1: \(1+\sqrt{6+2\sqrt{5}}=\sqrt{5}+2\)
2: \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5}\)
3: \(\sqrt{7+4\sqrt{3}}=2+\sqrt{3}\)
Đúng 0
Bình luận (0)
Bài 4: Liên hệ giữa phép chia và phép khai phương
Các câu hỏi tương tự
Tính: \(\left(\sqrt{\frac{\text{1}}{7}}-\sqrt{\frac{\text{1}\text{6}}{7}}\text{+}\sqrt{7}\right):\sqrt{7}\)
Tính:Asqrt{4+2sqrt{3}}-sqrt{4-2sqrt{3}}Bsqrt{9-4sqrt{5}}+sqrt{9+4sqrt{5}}Csqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}Dsqrt{5sqrt{3+5sqrt{48-10sqrt{7+4sqrt{3}}}}}Eleft(4+sqrt{15}right)left(sqrt{10}-sqrt{6}right)sqrt{4-sqrt{15}}(2 cách)Fdfrac{sqrt{17-12sqrt{2}}}{sqrt{3-2sqrt{2}}}-dfrac{sqrt{17}+12sqrt{2}}{sqrt{3+2sqrt{2}}}
Đọc tiếp
Tính:
A=\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
B=\(\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}\)
C=\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
D=\(\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
E=\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)(2 cách)
F=\(\dfrac{\sqrt{17-12\sqrt{2}}}{\sqrt{3-2\sqrt{2}}}-\dfrac{\sqrt{17}+12\sqrt{2}}{\sqrt{3+2\sqrt{2}}}\)
Tính
a) \(\frac{10}{\sqrt{5}}+\frac{8}{3 +\sqrt{5}}-\frac{\sqrt{15}-2\sqrt{5}}{\sqrt{3}-2}\)
b) \(\left(\frac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}-\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{5}-\sqrt{7}}\)
sqrt{3}×sqrt{27}-sqrt{144}:sqrt{36}
left(2sqrt{9}+3sqrt{36}right):4
sqrt{7}-sqrt{8-2sqrt{7}}
dfrac{sqrt{4-2sqrt{3}}}{sqrt{6}-sqrt{2}}
dfrac{5+3sqrt{5}}{sqrt{5}}+dfrac{3+sqrt{3}}{sqrt{3}+1}-left(sqrt{5}+3right)
sqrt{27}+5sqrt{12}-2sqrt{3}11sqrt{3}
Đọc tiếp
\(\sqrt{3}×\sqrt{27}-\sqrt{144}:\sqrt{36}\)
\(\left(2\sqrt{9}+3\sqrt{36}\right):4\)
\(\sqrt{7}-\sqrt{8-2\sqrt{7}}\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
\(\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(\sqrt{27}+5\sqrt{12}-2\sqrt{3}=11\sqrt{3}\)
bài 1 : giải pt
a,sqrt{dfrac{2x^2-4x+2}{6}}1
b, dfrac{6}{x-4}sqrt{2}
c,sqrt{dfrac{20}{2x^2-8x+8}}sqrt{5}
bài 2 : tính
a, sqrt{3+sqrt{5}}-sqrt{3-sqrt{5}}-sqrt{2}
b,left(sqrt{12}+sqrt{75}+sqrt{27}right):sqrt{15}
c, left(12sqrt{20}-8sqrt{200}+7sqrt{450}right):sqrt{10}
Đọc tiếp
bài 1 : giải pt
a,\(\sqrt{\dfrac{2x^2-4x+2}{6}}=1\)
b, \(\dfrac{6}{x-4}=\sqrt{2}\)
c,\(\sqrt{\dfrac{20}{2x^2-8x+8}}=\sqrt{5}\)
bài 2 : tính
a, \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
b,\(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
c, \(\left(12\sqrt{20}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
1
a. frac{10+2sqrt{10}}{sqrt{5}+sqrt{2}}+frac{8}{1-sqrt{5}} b.frac{2sqrt{8}-sqrt{12}}{sqrt{18}-sqrt{48}}-frac{sqrt{5}+sqrt{27}}{sqrt{30}+sqrt{162}} c. sqrt{frac{2-sqrt{3}}{2+sqrt{3}}}+sqrt{frac{2+sqrt{3}}{2-sqrt{3}}}
d. frac{sqrt{3-sqrt{5}}.left(3+sqrt{5}right)}{sqrt{10}+sqrt{2}} e. frac{1}{sqrt{2}+sqrt{2+sqrt{3}}}+frac{1}{sqrt{2}-sqrt{2-sqrt{3}}} f. frac{left(sqrt{5}+2right)^2-8sqrt{5}}{2sqrt{5}-4}
Đọc tiếp
1
a. \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\) b.\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\) c. \(\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)
d. \(\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\) e. \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\) f. \(\frac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}\)
bài 1 rút gọn
a \(A=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
b\(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)
c\(C=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\) d\(D=\sqrt{2+\sqrt{3}}+\sqrt{14-5\sqrt{3}}+\sqrt{2}\)
tính:a) sqrt{dfrac{1}{8}}.sqrt{2}.sqrt{125}.sqrt{dfrac{1}{5}}b)sqrt{sqrt{2}-1}.sqrt{sqrt{2}+1}c) sqrt{11-6sqrt{2}}.sqrt{11+6sqrt{2}}d) sqrt{12-6sqrt{3}}.sqrt{dfrac{1}{3-sqrt{3}}}e) dfrac{sqrt{15}-sqrt{6}}{sqrt{35}-sqrt{14}}f) dfrac{2sqrt{15}-2sqrt{10}+sqrt{6}-3}{2sqrt{5}-2sqrt{10}-sqrt{3}+sqrt{6}}g) left(dfrac{1}{5-2sqrt{6}}+dfrac{2}{5+2sqrt{6}}right)left(15+2sqrt{6}right)
Đọc tiếp
tính:
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}\)
d) \(\sqrt{12-6\sqrt{3}}.\sqrt{\dfrac{1}{3-\sqrt{3}}}\)
e) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
f) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
g) \(\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
Bài 1: rút gọn rồi tính giá trị biểu thức:
Adfrac{2bsqrt{x^2-1}-sqrt{x+1}}{x-2sqrt{x-1}} với x3; ysqrt{2}
Bài 2: Trục căn thức ở mẫu
a/dfrac{25}{5-2sqrt{3}} b/dfrac{8}{sqrt{5}+2} c/dfrac{6}{2sqrt{3}-sqrt{7}} d/dfrac{9-2sqrt{3}}{3sqrt{6}-2sqrt{2}} e/dfrac{1}{sqrt{2}+sqrt{3}-sqrt{5}}
Đọc tiếp
Bài 1: rút gọn rồi tính giá trị biểu thức:
A=\(\dfrac{2b\sqrt{x^2-1}-\sqrt{x+1}}{x-2\sqrt{x-1}}\) với x=3; y=\(\sqrt{2}\)
Bài 2: Trục căn thức ở mẫu
a/\(\dfrac{25}{5-2\sqrt{3}}\) b/\(\dfrac{8}{\sqrt{5}+2}\) c/\(\dfrac{6}{2\sqrt{3}-\sqrt{7}}\) d/\(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\) e/\(\dfrac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\)