\(\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\dfrac{1}{\sqrt{7}}-\dfrac{4}{\sqrt{7}}+\dfrac{7}{\sqrt{7}}\right)\times\dfrac{1}{\sqrt{7}}=\dfrac{1}{\sqrt{7}}\left(1-4+7\right)\times\dfrac{1}{\sqrt{7}}=\dfrac{1}{7}\times4=\dfrac{4}{7}\)
So sánh:
a) \(4\sqrt{7}\) và \(3\sqrt{13}\)
b) \(3\sqrt{12}\) và \(2\sqrt{16}\)
c) \(\dfrac{1}{4}\sqrt{84}\) và \(6\sqrt{\dfrac{1}{7}}\)
d) \(3\sqrt{12}\) và \(2\sqrt{16}\)
e) \(\dfrac{1}{2}\sqrt{\dfrac{17}{2}}\) và \(\dfrac{1}{3}\sqrt{19}\)
Thực hiện phép tính:
a) \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}\)
b) \(\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}\)
c) \(\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}\right):\sqrt{7}\)
rút gon bieu thức
\(\left(3\sqrt{2}+\sqrt{6}\right).\sqrt{6-3\sqrt{3}}\)
\(\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(\sqrt{\dfrac{13}{4}+\sqrt{3}}-\sqrt{\dfrac{7}{4}-\sqrt{3}}\)
\(\sqrt{\dfrac{289+4\sqrt{72}}{16}}+\sqrt{\dfrac{129}{16}+\sqrt{2}}\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{\sqrt{8}+3}\)
\(\sqrt{16-6\sqrt{7}}+\sqrt{10-2\sqrt{21}}\)
Tính
\(A=\sqrt{20}-3\sqrt{8}+5\sqrt{45}\)
\(B=\dfrac{30}{\sqrt{7}-1}+\dfrac{15}{\sqrt{7}+2}\)
\(C=\left(3-\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3+\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
\(D=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(E=\sqrt{7-4\sqrt{3}}-\sqrt{3+2\sqrt{3}}\)
tính \(\left(3\sqrt{2}\right)^2\sqrt{\dfrac{2}{16-6\sqrt{7}}}\)
Tính giá trị các biểu thức sau
1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}\)
2.\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+\dfrac{1}{5\sqrt{4}+4\sqrt{5}}+\dfrac{1}{6\sqrt{5}+5\sqrt{6}}+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\)
giúp mk vs ạ
\(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\) (ĐK x>0; x\(\ne9\))
a)Rút gọn A và B
b) Tìm các giá trị của x để giá trị biểu thức A lớn hơn giá trị biểu thức B
Thực hiện các phép tính sau:
a) \(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-2\sqrt{7}}{4}+\dfrac{6}{\sqrt{7}-2}-\dfrac{5}{4+\sqrt{7}}\)
b) \(\dfrac{1}{\sqrt{3}+\sqrt{2}-\sqrt{5}}-\dfrac{1}{\sqrt{3}+\sqrt{2}+\sqrt{5}}\)
Rút gọn:
A= \(\dfrac{1}{\sqrt{7-\sqrt{24}}+1}-\dfrac{1}{\sqrt{7+\sqrt{24}}+1}\)
B=\(\dfrac{2}{\sqrt{8-2\sqrt{5}}}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}-\dfrac{3}{\sqrt{7+2\sqrt{10}}}\)
