Bài 2: Căn thức bậc hai và hằng đẳng thức căn bậc hai của bình phương
Các câu hỏi tương tự
Rút gọn\(B=\frac{3\sqrt{8}+2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\\ C=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
đưa nhân tử ra ngoài dấu căn:
a, \(\sqrt{5\left(1-\sqrt{2}\right)^2}\)
b, \(\sqrt{27\left(2-\sqrt{5}\right)^2}\)
c, \(\sqrt{\dfrac{2}{\left(3-\sqrt{10}\right)^2}}\)
d, \(\sqrt{\dfrac{5\left(1-\sqrt{3}\right)^2}{4}}\)
a,\(\left(\sqrt{2}+2\right)\sqrt{2}-2\sqrt{2}\)
b, \(\sqrt{(\sqrt{5}-\sqrt{10})^2}-\sqrt{10(\sqrt{2+1)}^2}\)
c,\((\sqrt{2}+\sqrt{3})^2.\sqrt{49-20\sqrt{6}}\)
d, \(\dfrac{2}{\sqrt{8-\sqrt{60}}}-\sqrt{\dfrac{\sqrt{18}+\sqrt{27}}{\sqrt{3}+\sqrt{2}}}\)
Rút gọn :
a) left(sqrt{6}+sqrt{2}right).left(sqrt{3}-2right)left(sqrt{2+sqrt{3}}right)
b) sqrt{2}.left(sqrt{2-sqrt{3}}right).left(sqrt{3}+1right)
c) left(sqrt{10}-sqrt{6}right).left(sqrt{4-sqrt{15}}right)
d)left(sqrt{3}-sqrt{12}right).left(sqrt{5+2sqrt{6}}right)
e) sqrt{2-sqrt{3}}.left(sqrt{6}-sqrt{2}right).left(2+sqrt{3}right)
f) sqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}
Đọc tiếp
Rút gọn :
a) \(\left(\sqrt{6}+\sqrt{2}\right).\left(\sqrt{3}-2\right)\left(\sqrt{2+\sqrt{3}}\right)\)
b) \(\sqrt{2}.\left(\sqrt{2-\sqrt{3}}\right).\left(\sqrt{3}+1\right)\)
c) \(\left(\sqrt{10}-\sqrt{6}\right).\left(\sqrt{4-\sqrt{15}}\right)\)
d)\(\left(\sqrt{3}-\sqrt{12}\right).\left(\sqrt{5+2\sqrt{6}}\right)\)
e) \(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
f) \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
cho A=\(\left(\dfrac{\sqrt{3}}{x^2+x\sqrt{3}+3}+\dfrac{3}{x^3-\sqrt{27}}\right)\)\(\left(\dfrac{x}{\sqrt{3}}+\dfrac{\sqrt{3}}{x}+1\right)\)
a) rút gọn A
1A)rút gọn biểu thức
a)\(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
b)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\)
1B)thực hiện phép tính
a)\(\sqrt{\left(\sqrt[2]{2}-3\right)^2}+\sqrt[2]{2}\)
b)\(\sqrt{\left(\sqrt{10}-3\right)^2}+\sqrt{\left(\sqrt{10}-4\right)^2}\)
Rút gọn
a.\(\dfrac{\sqrt{7}-5}{2}-\dfrac{6}{\sqrt{7}-2}+\dfrac{1}{3+\sqrt{7}}+\dfrac{3}{5+2\sqrt{7}}\)
b.\(\left(\sqrt{10}+\sqrt{2}\right).\left(6-2\sqrt{5}\right).\sqrt{3+\sqrt{5}}\)
TP
23 tháng 6 2023 lúc 8:42
Rút gọn:
A = \(\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}+\sqrt{13-4\sqrt{3}}-\sqrt{22+12\sqrt{2}}\)
\(a,2\sqrt{20}-\sqrt{50}+3\sqrt{80}-\sqrt{320}\)
\(b,\sqrt{32}-\sqrt{50}+\sqrt{18}\)
\(c,3\sqrt{3}+4\sqrt{2}-5\sqrt{27}\)
\(d,\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)
e,\(\left(2+\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)