a) \(\sqrt{\dfrac{289}{225}}=\dfrac{\sqrt{289}}{\sqrt{225}}=\dfrac{17}{15}\)
b) \(\sqrt{2\dfrac{14}{25}}=\dfrac{\sqrt{64}}{\sqrt{25}}=\dfrac{8}{5}\)
c) \(\sqrt{\dfrac{0,25}{9}}=\dfrac{\sqrt{0,25}}{\sqrt{9}}=\dfrac{0,5}{3}=\dfrac{1}{6}\)
d) \(\sqrt{\dfrac{8,1}{1,6}}=\dfrac{\sqrt{81}}{\sqrt{16}}=\dfrac{9}{4}\)
tính
a.\(\sqrt{\dfrac{289}{225}}\)
b.\(\sqrt{2\dfrac{14}{25}}\)
c.\(\sqrt{\dfrac{0,25}{9}}\)
d.\(\sqrt{\dfrac{8,1}{1,6}}\)
bài 1: rut gọn
a, \(\sqrt{5\left\{1-a\right\}^2}\) với a>1
b,\(\sqrt{\dfrac{9\left[a^2+2a+1\right]}{144}}\)
c,\(\dfrac{2}{x-5}\times\sqrt{\dfrac{x^2\times10x+25}{64}}\)
d \(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\) với x≥0 và x≠1
Áp dụng quy tắc khai phương một thương, hãy tính :
a) \(\sqrt{\dfrac{9}{169}}\)
b) \(\sqrt{\dfrac{25}{144}}\)
c) \(\sqrt{1\dfrac{9}{16}}\)
d) \(\sqrt{2\dfrac{7}{81}}\)
1. Áp dụng quy tắc khai phương một thương, hãy tính:
a, \(\sqrt{\dfrac{36}{121}}\) b, \(\sqrt{\dfrac{9}{16}:\dfrac{25}{36}}\) c, \(\sqrt{0,0169}\)
d,\(\dfrac{\sqrt{15}}{\sqrt{735}}\) e, \(\sqrt{\dfrac{81}{8}:\sqrt{3\dfrac{1}{8}}}\) g, \(\dfrac{\sqrt{12,5}}{\sqrt{0,5}}\)
2. Tính:
a,\(\sqrt{\dfrac{25}{144}}\) b,\(\sqrt{2\dfrac{7}{81}}\) c,\(\sqrt{\dfrac{2,25}{16}}\) d, \(\sqrt{\dfrac{1,21}{0,49}}\)
3. Áp dụng quy tắc chia hai căn bậc hai, hãy tính:
a, \(\sqrt{18}:\sqrt{2}\) b, \(\sqrt{45}:\sqrt{80}\)
c, (\(\sqrt{20}-\sqrt{45}+\sqrt{5}\) ) : \(\sqrt{5}\) d, \(\dfrac{\sqrt{8^2}}{\sqrt{4^5.2^3}}\)
4. Khẳng định nào sau đây là đúng?
A. \(\sqrt{\dfrac{3}{\left(-5\right)^2}}=-\dfrac{\sqrt{3}}{5}\) B. \(\left(\sqrt{\dfrac{-3}{-5}}\right)^2=\dfrac{3}{5}\)
5. Tính.
a, \(\sqrt{2\dfrac{7}{81}}:\dfrac{\sqrt{6}}{\sqrt{150}}\) b, \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right):\sqrt{3}\)
c, \(\left(\sqrt{\dfrac{1}{5}-\sqrt{\dfrac{9}{5}}+\sqrt{5}}\right):\sqrt{5}\) d, \(\sqrt{\dfrac{2+\sqrt{3}}{\sqrt{2}}}\)
6. So sánh
a, So sánh \(\sqrt{144-49}\) và \(\sqrt{144}-\sqrt{49}\);
b, Chứng minh rằng , với hai số a,b thỏa mãn a> b> 0 thì \(\sqrt{a}-\sqrt{b}< \sqrt{a-b}\)
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/\(\dfrac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\)
Rút gọn:
a, \(\sqrt{\dfrac{4}{9-4\sqrt{5}}}\) -\(\sqrt{\dfrac{4}{9+4\sqrt{5}}}\)
b, \(\dfrac{\sqrt{8-4\sqrt{3}}}{\sqrt{2}}\)
c, \(\sqrt{14-8\sqrt{3}}\)-\(\sqrt{24-12\sqrt{3}}\)
d, \(\sqrt{2-\sqrt{3}}\)\(\times\)\(\left(\sqrt{5}+\sqrt{2}\right)\)
1. Tìm x để bt có nghĩa
A=\(\dfrac{\sqrt{2x+3}}{\sqrt{x-3}}\)
B=\(\sqrt{\dfrac{2x+3}{x-3}}\)
C=\(\sqrt{-\dfrac{5}{x+2}}\)
D=\(\sqrt{-x}+\dfrac{1}{x+3}\)
2. Rút gọn bt
A=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-1}}{2}};\left(a>1\right)\)
B=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}};\left(a\ge\sqrt{b};b\ge0\right)\)
C=\(\left(1+\dfrac{a+\sqrt{a}}{a+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}+1}\right);\left(a\ge0,a\ne1\right)\)
D=\(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}};\left(x>0\right)\)
tính \(B=\dfrac{1}{\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{10}}+......+\dfrac{1}{\sqrt{220}+\sqrt{225}}\)
Tính:
a) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
b) \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
