rút gon:
A=\(\dfrac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}-\sqrt{3}}-\dfrac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}+\sqrt{3}}\)
B=\(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
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 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: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(=2-\sqrt{3}+\dfrac{1}{3}\sqrt{3}-1+\dfrac{1}{3}\sqrt{3}\)
\(=\dfrac{3-\sqrt{3}}{3}\)
Rút gọn:
a)\(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}+\dfrac{6}{2-\sqrt{10}}\)
b)\(\dfrac{6}{\sqrt{5}-1}+\dfrac{7}{1-\sqrt{3}}-\dfrac{2}{\sqrt{3}-\sqrt{5}}\)
c)\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right)\div\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
d)\(\sqrt{2}+\dfrac{1}{\sqrt{5+2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
e)\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\times\left(\sqrt{6}+11\right)\)
Lm nhanh giúp mk nhé, mk đang cần gấp!
Bạn chia nhỏ ra để nhận được câu tl sớm nhất nhé!Bạn đặt câu hỏi free mà để dày cộp như này khum ai dám làm =(((
Rút gọn biểu thức
a)\(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}+\dfrac{3}{\sqrt{6}+\sqrt{5}}\)
b)\(\dfrac{3}{\sqrt{5}-\sqrt{2}}-\dfrac{1}{\sqrt{5-\sqrt{24}}}-\dfrac{\sqrt{2}}{\sqrt{4+\sqrt{15}}}\)
Help me plsssssssssssssss
\(a,=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{6-2}+\dfrac{3.\left(\sqrt{6}-\sqrt{5}\right)}{6-5}\\ =\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}+3\left(\sqrt{6}-\sqrt{5}\right)\\ =\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}+3\sqrt{6}-3\sqrt{5}\\ =4\sqrt{6}-2\sqrt{5}\)
\(b,=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}-\dfrac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\sqrt{4+\sqrt{15}}}\\ =\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}-\dfrac{1}{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}-\dfrac{2}{\sqrt{8+2.\sqrt{3}.\sqrt{5}}}\\ =\sqrt{5}+\sqrt{2}-\dfrac{1}{\left|\sqrt{3}-\sqrt{2}\right|}-\dfrac{2}{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}\\ =\sqrt{5}+\sqrt{2}-\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{2}{\left|\sqrt{5}+\sqrt{3}\right|}\)
\(=\sqrt{5}+\sqrt{2}-\dfrac{\sqrt{3}+\sqrt{2}}{3-2}-\dfrac{2.\left(\sqrt{5}-\sqrt{3}\right)}{5-3}\\ =\sqrt{5}+\sqrt{2}-\sqrt{3}-\sqrt{2}-\dfrac{2.\left(\sqrt{5}-\sqrt{3}\right)}{2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{3}\\ =0\)
a: \(=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}+\dfrac{3\left(\sqrt{6}-\sqrt{5}\right)}{1}\)
\(=\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}+3\sqrt{6}-3\sqrt{5}\)
\(=-2\sqrt{5}+4\sqrt{6}\)
b: \(=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
\(=\sqrt{5}+\sqrt{2}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{2}\)
=2căn 5-2căn 3
rút gọn
a) \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}\)+\(\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
b)\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
c)\(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\times\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)
d)\(\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}+\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
a) Ta có: \(\dfrac{2-\sqrt{2}}{1-\sqrt{2}}+\dfrac{\sqrt{2}-\sqrt{6}}{\sqrt{3}-1}\)
\(=\dfrac{-\sqrt{2}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\dfrac{-\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\)
\(=-2\sqrt{2}\)
b) Ta có: \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
\(=\sqrt{3}+2+\sqrt{2}-2-\sqrt{3}\)
\(=\sqrt{2}\)
c) Ta có: \(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)
\(=\left(\dfrac{-\sqrt{5}\left(2-\sqrt{5}\right)}{2-\sqrt{5}}-2\right)\left(\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}-2\right)\)
\(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
\(=-\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)=-1\)
d) Ta có: \(\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}+\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
\(=\left(\sqrt{2}-\sqrt{3}\right)^2+\left(\sqrt{3}+\sqrt{2}\right)^2\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
rút gọn
d,\(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}-\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}\) e,\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\) f,\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
d: \(=\sqrt{5}\left(\sqrt{3}-1\right)-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
=căn 5-1/2*căn 5
=1/2*căn 5
e: \(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\)
f:=2+căn 3+căn 2-2-căn 3=căn 2
Trục căn thức ở mẫu và rút gọn
a,\(\dfrac{\sqrt{2}}{\sqrt{5}-\sqrt{3}}\) b,\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
c,\(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}\) d,\(\dfrac{2\sqrt{6}-\sqrt{10}}{4\sqrt{3}-2\sqrt{5}}\)
Câu 1 :Tính : B = ( 3 - \(\sqrt{5}\)) ( \(\sqrt{5}\) + 3 )
Câu 2 : Rút gọn : \(\dfrac{\sqrt{5}+1}{3-2\sqrt{2}}-\dfrac{\sqrt{10}}{\sqrt{5}-2}+3\left(\sqrt{2}-\sqrt{5}\right)\)
Câu 3: \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\)
a, Rút gọn biểu thức a
b, Tính giá trị của A khi x + \(\dfrac{2}{2+\sqrt{3}}\)
\(1,B=9-5=4\\ 2,\dfrac{\sqrt{5}+1}{3-2\sqrt{2}}-\dfrac{\sqrt{10}}{\sqrt{5}-2}+3\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(\sqrt{5}+1\right)\left(3+2\sqrt{2}\right)-\sqrt{10}\left(\sqrt{5}+2\right)+3\sqrt{2}-3\sqrt{5}\\ =3\sqrt{5}+2\sqrt{10}+3+2\sqrt{2}-5\sqrt{2}-2\sqrt{10}+3\sqrt{2}-3\sqrt{5}=3\)
\(3,\\ a,\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}+\sqrt{y}}{1+\sqrt{xy}}\right):\left(1+\dfrac{x+y+2xy}{1-xy}\right)\left(x,y\ge0;xy\ne1\right)\\ =\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{1-xy}:\dfrac{1-xy+x+y+2xy}{1-xy}\\ =\dfrac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}+\sqrt{y}-y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{1+x+y+xy}\\ =\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+x\right)+y\left(1+x\right)}=\dfrac{2\left(\sqrt{x}+\sqrt{y}\right)}{\left(1+y\right)\left(1+x\right)}\)
\(b,x=\dfrac{2}{2+\sqrt{3}}=\dfrac{2\left(2-\sqrt{3}\right)}{1}=4-2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}-1\)
Thay vào BT
\(=\dfrac{2\left(\sqrt{3}-1+\sqrt{y}\right)}{\left(1+y\right)\left(1+4-2\sqrt{3}\right)}=\dfrac{2\sqrt{3}-2+2\sqrt{y}}{\left(1+y\right)\left(3-2\sqrt{3}\right)}\\ =\dfrac{2\sqrt{3}-2+2\sqrt{y}}{3-2\sqrt{3}+3y-2y\sqrt{3}}\)
Rút gọn
\(A=\dfrac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(B=\dfrac{1}{\sqrt{1}-\sqrt{2}}+\dfrac{1}{\sqrt{2}-\sqrt{3}}+....+\dfrac{1}{\sqrt{n-1}-\sqrt{n}}\) (n thuộc N, n>=2)