\(=\left(4-\sqrt{7}\right)\sqrt{\left(4+\sqrt{7}\right)^2}=\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)=16-7=9\)
\(=\left(4-\sqrt{7}\right)\sqrt{\left(4+\sqrt{7}\right)^2}=\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)=16-7=9\)
\(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)
\(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)
\(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)
\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
Chứng minh đẳng thức
\(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
a) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)
b) \(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\)
\(\sqrt{\left(4-\sqrt{7}\right)^2}+\dfrac{1}{2}\sqrt{28}\)
Cho hai biểu thức:
A= \(\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)}^2\)
B= \(\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\left(x>0;x\ne9\right)\)
a) Rút gọn A,B
b) Tìm các giá trị của x để A>B?
Help !!!
Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}.\)
\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)
\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
\(\left(\sqrt{45}-2\sqrt{10}+\sqrt{5}\right)\sqrt{5}+5\sqrt{8}\)
Tính
a : \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b : \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
c : \(\sqrt{\left(2\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
d : \(\sqrt{52-16\sqrt{3}}+\sqrt{\left(4\sqrt{3}-7\right)^2}\)