i,\(\sqrt{12,1.360}\)
k,\(\sqrt{0,4}.\sqrt{6,4}\)
l,-0,4\(\sqrt{\left(-0,4\right)^2}\)
m,\(\sqrt{2^4.\left(-7\right)^2}\)
f)\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\)- \(\dfrac{\sqrt{6}-3}{\sqrt{2}-\sqrt{3}}\)
g)\(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0,4}\right)\)
giải chi tiết cụ thể giúp mk với ạ
1 nhân chia căn bậc hai
a/\(\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\sqrt{3\dfrac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{0,2}\right)\)
b/ \(\left(\dfrac{3x}{2}\sqrt{\dfrac{x}{2y}}-0,4\sqrt{\dfrac{2}{xy}}+\dfrac{1}{3}\sqrt{\dfrac{xy}{2}}\right):\dfrac{4}{15}\sqrt{\dfrac{2x}{3y}}\)
2 Cộng trừ căn bậc hai
a/ \(0,1\sqrt{200}-2\sqrt{0,08}+4\sqrt{0,5}+0,4\sqrt{50}\)
b/ \(\dfrac{2}{3}x\sqrt{9x}+6x\sqrt{\dfrac{x}{4}-x^2}\sqrt{\dfrac{1}{x}}\)
Bài 2:
a: \(=\sqrt{2}-\dfrac{2}{5}\sqrt{2}+2\sqrt{2}+2\sqrt{2}=\dfrac{23}{5}\sqrt{2}\)
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) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
Tính:
a) \(\sqrt{\left(0,1\right)^2}\)
b) \(\sqrt{\left(-0,3\right)^2}\)
c) \(-\sqrt{\left(-1,3\right)^2}\)
d) \(-0,4\sqrt{\left(-0,4\right)^2}\)
Rút gọn:
\(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\)
ta có : \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\)
\(=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-\dfrac{3\sqrt{10}}{5}\right)\)
\(=2\sqrt{5}-6-2+\dfrac{6\sqrt{5}}{5}=\dfrac{16}{5}\sqrt{5}-8\)
Tính
A=\(\sqrt{0,36}-\left(\sqrt{25}-\sqrt{49}:\sqrt{4}\right).\sqrt{0,4}-\sqrt{10}^2\)
\(A=\frac{\left(0,4-\frac{2}{\sqrt{81}}+\frac{2}{11}\right)}{\left(1,4-\frac{7}{\sqrt{81}}+\frac{7}{11}\right)}\)
\(A=\frac{\left(0,4-\frac{2}{\sqrt{81}}+\frac{2}{11}\right)}{1,4-\frac{7}{\sqrt{81}}+\frac{7}{11}}=\frac{2\left(0,2-\frac{1}{\sqrt{81}}+\frac{1}{11}\right)}{7\left(0,2-\frac{1}{\sqrt{81}}+\frac{1}{11}\right)}=\frac{2}{7}\)
Ta có: A=\(\frac{\left(0,4-\frac{2}{\sqrt{81}}+\frac{2}{11}\right)}{\left(1,4-\frac{7}{\sqrt{81}}+\frac{7}{11}\right)}\)
=\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}\)
=\(\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\)
=> A=\(\frac{2}{7}\)
* Tính:
a.\(\dfrac{-4}{3}.\sqrt{\left(-0,4\right)^2}\)
b.\(\sqrt[3]{\dfrac{3}{4}}.\sqrt[3]{\dfrac{9}{16}}\)
c.\(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
a) Ta có: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}\)
\(=-\dfrac{4}{3}\cdot0.4\)
\(=\dfrac{-1.6}{3}=-\dfrac{8}{15}\)
b) Ta có: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)
\(=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)
c) Ta có: \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
\(=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{7}\)
\(=\dfrac{6}{7}\)
* Tính
a. \(\dfrac{-4}{3}.\sqrt{\left(-0,4\right)^2}\)
b. \(\sqrt[3]{\dfrac{3}{4}}.\sqrt[3]{\dfrac{9}{16}}\)
c. \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
a: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}=\dfrac{-4}{3}\cdot\dfrac{2}{5}=\dfrac{-8}{15}\)
b: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\dfrac{3}{4}\)