Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)
\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)
\(=-\sqrt{5}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
Tính:
\(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)
\(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+1-2+\sqrt{3}\)
\(=2\sqrt{3}-1\)
f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+2\)
=3
a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)
\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)
\(=3\sqrt{5}+12\sqrt{2}\)
b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)
\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)
\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)
\(=9+3\sqrt{5}-4\sqrt{5}+4\)
\(=13-\sqrt{5}\)
Tính:
\(A=\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)
\(B=\dfrac{1}{\sqrt{2}-1}+\dfrac{14}{3+\sqrt{2}}\)
\(C=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(D=\sqrt{\left(1-\sqrt{2}\right)^2}-3\sqrt{18}+4\sqrt{\dfrac{1}{2}}\)
1)tính
a)\(\left(\dfrac{1}{5}\sqrt{500}-3\sqrt{45}+5\sqrt{20}\right):\sqrt{5}\)
b)\(\left(\dfrac{\sqrt{3}+1}{\sqrt{3}-1}-\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\right).\sqrt{\dfrac{1}{48}}\)
c)\(\left(\dfrac{2\sqrt{3}+3}{\sqrt{3}+2}+\dfrac{2\sqrt{2}}{\sqrt{2}+1}\right):\left(\sqrt{12}+\sqrt{18}\right)\)
a) bấm mày
b) qui đồng trong ngặc trước rồi thu gọn
c) trong ngặc : khử phân số thứ nhất \(\Rightarrow\) qui đồng \(\Rightarrow\) giải bình thường
a: \(=\dfrac{\left(2\sqrt{5}-9\sqrt{5}+10\sqrt{5}\right)}{\sqrt{5}}=2-9+10=10-7=3\)
b: \(=\dfrac{4+2\sqrt{3}-4+2\sqrt{3}}{2}\cdot\dfrac{1}{4\sqrt{3}}\)
\(=\dfrac{8\sqrt{3}}{8\sqrt{3}}=1\)
c: \(=\dfrac{\left(\sqrt{3}+4-2\sqrt{2}\right)}{2\sqrt{3}+3\sqrt{2}}\simeq0.377\)
Tính
\(A=\sqrt{20}-3\sqrt{8}+5\sqrt{45}\)
\(B=\dfrac{30}{\sqrt{7}-1}+\dfrac{15}{\sqrt{7}+2}\)
\(C=\left(3-\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3+\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
\(D=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(E=\sqrt{7-4\sqrt{3}}-\sqrt{3+2\sqrt{3}}\)
1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)
2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)
3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)
4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)
a, \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)
b, \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)
c, \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)
a) \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)
\(=\sqrt{10^2\cdot2}-\sqrt{4^2\cdot2}+\sqrt{6^2\cdot2}\)
\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)
\(=\left(10-4+6\right)\sqrt{2}\)
\(=12\sqrt{2}\)
b) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)
\(=4\cdot2\sqrt{5}-3\cdot5\sqrt{5}+5\cdot3\sqrt{5}-3\sqrt{5}\)
\(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}\)
\(=\left(8-15+15-3\right)\sqrt{5}\)
\(=5\sqrt{5}\)
c) \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)
\(=\left(2\cdot2\sqrt{2}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\cdot2\sqrt{5}-2\sqrt{2}\right)\)
\(=\left(3\sqrt{5}-3\sqrt{2}\right)\left(72-10\sqrt{5}-2\sqrt{2}\right)\)
1) Rút gọn các biểu thức sau:
a) \(\sqrt[3]{27}\) - \(\sqrt[3]{-8}\) - \(\sqrt[3]{125}\)
b) \(\sqrt{20}\) - \(\sqrt{45}\) + 3\(\sqrt{18}\) + \(\sqrt{72}\)
c) 2\(\sqrt{5}\) + \(\sqrt{\left(1-\sqrt{5}\right)^2}\)
d) \(\dfrac{1}{\sqrt{3}+1}\) + \(\dfrac{1}{\sqrt{3}-1}\) - 2\(\sqrt{3}\)
e) \(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}\) - \(\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\) với a ≥ 0 , b≥0 , a ≠ b
* Rút gọn biểu thức
a. \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}\)
b. \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
c. \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
d. \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=2\sqrt{2}+\sqrt{3}\)
c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=3\sqrt{5}-1\)
d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)
Tính :
a) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\)
b) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21+12\sqrt{3}}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
d) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
e) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
f) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
g) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)-\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
