\(\dfrac{5}{\sqrt{11}-\sqrt{6}}-\dfrac{2}{\sqrt{6}-2}+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\)
thực hiện phép tính
giúp tuui nha
thực hiện phép tính
\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{16}-2}-\dfrac{12}{3-\sqrt{16}}\right).\left(\sqrt{6}+11\right)\)
Lời giải:
\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{16}-2}-\frac{12}{3-\sqrt{16}}\right).(\sqrt{6}+11)=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4}{4-2}-\frac{12}{3-4}\right)(\sqrt{6}+11)\)
\(=\left(\frac{15(\sqrt{6}-1)}{6-1}+2+12\right)(\sqrt{6}+11)=(3\sqrt{6}-3+14)(\sqrt{6}+11)\)
\(=(3\sqrt{6}+11)(\sqrt{6}+11)\)
thực hiện phép tính ( rút gọn biểu thức )
a) \(\left(\dfrac{3+2\sqrt{3}}{\sqrt{3}+2}-\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
b) \(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\)
a)
\(\left(\dfrac{3+2\sqrt{3}}{\sqrt{3}+2}-\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =\left(\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\left(\sqrt{3}+2\right)}-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =3-2\\ =1\)
b)
\(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\\ =\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{-\left(\sqrt{11}-1\right)}\right)\left(2+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{\sqrt{11}+1}\right)\\ =\left(2-\sqrt{11}\right)\left(2+\sqrt{11}\right)\\ =4-11\\ =-7\)
a: \(=\left(\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{2+\sqrt{3}}-\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
=(căn 3-căn 2)(căn 3+căn 2)
=3-2=1
b: \(=\left(2-\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\sqrt{11}-1}\right)\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}+1\right)}{\sqrt{11}+1}\right)\)
=(2-căn 11)(2+căn 11)
=4-11
=-7
6) Thực hiện các phép tính
a. \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
b. \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
c. \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)
Mng giúp mik vs ạ >.<!!
b, \(\dfrac{2}{\sqrt{5}+2}+\dfrac{2}{2-\sqrt{5}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{2\left(\sqrt{5}+2\right)}{5-4}\)
\(=2\sqrt{5}-4-2\sqrt{5}-4=-8\)
a, \(\sqrt{2}\left(\sqrt{8}+\sqrt{32}-\sqrt{98}\right)\)
\(=\sqrt{2}\left(2\sqrt{2}+4\sqrt{2}-7\sqrt{2}\right)\)
\(=\sqrt{2}.\left(-\sqrt{2}\right)=-2\)
c, \(\left(2+\sqrt{3}\right)\sqrt{11-6\sqrt{2}}\)
\(=\left(2+\sqrt{3}\right)\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(=\left(2+\sqrt{3}\right)\left(3-\sqrt{2}\right)\)
\(=6-2\sqrt{2}+3\sqrt{3}-\sqrt{6}\)
Trục căn thức và thực hiện phép tính:
a, \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)
b, \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
a: Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}-11\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}-11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}-11\right)\)
\(=127-22\sqrt{6}\)
b: Ta có: \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
=-1+5
=4
thực hiện phép tính
a)\(\dfrac{3}{5}\)-\(\dfrac{1}{2}\)\(\sqrt{1\dfrac{11}{25}}\)
b)(5+2\(\sqrt{6}\))(5-2\(\sqrt{6}\))
c)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)+\(\sqrt{4-2\sqrt{3}}\)
d)\(\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)(với x,y>0)
\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)
\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)
\(c,\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\\ =\left|2-\sqrt{3}\right|+\sqrt{\sqrt{3^2}-2\sqrt{3}+1}\\ =2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\\ =2-\sqrt{3}+\left|\sqrt{3}-1\right|\\ =2-\sqrt{3}+\sqrt{3}-1\\ =1\)
\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)
\(\dfrac{6}{\sqrt{5}+\sqrt{11}}\)-\(\dfrac{11+\sqrt{11}}{\sqrt{11}+1}\)+\(\dfrac{1}{2\sqrt{5}}\)
`6/(sqrt11+sqrt5)-(11+sqrt11)/(sqrt11+1)+1/(2sqrt5)`
`=(6(sqrt11-sqrt5))/(11-5)-(sqrt11(sqrt11+1))/(sqrt11+1)+sqrt5/10`
`=sqrt11-sqrt5-sqrt11+sqrt5/10`
`=sqrt5/10-sqrt5=(-9sqrt5)/10`
\(\dfrac{6}{\sqrt{11}+\sqrt{5}}-\dfrac{11+\sqrt{11}}{\sqrt{11}+1}+\dfrac{1}{2\sqrt{5}}\)
\(=\sqrt{11}-\sqrt{5}-\sqrt{11}+\dfrac{1}{10}\sqrt{5}\)
\(=-\dfrac{9}{10}\sqrt{5}\)
tính:
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}\)
d) \(\sqrt{12-6\sqrt{3}}.\sqrt{\dfrac{1}{3-\sqrt{3}}}\)
e) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
f) \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)
g) \(\left(\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)\)
a) \(\sqrt{\dfrac{1}{8}}\cdot\sqrt{2}\cdot\sqrt{125}\cdot\sqrt{\dfrac{1}{5}}\) = \(\sqrt{\dfrac{1}{8}\cdot2}.\sqrt{125\cdot\dfrac{1}{5}}=\sqrt{\dfrac{1}{4}}.\sqrt{25}=\dfrac{1}{2}\cdot5=2,5\)
b)\(\sqrt{\sqrt{2}-1}.\sqrt{\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2-1}=1\)
c) \(\sqrt{11-6\sqrt{2}}.\sqrt{11+6\sqrt{2}}=\sqrt{\left(11-6\sqrt{2}\right)\left(11+6\sqrt{2}\right)}=\sqrt{121-72}=\sqrt{49}=7\)
chứng minh đẳng thúc
\(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}=4+\sqrt{11}-3\sqrt{7}\)
\(VT=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
Thực hiện phép tính và thu gọn biểu thức:
B= \(\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
Thực hiện phép tính:
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)