rút gọn các biểu thức sau:
1, (7\(\sqrt{18}\)+3\(\sqrt{27}\)+2\(\sqrt{12}\)):\(\sqrt{13}\)
2, \(\dfrac{\left(5\sqrt{2}+2\sqrt{5}\right)\left(\sqrt{3}-3\sqrt{2}_{ }\right)}{\sqrt{21}}\)
rút gọn các biểu thức sau:
\(\dfrac{1}{2}\sqrt{20}+5\)
\(\sqrt{16}+\sqrt{64}\)
\(\sqrt{20}-\sqrt{45}+3\sqrt{18}\)
\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2}\)
A= \(\dfrac{2}{\sqrt{7}-5}-\dfrac{2}{\sqrt{7}+5}\)
B=\(\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}+\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}\)
mình cần gấp á. tại vì mình khá là ngu toán nên giúp mik vs
a) \(\dfrac{1}{2}\sqrt{20}+5=\dfrac{1}{2}\cdot2\sqrt{5}+5=5+\sqrt{5}\)
b) \(\sqrt{16}+\sqrt{64}=4+8=12\)
c) \(\sqrt{20}-\sqrt{45}+3\sqrt{18}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}=9\sqrt{2}-\sqrt{5}\)
d) \(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{2}=2-\sqrt{2}+\sqrt{2}=2\)
Rút gọn các biểu thức sau:
\(a.A=2\sqrt{3}-\sqrt{75}+2\sqrt{12}\)
\(b.B=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(c.C=\left(\dfrac{x+2\sqrt{x}}{x-2\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}+1}\) (x > 0;x ≠ 4)
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 các biểu thức sau:
a. \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}\) - \(\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
b.\(\dfrac{1}{4-3\sqrt{2}}\) - \(\dfrac{1}{4+3\sqrt{2}}\)
c.\(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right)\): \(\sqrt{28}\)
d.\(\dfrac{3}{\sqrt{6}-\sqrt{3}}\)+\(\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)
\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)
\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)
b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)
Rút gọn các biểu thức sau:
j) \(\left(\dfrac{1}{\sqrt{7-2\sqrt{10}}}-\dfrac{\sqrt{2}}{\sqrt{10}+2}+1\right):\left(\sqrt{2}+1\right)^2\)
k) \(\sqrt{5}\left(\sqrt{6}+1\right):\dfrac{\sqrt{2\sqrt{3}+\sqrt{2}}}{\sqrt{2\sqrt{3}}-\sqrt{2}}\)
o) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
p) \(\left(\sqrt{5}+3\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
j.
\(J=\left[\frac{1}{\sqrt{(\sqrt{5}-\sqrt{2})^2}}-\frac{\sqrt{2}}{\sqrt{2}(\sqrt{5}+\sqrt{2})}+1\right].\frac{1}{(\sqrt{2}+1)^2}\)
\(=\left(\frac{1}{\sqrt{5}-\sqrt{2}}-\frac{1}{\sqrt{5}+\sqrt{2}}+1\right).\frac{1}{(\sqrt{2}+1)^2}\)
\(=[\frac{\sqrt{5}+\sqrt{2}-(\sqrt{5}-\sqrt{2})}{(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})}+1].\frac{1}{(\sqrt{2}+1)^2}=(\frac{2\sqrt{2}}{3}+1).\frac{1}{(\sqrt{2}+1)^2}=\frac{3+2\sqrt{2}}{3}.\frac{1}{3+2\sqrt{2}}=\frac{1}{3}\)
k. Đề sai sai, bạn xem lại
o.
\(O=(4+\sqrt{15})(\sqrt{5}-\sqrt{3}).\sqrt{2}.\sqrt{4-\sqrt{15}}\)
\(=(4+\sqrt{15}(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})=(4+\sqrt{15})(8-2\sqrt{15})\)
\(=2(4+\sqrt{15})(4-\sqrt{15})=2(16-15)=2\)
p: Ta có: \(\left(3+\sqrt{5}\right)\cdot\left(\sqrt{10}-\sqrt{2}\right)\cdot\sqrt{3-\sqrt{5}}\)
\(=\left(3+\sqrt{5}\right)\cdot\left(6-2\sqrt{5}\right)\)
\(=18-6\sqrt{5}+6\sqrt{5}-20\)
=-2
rút gọn biểu thức
a) \(\left(\sqrt{7}-\sqrt{2}\right).\left(\sqrt{9+2\sqrt{14}}\right)\)
b) \(\sqrt{\sqrt{13}-\sqrt{3-\sqrt{13}}-4\sqrt{3}}\)
c) \(\sqrt{80-\sqrt{321-16\sqrt{5}}-\sqrt{226-80\sqrt{5}-\sqrt{89-25\sqrt{5}}}}\)
d) \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
e) \(\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}+\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}\)
f) \(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3}+\sqrt{5}}+\dfrac{\sqrt{2}}{2\sqrt{2}-\sqrt{3}-\sqrt{5}}\)
g) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
Rút gọn các biểu thức :
a) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\)
c) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)
d) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
a) \(\left(2\sqrt{3}+\sqrt{5}\right)\sqrt{3}-\sqrt{60}\) = \(6+\sqrt{15}-2\sqrt{15}\)
= \(6-\sqrt{15}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\sqrt{250}\) = \(5\sqrt{10}+10-5\sqrt{10}\) = \(10\)
c) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\) = \(14-2\sqrt{21}-7+2\sqrt{21}\)
= \(7\)
d) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)
= \(33-3\sqrt{22}-11+3\sqrt{22}\) = \(22\)
a)(2√3+√5)√3-√60
=6+√15-2√15
=6-√15
b)(5√2+2√5)√5-√250
=5√10+10-5√10
=10
c)(√28-√12-√7)√7+2√21
=14-2√21-7+2√21
=7
d)(√99-√18-√11)√11+3√22
=33-3√22-11+3√22
=22
Rút gọn biểu thức sau:
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
\(=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
* 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) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)