\(B=\sqrt{\left(5+3\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{11}-3\sqrt{2}\right)^2}+\frac{11}{\sqrt{11}}\)
Tính nhanh:
\(\frac{3-3^2+3^3-3^4+...+3^{99}}{\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}}.\left(11-\sqrt{91}\right)\left(11-\sqrt{95}\right)\left(11+\sqrt{99}\right)\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)...\left(11-\sqrt{113}\right)\left(11-\sqrt{104}\right)\)
Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)
\(=0\)
Do đó biểu thức trên đầu bài bằng 0
bạn ơi, trong dãy này không có số \(\sqrt{121}\)đâu
Tính:
A) \(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}-2\right)^2\)
B) \(\left(11-4\sqrt{3}\right)\left(11-4\sqrt{3}\right)\)
C) \(\left(1+\sqrt{2018}\right)\left(\sqrt{2019}-2\sqrt{2018}\right)\)
D)
\(\left(\sqrt{2}-1\right)^2+\frac{3}{2}\sqrt{\left(-2\right)}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{25}}.2\)
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
a)\(\sqrt{\left(2\sqrt{2}-3\right)^2+\sqrt{15}}\)
b)\(\sqrt{\left(\sqrt{10}-3\right)}^2+\sqrt{\left(\sqrt{10}-4\right)^2}\)
c)\(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
d)\(\sqrt{11}+6\sqrt{2}+\sqrt{11-6\sqrt{2}=6}\)
b: =căn 10-3+4-căn 10=1
a: \(=\sqrt{11-4\sqrt{6}+\sqrt{15}}\)
Tính:
a) \(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5\)
b) \(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2\)
c) \(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)\)
d) \(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
e) \(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)
\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)
b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)
\(=\left[3-4\right]^2=1\)
c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)
\(=121-48=73\)
d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)
\(=-2\sqrt{2}+\dfrac{4\sqrt{2}+6\sqrt{2}}{5}\)
\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)
e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)
\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)
\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)
Rút gọn
1) \(E=\left(\sqrt{11}-3\right)\left(\sqrt{13-\sqrt{6}+2\sqrt{30-\sqrt{54}}}+\sqrt{11}-\sqrt{10-\sqrt{6}}\right)\)
2) \(F=\frac{\left(\sqrt{3-\sqrt{5}}-1\right)\left(\sqrt{3-\sqrt{5}}\left(3-\sqrt{5}\right)+1\right)}{4-\sqrt{5}-\sqrt{3-\sqrt{5}}}+\sqrt{5}\)
\(Tính\)
\(a.\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\sqrt{27-9\sqrt{5}}\)
\(b.\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{4+\sqrt{3}}{5-2\sqrt{3}}}\)
\(c.\frac{3-4\sqrt{3}}{\sqrt{6}-\sqrt{2}-\sqrt{5}}\)
\(d.\left(\sqrt{11}-\sqrt{3}\right)\left(\sqrt{13-\sqrt{6}+2\sqrt{30-\sqrt{45}}}+\sqrt{11}-\sqrt{10-\sqrt{6}}\right)\)
\(e.\frac{\sqrt{4+\sqrt{5}}+\sqrt{4-\sqrt{5}}}{\sqrt{4}+\sqrt{11}}-\frac{\sqrt{20-4\sqrt{23}}}{\sqrt{5+\sqrt{2}}-\sqrt{5-\sqrt{2}}}\)
\(\left(1+\frac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(\frac{11+\sqrt{11}}{1+\sqrt{11}}+1\right)\)
mọi người giúp em bài này với,em đang cần gấp ạ
bài 2:rút gọn các biểu thức sau
a)A=\(\sqrt{5-\sqrt{21}}+\sqrt{5+\sqrt{21}}\)
b)B=\(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
c)C=\(\left(1+\frac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(\frac{11+\sqrt{11}}{1+\sqrt{11}}+1\right)\)
d)D=\(\frac{\sqrt{2}}{\sqrt{2}-\sqrt{3}}-\frac{\sqrt{2}}{\sqrt{2}+\sqrt{3}}\)
e)E=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
a, \(\sqrt{2}A=\sqrt{10-2\sqrt{3.7}}+\sqrt{10+2\sqrt{3.7}}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{7}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{3}\right|+\left|\sqrt{7}+\sqrt{3}\right|\)
\(=\sqrt{7}-\sqrt{3}+\sqrt{3}+\sqrt{7}=2\sqrt{7}\)
\(\Rightarrow A=\sqrt{14}\)
b, \(B=\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}+\frac{\sqrt{5}}{2}=\frac{3\sqrt{5}}{2}\)
c, \(C=\left(1-\sqrt{11}\right)\left(\sqrt{11}+1\right)=1-11=-10\)
d, \(D=\frac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}\right)}{2-3}-\frac{\sqrt{2}\left(\sqrt{2}-\sqrt{3}\right)}{2-3}\)
\(=-2-\sqrt{6}+2-\sqrt{6}=-2\sqrt{6}\)