* Thực hiện phép tính
a, A= \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
b, B= \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c, C= \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
thực hiện phép tính
a, \(\dfrac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
b, \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
c, \(\sqrt{2-\sqrt{3}}.\left(\sqrt{5}+\sqrt{2}\right)\)
d, \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
e, \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
f, \(\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}\)
g, \(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
h, \(\dfrac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}\)
i, \(\dfrac{\left(\sqrt{5+2}\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
k, \(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}}\)
l, \(\dfrac{4}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)
m, \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
n, \(\dfrac{\sqrt{3}}{1-\sqrt{\sqrt{3+1}}}+\dfrac{\sqrt{3}}{1+\sqrt{\sqrt{3+1}}}\)
Thực hiện các phép tính sau:
a, A=\(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
b, B=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c, C=\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
a, A= \(\frac{\sqrt{48-12\sqrt{7}}}{2}-\frac{\sqrt{48+12\sqrt{7}}}{2}\)
= \(\frac{\sqrt{\left(\sqrt{42}-\sqrt{6}\right)^2}}{2}-\frac{\sqrt{\left(\sqrt{42}+\sqrt{6}\right)^2}}{2}\)
= \(\frac{-2\sqrt{6}}{2}\)
= \(-\sqrt{6}\)
\(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)\(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
Thực hiện phép tính:
a) \(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
b) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
c)\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}=A\\ \Rightarrow\sqrt{2}A=\sqrt{6+2\sqrt{5}+}\sqrt{6-\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\sqrt{5}+1+\sqrt{5}-1\\ =2\sqrt{5}\\ \Rightarrow A=\sqrt{2}.\sqrt{5}=\sqrt{10}\)
Tính
a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
b) \(\sqrt{4+\sqrt{7}} -\sqrt{4-\sqrt{7}}\)
c) \(\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
a: =2-căn 3-2-căn 3
=-2căn 3
b: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\right)\)
\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+1-\sqrt{7}+1\right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
c: \(A=\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
=>\(A^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}+2\cdot\sqrt{16-10+2\sqrt{5}}\)
\(\Leftrightarrow A^2=8+2\left(\sqrt{5}+1\right)=10+2\sqrt{5}\)
=>\(A=\sqrt{10+2\sqrt{5}}\)
Bài 1: Thực hiện các phép tính
a) A= \(\sqrt{12-2\sqrt{7}}-\sqrt{12+2\sqrt{7}}\)
b) B= \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c) C= \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
a: \(A^2=12-2\sqrt{7}+12+2\sqrt{7}-2\cdot\sqrt{116}\)
\(\Leftrightarrow A^2=24-4\sqrt{29}\)
hay \(A=\sqrt{24-4\sqrt{29}}\)
c: \(C=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
THỰC HIỆN PHÉP TÍNH
1,\(\sqrt{3+\sqrt{5}}.\sqrt{2}\)
2,\(\sqrt{3-\sqrt{5}.\sqrt{8}}\)
3,\((\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\sqrt{\dfrac{4}{3})}.\sqrt{12}\)
4,\((\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}):\sqrt{7}\)
5, \(\sqrt{36-12\sqrt{5}}:\sqrt{6}\)
6,\(\sqrt{3-\sqrt{5}:}\sqrt{2}\)
1: \(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
3) \(\left(\sqrt{\dfrac{3}{4}}-\sqrt{3}+5\cdot\sqrt{\dfrac{4}{3}}\right)\cdot\sqrt{12}\)
\(=\left(\dfrac{\sqrt{3}}{2}-\dfrac{2\sqrt{3}}{2}+5\cdot\dfrac{2}{\sqrt{3}}\right)\cdot\sqrt{12}\)
\(=\dfrac{17\sqrt{3}}{6}\cdot2\sqrt{3}\)
\(=\dfrac{34\cdot3}{6}=\dfrac{102}{6}=17\)
thực hiện phép tính :
a) A=\(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
b) B =\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
c) C =\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
giúp mjk bài này với :) mjk đang cần gấp lắm
a) \(A=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\Rightarrow A^2=12-3\sqrt{7}+12+3\sqrt{7}-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}\Rightarrow A^2=24-2\sqrt{144-63}\Rightarrow A^2=24-18\Rightarrow A^2=6\Rightarrow A=\pm\sqrt{6}\)Ta có \(12-3\sqrt{7}< 12+3\sqrt{7}\Rightarrow\sqrt{12-3\sqrt{7}}< \sqrt{12+3\sqrt{7}}\Rightarrow\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}< 0\Rightarrow A< 0\)Vậy A=-6
b) \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\Rightarrow B^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\Rightarrow B^2=8+2\sqrt{16-10-2\sqrt{5}}\Rightarrow B^2=8+2\sqrt{5-2\sqrt{5}+1}\Rightarrow B^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\Rightarrow B^2=8+2\sqrt{5}-2\Rightarrow B=\pm\sqrt{5+2\sqrt{5}+1}\Rightarrow B=\pm\left(\sqrt{5}+1\right)\)Ta có B>0⇒B=\(\sqrt{5}+1\)
c) \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\Rightarrow C^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\Rightarrow C^2=6+2\sqrt{9-5}\Rightarrow C^2=6+4=10\Rightarrow C=\pm\sqrt{10}\)Ta có C>0⇒C=\(\sqrt{10}\)
Thực hiện phép tính
a)\(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
b)\(\left(\sqrt[3]{25}-\sqrt[3]{10}+\sqrt[3]{4}\right)\)\(\left(\sqrt[3]{5}+\sqrt[3]{2}\right)\)
`a)\root[3]{135}/\root[3]{5}-\root[3]{54}.\root[3]{4}`
`=\root[3]{135/5}-\root[3]{54.4}`
`=\root[3]{27}-\root[3]{216}`
`=3-6=-3`
`b)(\root[3]{25}-\root[3]{10}+\root[3]{4})(\root[3]{5}+\root[3]{2})`
`=5+\root[3]{50}-\root[3]{50}-\root[3]{20}+\root[3]{20}+2`
`=7`.