CM A thuoc Z va B thuoc Z voi :
A = \(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
B = \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
Những câu hỏi liên quan
Chứng minh: A\(\in\)Z và B\(\in\)Z với
A=\(\sqrt{6-2\sqrt{5}}\)\(-\)\(\sqrt{6+2\sqrt{5}}\)
B=\(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}\)\(-\)\(\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}-1-\sqrt{5}-1=-2\)
Vậy \(A\in Z\)
Làm tương tự với B.
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Bình luận (0)
C/m A, B ∈ Z, với:
A= \(\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
B= \(\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
Help me
*\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}-1-\sqrt{5}+1=2\)
\(\Rightarrow A\in Z\)
* \(B=\dfrac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-2\sqrt{2}}}-\dfrac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\) \(=\dfrac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\dfrac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\) \(=\dfrac{\sqrt{2}-1}{3-2\sqrt{2}}-\dfrac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\) \(=\dfrac{3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}}{9-8}\)
\(=2\)
\(\Rightarrow B\in Z\)
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Bình luận (1)
Tính giá trị các biểu thức sau:
a) Asqrt{frac{2+sqrt{3}}{2-sqrt{3}}}+sqrt{frac{2-sqrt{3}}{2+sqrt{3}}}
b) Afrac{sqrt{3-2sqrt{2}}}{sqrt{17-12sqrt{2}}}-frac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}
c) Afrac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}+frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}
c) Afrac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}+frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}-frac{sqrt{5}+1}{sqrt{5}-1}
Đọc tiếp
Tính giá trị các biểu thức sau:
a) \(A=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)
b) \(A=\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
c) \(A=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
c) \(A=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
a/ \(A=\frac{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{2-\sqrt{3}}+\frac{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}{2+\sqrt{3}}\)
\(A=\frac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4}{1}=4\)
b/\(A=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
\(A=\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(A=\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{9-8}\)
\(A=3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}=8\)
c/ \(A=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{5-3}\)
\(A=\frac{5+2\sqrt{15}+3+5-2\sqrt{15}+3}{2}=8\)
d/ theo câu c có \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=8\)
\(\Rightarrow A=8-\frac{\left(\sqrt{5}+1\right)^2}{5-1}=\frac{32-5-2\sqrt{5}-1}{4}=\frac{2\left(13-\sqrt{5}\right)}{4}=\frac{13-\sqrt{5}}{2}\)
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Bình luận (1)
Chứng minh A,B là số nguyên với:
A = \(\sqrt{6-2\sqrt{5}}\)- \(\sqrt{6+2\sqrt{5}}\)
B= \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17}-12\sqrt{2}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
Cách 1 :\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\sqrt{5}^2-2\sqrt{5}+\sqrt{1}^2}-\sqrt{\sqrt{5}^2+2\sqrt{5}+\sqrt{1}^2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{1}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\)
\(=|\sqrt{5}-\sqrt{1}|-|\sqrt{5}+\sqrt{1}|=\sqrt{5}-\sqrt{1}-\sqrt{5}-\sqrt{1}=-2\)
Cách 2 \(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(< =>A^2=6-2\sqrt{5}-6-2\sqrt{5}+2\sqrt{36-20}\)
\(< =>A^2=8-2\sqrt{5}-2\sqrt{5}=8-2\left(2\sqrt{5}\right)=8-4\sqrt{5}\)
<=>...
\(B=\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{2}+\sqrt{1}}{\sqrt{17+12\sqrt{2}}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\sqrt{17+12\sqrt{2}}-\left(\sqrt{2}+1\right)\sqrt{17-12\sqrt{2}}}{\sqrt{17^2-\left(12\sqrt{2}\right)^2}}\)
tự làm tiếp đi , mình lười viết
Cho hđt:
sqrt{apmsqrt{b}}sqrt{frac{a+sqrt{a^2-b}}{2}}pmsqrt{frac{a-sqrt{a^2-b}}{2}} (a,b0 và a^2-b0)
Áp dụng kq để rút gọn:
a.frac{2+sqrt{3}}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{2-sqrt{3}}{sqrt{2}-sqrt{2-sqrt{3}}}
b. frac{sqrt{3-2sqrt{2}}}{sqrt{17-12sqrt{2}}}-frac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}
c. sqrt{frac{2sqrt{10}+sqrt{30}-2sqrt{2}-sqrt{6}}{2sqrt{10}-2sqrt{2}}}:frac{2}{sqrt{3}-1}
Đọc tiếp
Cho hđt:
\(\sqrt{a\pm\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\) (a,b>0 và \(a^2-b>0\))
Áp dụng kq để rút gọn:
\(a.\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b. \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
c. \(\sqrt{\frac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}}:\frac{2}{\sqrt{3}-1}\)
1/ frac{2}{3-sqrt{7}}sqrt{frac{6sqrt{2}-2sqrt{14}}{3sqrt{2}+sqrt{14}}}2/ sqrt{6+2sqrt{sqrt{5}-sqrt{13-sqrt{48}}}}3/ frac{sqrt{3-2sqrt{2}}}{sqrt{17-12sqrt{2}}}-frac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}4/ frac{24}{sqrt{7}+1}+frac{4}{3+sqrt{7}}-frac{3}{sqrt{7}+2}left(4-sqrt{7}right)5/ sqrt{7-3sqrt{5}}left(7+3sqrt{5}right)left(3sqrt{2}+sqrt{10}right)
Đọc tiếp
1/ \(\frac{2}{3-\sqrt{7}}\sqrt{\frac{6\sqrt{2}-2\sqrt{14}}{3\sqrt{2}+\sqrt{14}}}\)
2/ \(\sqrt{6+2\sqrt{\sqrt{5}-\sqrt{13-\sqrt{48}}}}\)
3/ \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
4/ \(\frac{24}{\sqrt{7}+1}+\frac{4}{3+\sqrt{7}}-\frac{3}{\sqrt{7}+2}\left(4-\sqrt{7}\right)\)
5/ \(\sqrt{7-3\sqrt{5}}\left(7+3\sqrt{5}\right)\left(3\sqrt{2}+\sqrt{10}\right)\)
1. So sánh:
a. sqrt{18}+sqrt{19} và 9
b. frac{16}{sqrt{2}}và sqrt{5}.sqrt{25}
2. Cho Hđt sqrt{apmsqrt{b}}sqrt{frac{a+sqrt{a^2-b}}{2}}pmsqrt{frac{a-sqrt{a^2-b}}{2}}vs left(a,b0,a^2-b0right)
Áp dụng kết quả để rút gọn:
a. frac{2+sqrt{3}}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{2-sqrt{3}}{sqrt{2}-sqrt{2-sqrt{3}}}
b. frac{sqrt{3-2sqrt{2}}}{sqrt{17-12sqrt{2}}}-frac{sqrt{3+2sqrt{2}}}{sqrt{17+12sqrt{2}}}
c. sqrt{frac{2sqrt{10}+sqrt{30}-2sqrt{2}-sqrt{6}}{2sqrt{10}-2sqrt{2}}}:frac{2}{sqrt{3}-1}
Đọc tiếp
1. So sánh:
a. \(\sqrt{18}+\sqrt{19}\) và 9
b. \(\frac{16}{\sqrt{2}}\)và \(\sqrt{5}.\sqrt{25}\)
2. Cho Hđt \(\sqrt{a\pm\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\)vs \(\left(a,b>0,a^2-b>0\right)\)
Áp dụng kết quả để rút gọn:
a. \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
b. \(\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
c. \(\sqrt{\frac{2\sqrt{10}+\sqrt{30}-2\sqrt{2}-\sqrt{6}}{2\sqrt{10}-2\sqrt{2}}}:\frac{2}{\sqrt{3}-1}\)
bài 1:rút gọn biểu thức:
a)sqrt{17-3sqrt{32}}+sqrt{17+3sqrt{32}}
b)sqrt{5+2sqrt{6}}-sqrt{5-2sqrt{6}}
c)left(frac{2}{sqrt{3}-1}+frac{3}{sqrt{3}-2}+frac{15}{3-sqrt{3}}right)timesfrac{1}{sqrt{3}+5}
d)2sqrt{40sqrt{12}}+3sqrt{5sqrt{48}}-2sqrt{sqrt{75}}-4sqrt{15sqrt{27}}
bài 2:giải phương trình
a)sqrt{9-12chi+4chi^2}4
b)sqrt{chi^2-2chi+1}+sqrt{chi^2-6chi+9}1
GIÚP EM VỚI EM ĐANG CẦN GẤP Ạ!!!!!!!!!!!!!!!!!
Đọc tiếp
bài 1:rút gọn biểu thức:
a)\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
b)\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
c)\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right)\times\frac{1}{\sqrt{3}+5}\)
d)\(2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}\)
bài 2:giải phương trình
a)\(\sqrt{9-12\chi+4\chi^2}=4\)
b)\(\sqrt{\chi^2-2\chi+1}+\sqrt{\chi^2-6\chi+9}=1\)
GIÚP EM VỚI EM ĐANG CẦN GẤP Ạ!!!!!!!!!!!!!!!!!
Thực hiện phép tính:
a)frac{sqrt{9-6sqrt{2}}-sqrt{6}}{sqrt{3}}
b)sqrt{frac{3-2sqrt{2}}{7-12sqrt{2}}}-sqrt{frac{3+2sqrt{2}}{17+12sqrt{2}}}
c)sqrt{6-sqrt{6-sqrt{25-sqrt{96}}}}
d)frac{2+sqrt{3}}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{2-sqrt{3}}{sqrt{2}-sqrt{2-sqrt{3}}}
e)frac{sqrt{3-sqrt{5}}}{sqrt{2}}
f)frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}+frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}
g)sqrt{left|40sqrt{2}-57right|}-sqrt{40sqrt{2}+57}
h) frac{sqrt{2+sqrt{3}}+sqrt{2-sqrt{3}}}{sqrt{2+sqrt{3}}-sqrt{2-sqrt{3...
Đọc tiếp
Thực hiện phép tính:
a)\(\frac{\sqrt{9-6\sqrt{2}}-\sqrt{6}}{\sqrt{3}}\)
b)\(\sqrt{\frac{3-2\sqrt{2}}{7-12\sqrt{2}}}-\sqrt{\frac{3+2\sqrt{2}}{17+12\sqrt{2}}}\)
c)\(\sqrt{6-\sqrt{6-\sqrt{25-\sqrt{96}}}}\)
d)\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
e)\(\frac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}\)
f)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
g)\(\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{40\sqrt{2}+57}\)
h) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}\)