rút gọn biểu thức sau
a. \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-4}\)
b. \(\dfrac{a^2\sqrt{b}-\sqrt{ab^3}}{\sqrt{a^3b^2}-b^2}\)
c. \(\dfrac{a^3-2\sqrt{2}}{a-\sqrt{2}}\)
d. \(18-\sqrt{8}+\dfrac{1}{4}\sqrt{2}\)
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(b.\dfrac{\sqrt{\dfrac{5}{3}}+\sqrt{\dfrac{3}{5}}-2}{\sqrt{\dfrac{5}{3}}-\sqrt{\dfrac{3}{5}}}\)
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}\)
\(=\dfrac{1}{\sqrt{2}+1}=\sqrt{2}-1\)
Rút gọn các biểu thức sau :
a) A= \(\sqrt{18}\) . \(\sqrt{2}\) - \(\sqrt{48}\) : \(\sqrt{3}\)
b)B= \(\dfrac{8}{\sqrt{5}-1}\) + \(\dfrac{8}{\sqrt{5}+1}\)
a) \(A=\sqrt{18}.\sqrt{2}-\sqrt{48}:\sqrt{3}=\sqrt{18.2}-\sqrt{48:3}\)
\(=\sqrt{36}-\sqrt{16}=6-4=2\)
b) \(B=\dfrac{8}{\sqrt{5}-1}+\dfrac{8}{\sqrt{5}+1}=\dfrac{8\sqrt{5}+8+8\sqrt{5}-8}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\dfrac{16\sqrt{5}}{4}=4\sqrt{5}\)
Thực hiện phép tính ( rút gọn biểu thức )
a) \(\dfrac{\sqrt{2}}{2\sqrt{2}-3}\)+\(\dfrac{1}{3+2\sqrt{2}}\) b) \(\dfrac{1}{\sqrt{10}+\sqrt{6}}\)+\(\dfrac{1}{\sqrt{6}-\sqrt{10}}\)
c) \(\dfrac{-2}{3\sqrt{8}}\)+\(\dfrac{1}{3-2\sqrt{2}}\)
a: \(=\dfrac{\sqrt{2}\left(2\sqrt{2}+3\right)+2\sqrt{2}-3}{-1}\)
\(=\dfrac{4+3\sqrt{2}+2\sqrt{2}-3}{-1}=-1-5\sqrt{2}\)
b: \(=\dfrac{1}{\sqrt{10}+\sqrt{6}}-\dfrac{1}{\sqrt{10}-\sqrt{6}}\)
\(=\dfrac{\sqrt{10}-\sqrt{6}-\sqrt{10}-\sqrt{6}}{4}=\dfrac{-2\sqrt{6}}{4}=-\dfrac{\sqrt{6}}{2}\)
c: \(\dfrac{-2}{3\sqrt{8}}+\dfrac{1}{3-2\sqrt{2}}\)
\(=\dfrac{-2\left(3-2\sqrt{2}\right)+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{-6+4\sqrt{2}+6\sqrt{2}}{6\sqrt{2}\left(3-2\sqrt{2}\right)}\)
\(=\dfrac{10\sqrt{2}-6}{6\sqrt{2}\left(3-2\sqrt{2}\right)}=\dfrac{10-3\sqrt{2}}{6\left(3-2\sqrt{2}\right)}=\dfrac{18+11\sqrt{2}}{6}\)
Rút gọn các biểu thức sau :
a,\(\dfrac{\sqrt{6}+\sqrt{10}}{\sqrt{21}+\sqrt{35}}\)
b,\(\dfrac{\sqrt{405}+3\sqrt{27}}{3\sqrt{3}+\sqrt{45}}\)
c,\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}-\sqrt{6}-\sqrt{9}-\sqrt{12}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
d, D=\(\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\) \(\left(vớix\ne y,x\ne-y\right)\)
d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)
\(=\dfrac{3}{x-y}\)
rút gọn các biểu thức sau: (giả thiết các biểu thức chữ đều có nghĩa)
a) \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)
b) \(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\)
c) \(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}\)
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
rút gọn biểu thức :
A= \(\dfrac{\sqrt{4+\sqrt{3}}+\sqrt{4-\sqrt{3}}}{\sqrt{4+\sqrt{13}}}+\sqrt{27-10\sqrt{2}}\).
B= \(\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\).
C= \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\).
Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+2\sqrt{5}}{3\sqrt{5}-1}\)
=1
Rút gọn các biểu thức sau :
a) \(\sqrt{18\left(\sqrt{2}-\sqrt{3}\right)^2}\) - \(\sqrt{54}\)
b) \(\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\) - a\(\sqrt{\dfrac{1}{a}}\)
c) ( \(\sqrt{28}\) - 2\(\sqrt{3}\) +7) \(\sqrt{7}\) +\(\sqrt{84}\)
a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)
=3căn 6-6-3căn 6=-6
b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)
\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
Tính giá trị của biểu thức: \(M=\dfrac{1+ab}{a+b}-\dfrac{1-ab}{a-b}\) với \(b=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\); \(a=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
Ta có: \(b=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(=\dfrac{2\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}{3\left(3\sqrt{2}-2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{2}{3}\)
Ta có: \(a=\sqrt{4+2\sqrt{2}}\cdot\sqrt{2+\sqrt{2+\sqrt{2}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+2\sqrt{2}}\cdot\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{2\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\)
=2
Thay a=2 và \(b=\dfrac{2}{3}\) vào M, ta được:
\(M=\dfrac{1+2\cdot\dfrac{2}{3}}{2+\dfrac{2}{3}}-\dfrac{1-2\cdot\dfrac{2}{3}}{2-\dfrac{2}{3}}\)
\(=\dfrac{7}{8}+\dfrac{1}{4}\)
\(=\dfrac{7}{8}+\dfrac{2}{8}=\dfrac{9}{8}\)
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{5}-2}{5+2\sqrt{5}}-\dfrac{1}{2+\sqrt{5}}+\dfrac{1}{\sqrt{5}}\)
\(b.\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(c.\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+3}\)
b: Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(=2-\sqrt{3}+\dfrac{1}{3}\sqrt{3}-1+\dfrac{1}{3}\sqrt{3}\)
\(=\dfrac{3-\sqrt{3}}{3}\)