tháng 8 2019
1/ \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
=\(\frac{\left(\sqrt{15}-\sqrt{5}\right)\cdot\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)}+\frac{\left(5-2\sqrt{5}\right)\cdot\left(2\sqrt{5}+4\right)}{\left(2\sqrt{5}-4\right)\cdot\left(2\sqrt{5}+4\right)}\)
=\(\frac{2\sqrt{5}}{2}+\frac{2\sqrt{5}}{4}\)
=\(\sqrt{5}+\frac{\sqrt{5}}{2}\)
=\(\frac{3\sqrt{5}}{2}\)
2/\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
=\(\frac{\left(\sqrt{15}-\sqrt{12}\right)\cdot\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\cdot\left(\sqrt{5}+2\right)}+\frac{\left(6+2\sqrt{6}\right)\cdot\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+2\right)\cdot\left(\sqrt{3}-2\right)}\)
=\(\frac{\sqrt{3}}{1}+\frac{2\sqrt{3}}{1}\)
=\(3\sqrt{3}\)
3/\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
=\(\frac{\sqrt{3}\cdot\left(3+2\sqrt{3}\right)}{3}+\frac{\left(2+\sqrt{2}\right)\cdot\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\cdot\left(\sqrt{2}-1\right)}-\left(2+\sqrt{3}\right)\)
=\(\frac{6+3\sqrt{3}}{3}+\sqrt{2}-\left(2-\sqrt{3}\right)\)
=\(\frac{3\cdot\left(2+\sqrt{3}\right)}{3}+\sqrt{2}-\left(2+\sqrt{3}\right)\)
=\(2+\sqrt{3}+\sqrt{2}-2-\sqrt{3}\)
=\(\sqrt{2}\)
Câu số 4 bạn có chắc là đúng đề bài không ạ ? Xem lại đề giúp mình nhé, cảm ơn bạn ^^
AH
Akai Haruma
Giáo viên
4 tháng 7 2019
Lời giải:
d)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+2\sqrt{12}}}}\)
\(=\sqrt{6+2\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}}=\sqrt{6+2\sqrt{5-(\sqrt{12}+1)}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}\)
\(=\sqrt{6+2(\sqrt{3}-1)}=\sqrt{4+2\sqrt{3}}=\sqrt{(\sqrt{3}+1)^2}\)
\(=\sqrt{3}+1\)
e)
\(\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}(2-\sqrt{3})}{2-\sqrt{3}}=\frac{2}{\sqrt{3}-1}+\sqrt{3}\)
\(=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\sqrt{3}=\frac{2(\sqrt{3}+1)}{3-1}+\sqrt{3}=\sqrt{3}+1+\sqrt{3}=2\sqrt{3}+1\)
Y
4 tháng 7 2019
d)
\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12+2\cdot\sqrt{12}\cdot1+1}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\\ =\sqrt{6+2\sqrt{4-\sqrt{12}}}\\ =\sqrt{6+2\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\left(\sqrt{3}-1\right)}\\ =\sqrt{6-2\sqrt{3}-2}\\ =\sqrt{4-2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
e)
\(\frac{2}{\sqrt{3}-1}-\frac{3-2\sqrt{3}}{2-\sqrt{3}}\\ =\frac{2}{\sqrt{3}-1}+\frac{\sqrt{3}\left(\sqrt{3}-2\right)}{\sqrt{3}-2}\\ =\frac{2}{\sqrt{3}-1}+\sqrt{3}\\ =\frac{2+\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\\ =\frac{5-\sqrt{3}}{\sqrt{3}-1}\\ =\frac{\left(5-\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}\right)^2-1}\\ -\frac{6\sqrt{3}-8}{2}=3\sqrt{3}-4\)
(bạn nhớ ktr đã nha)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 9 2020
\(=\frac{2\left(\sqrt{3}-1\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{2\left(\sqrt{3}+1\right)}{2-\sqrt{4-2\sqrt{3}}}=\frac{2\left(\sqrt{3}-1\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\frac{2\left(\sqrt{3}+1\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\frac{2\left(\sqrt{3}-1\right)}{2+\sqrt{3}+1}+\frac{2\left(\sqrt{3}+1\right)}{2-\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)}{3+\sqrt{3}}+\frac{2\left(\sqrt{3}+1\right)}{3-\sqrt{3}}\)
\(=\frac{2\left(\sqrt{3}-1\right)\left(3-\sqrt{3}\right)+2\left(\sqrt{3}+1\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}=\frac{16\sqrt{3}}{6}=\frac{8\sqrt{3}}{3}\)
27 tháng 7 2016
a) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
b)\(\frac{x-4}{2\left(\sqrt{x}+2\right)}\) (ĐK:x\(\ge0\))
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2}{2}\)
c)\(\frac{x-5\sqrt{x}+6}{3\sqrt{x}-6}\) (ĐK:x\(\ge0;x\ne4\))
\(=\frac{x-3\sqrt{x}-2\sqrt{x}+6}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)-2\left(\sqrt{x}-3\right)}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{3\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}-3}{3}\)
27 tháng 7 2016
b) Tử \(x-4=\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\) (hằng đăngt thức số 3 )