1.\(\sqrt{3+\sqrt{5}}\left(\sqrt{3-\sqrt[]{5}}-2\sqrt{2}\right)\)
2.\(\sqrt{13+4\sqrt{3}}-\sqrt{\left(3-2\sqrt{3}\right)^2}\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
Tính:
1) ( \(2\sqrt{5}-\sqrt{7}\) ) \(\left(2\sqrt{5}+\sqrt{7}\right)\)
2) \(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)\)
3) \(\sqrt{\left(\sqrt{7}-2\right)^2}+\sqrt{\left(\sqrt{7}+2\right)^2}\)
4) \(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
5) \(\left(\sqrt{5}-\sqrt{6}\right)^2\)
6) \(\left(\sqrt{3}-\sqrt{5}\right)^2\)
7) \(\left(2\sqrt{2}+\sqrt{3}\right)^2\)
\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)
1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)
4) \(=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\)
5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)
6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)
7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
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) \(F=\frac{\left(\sqrt{3-\sqrt{5}}-1\right)\left(\sqrt{3-\sqrt{5}}\left(3-\sqrt{5}\right)+1\right)}{4-\sqrt{5}-\sqrt{3-\sqrt{5}}}+\sqrt{5}\)
Thực hiện phép tính
1)\(\frac{\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}+\sqrt{2}}{\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}+\sqrt{5}}\)
2)\(\left(4+\sqrt{15}\right)\left(10-\sqrt{6}\right)-\sqrt{4-\sqrt{15}}\)
3)\(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
4)\(\frac{2\sqrt{3-\sqrt{5+\sqrt{13-\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
5)\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
Tính :
a) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)\)
b) \(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21+12\sqrt{3}}}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}\)
d) \(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\)
e) \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)
f) \(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{5-2\sqrt{6}}\right)}{9\sqrt{3}-11\sqrt{2}}\)
g) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)-\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
Tính
A = \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
B = \(\sqrt{\left(4-\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
C = \(\sqrt{\left(1-\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)
\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(C=\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)
\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)
\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}\)
C=\(\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)
rút gọn
g, \(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right).\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\) h,\(\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\sqrt{3\dfrac{1}{3}}\right).\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\dfrac{1}{5}}\right)\)
g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)
=-(căn 5+2)(căn 5-2)
=-(5-4)=-1
h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)
=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6
=4
1, \(\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)}^2\)
2, \(\sqrt{\left(\sqrt{3+1}\right)^2+\sqrt{\left(1-\sqrt{3}\right)^2}}\)
3, \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
4, \(\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)
5, \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
6, \(\sqrt{4+2\sqrt{3}-\sqrt{4-2\sqrt{3}}}\)
1/ \(=2+\sqrt{5}-\left|2-\sqrt{5}\right|=2+\sqrt{5}-\sqrt{5}+2=4\)
2/ bạn coi lại đề
3/ \(=\sqrt{2}+1-\left|1-\sqrt{2}\right|=\sqrt{2}+1-\sqrt{2}+1=2\)
4/ \(=\sqrt{3}+2-\left|\sqrt{3}-2\right|=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\)
5/ \(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\)
6/ \(=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)
Các bạn giúp mình với, tối nay mình nộp rồi.
Câu 6 sửa lại đề giúp mình như này nhé:
\(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\)
Tính:
\(A=\sqrt{27}-2\sqrt{48}+3\sqrt{75}\)
\(B=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}-3\right)^2}\)
\(C=\sqrt{\left(2\sqrt{3}+1\right)^2}+\sqrt{\left(2\sqrt{3}-5\right)^2}\)
\(D=\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(E=\dfrac{4}{\sqrt{5}-2}-\dfrac{32}{\sqrt{5}+1}\)
\(M=\dfrac{10}{3\sqrt{2}-4}+\dfrac{28}{3\sqrt{2}+2}\)
please help ;-;