\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
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 :
\(\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
b)\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
c)\(\left(\sqrt{5}+1\right)\left(\sqrt{7}+1\right)\left(\sqrt{35}+1\right)\left(34-4\sqrt{7}-6\sqrt{5}\right)\)
d) \(\left(\sqrt{7}+1\right)\left(2\sqrt{2}-1\right)\left(2\sqrt{14}-1\right)\left(55+12\sqrt{2}-7\sqrt{7}\right)\)
e)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
Tính:
\(A=2\sqrt{\left(-3\right)^6}+2\sqrt{\left(-2\right)^4}-4\sqrt{\left(-2\right)^6}\)
\(B=\sqrt{\left(\sqrt{2}-2\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(C=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(D=\sqrt{\left(5+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}-5\right)^2}\)
\(E=\sqrt{17^2-8^2}-\sqrt{3^2+4^2}\)
\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)
\(=54+8-32=30\)
\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)
\(=5-2\sqrt{2}\)
\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)
\(=2-2\sqrt{3}\)
\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)
\(=2\sqrt{6}\)
\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)
`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`
Tính:
1.\(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\) 4.\(\sqrt{\left(\sqrt{3}\right)^2+2.\left(\sqrt{3}\right).\left(1\right)+\left(1\right)^2}\)
2.\(\sqrt{\left(\sqrt{5}-\sqrt{6}\right)^2}\) 5.\(\sqrt{\left(\sqrt{5}\right)^2+2.\left(\sqrt{5}\right).\left(\sqrt{3}\right)+\left(\sqrt{3}\right)^2}\)
3.\(\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}\) 6.\(\sqrt{\left(\sqrt{6}\right)^2-2.\left(\sqrt{6}\right).\left(\sqrt{5}\right)+\left(\sqrt{5}\right)^2}\)
tính
1.\(\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}\)
2.\(3\sqrt{2}\left(4-\sqrt{2}\right)+3\left(1-2\sqrt{2}\right)^2\)
3.\(\dfrac{1}{2}\left(\sqrt{6}+\sqrt{5}\right)^2-\dfrac{1}{4}\sqrt{120}-\sqrt{\dfrac{15}{2}}\)
4.\(\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
5.\(\left(\sqrt{\sqrt{14}+\sqrt{5}}+\sqrt{\sqrt{14}-\sqrt{5}}\right)^2\)
6.\(\left(\sqrt{3}+1\right)^3-\left(\sqrt{3}-1\right)^3\)
7.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
8.\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
9.\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
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:
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) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
1> Rut gon
a)\(\sqrt{6-2\sqrt{2}+2\sqrt{3}-2\sqrt{6}}\)
b) \(\left(\sqrt{2}+1\right)\left(\left(\sqrt{2}\right)^2+1\right)\left(\left(\sqrt{2}\right)^4+1\right)\left(\left(\sqrt{2}\right)^8+1\right)\left(\left(\sqrt{2}\right)^{16}+1\right)\)
c)\(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
d) \(\sqrt{3-\frac{4\sqrt{5}}{3}}+\sqrt{3+\frac{4\sqrt{5}}{3}}\)
Gấp lắm . Giúp mình cảm ơn ạ
Bài 1
\(2\sqrt{\left(1+\sqrt{3}\right)^{ }3}-\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(\left(1+\sqrt{3}-\sqrt{5}\right).\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(\left(\sqrt[]{\dfrac{8}{3}}-\sqrt{5}\right)x\sqrt{6}\)
\(\left(5+4\sqrt{2}\right).\left(3+2\sqrt{1}+\sqrt{2}\right).\left(3-2\sqrt{1}+2\right)\)
\(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
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,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)
\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)
\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)
\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)