Chứng minh rằng B > 8
\(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+...+\dfrac{1}{\sqrt{24}}\)
Bài 1
a. Tìm điều kiện để căn thức bậc hai có nghĩa \(\sqrt{\dfrac{1}{2-x}}\)
b. \(\sqrt[3]{125}.\sqrt[3]{-216}-\sqrt[3]{512}.\sqrt[3]{\dfrac{1}{8}}\)
* Chứng minh
\(\dfrac{\sqrt{ab}-b}{b}-\sqrt{\dfrac{a}{b}}\) < 0 với a ≥ 0, b≥0
Chứng minh đẳng thức:
a) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{2b}{\sqrt{a}-\sqrt{b}}\)
b) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\dfrac{-3}{2}\)
Rút gọn:
A= \(\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)
B=\(\dfrac{2}{\sqrt{8-2\sqrt{5}}}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}-\dfrac{3}{\sqrt{7+2\sqrt{10}}}\)
Tính:
\(A=\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)
\(B=\dfrac{1}{\sqrt{2}-1}+\dfrac{14}{3+\sqrt{2}}\)
\(C=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(D=\sqrt{\left(1-\sqrt{2}\right)^2}-3\sqrt{18}+4\sqrt{\dfrac{1}{2}}\)
bà 1 rút gọn biểu thức :\(\sqrt{9ab}\) + 7\(\sqrt{\dfrac{a}{b}}\) - 5\(\sqrt{\dfrac{b}{a}}\) - 3ab \(\sqrt{\dfrac{1}{ab}}\)
bài 2 :cho a>0,b>0 chứng minh : \(\dfrac{a^2b}{a-b}\).\(\sqrt{\dfrac{8\left(a^2-2ab+b^2\right)}{75a^4b}}\) = \(\dfrac{2}{15}\) .\(\sqrt{6b}\)
chứng minh
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\dfrac{-3}{2}\)
* Rút gọn biểu thức
a. \(\sqrt{72}-5\sqrt{2}+3\sqrt{12}\)
b. \(6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\)
c. \(\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\)
d. \(\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\)
* Rút gọn các biểu thức
a. \(\sqrt{72}-5\sqrt{2}+3\sqrt{12}\)
b. \(6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\)
c. \(\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\)
d. \(\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\)