\(\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{3}+\sqrt{5}\right)\left(-\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\)
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) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)
\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}-\sqrt{2}\right)\)
\(\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(\left(\dfrac{1}{\sqrt{3}-\sqrt{2}}\right)\left(\dfrac{1}{\sqrt{3}-\sqrt{2}}\right)\)
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=\(\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)
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.\(\sqrt{\left(\sqrt{6}\right)^2-2.\left(\sqrt{6}\right).\left(\sqrt{5}\right)+\left(\sqrt{5}\right)^2}\)
Bài 2:
a) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\) b) \(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}\)
c) \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\) d) \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)
e) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\) f) \(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
Hộ mk vs ạ
a)\(\sqrt{\left(3-2\sqrt{2}\right)^2}\) + \(\sqrt{\left(3+2\sqrt{2}\right)^2}\) = \(\left(3-2\sqrt{2}\right)+\left(3+2\sqrt{2}\right)\) =\(3-2\sqrt{2}+3+2\sqrt{2}\) =\(6\)
b)\(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}=\left(5-2\sqrt{6}\right)-\left(5+2\sqrt{6}\right)=5-2\sqrt{6}-5-2\sqrt{6}\)\(=-4\sqrt{6}\)
c)\(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}=2\sqrt{5}\)
d)\(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}=\sqrt{2}+1-5+\sqrt{2}=2\sqrt{2}-4\)
e)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}=2-\sqrt{3}+\sqrt{3}-1=1\)
f)\(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}=3+\sqrt{2}-\sqrt{2}+1=4\)
a) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=3-2\sqrt{2}+3+2\sqrt{2}\)
=6
c) Ta có: \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}\)
\(=2\sqrt{5}\)
tính
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(\sqrt{\left(x-3\right)^2}\left(x>3\right)\)
\(\sqrt{\left(1-x\right)^2}\left(x>1\right)\)
\(\sqrt{9a^4}\)
\(\sqrt{100a^2}\left(a< 0\right)\)
\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\\ =\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\\ =-2+\sqrt{2}\)
\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\\ =2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\\ =3-\sqrt{7}\)
\(\sqrt{\left(x-3\right)^2}\\ =\left|x-3\right|\\ =x-3\left(vì.x>3\right)\)
\(\sqrt{\left(1-x\right)^2}\\ =\left|1-x\right|\\ =x-1\left(vì.x>1\right)\)
\(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}\\ =\left|3a^2\right|\\ =3a^2\)
\(\sqrt{100a^2}\\ =\sqrt{\left(10a\right)^2}\\ =\left|10a\right|\\ =-10a\left(vì.a< 0\right)\)
Lời giải:
a. $=|2-\sqrt{5}|+|2\sqrt{2}-\sqrt{5}|$
$=(\sqrt{5}-2)+(2\sqrt{2}-\sqrt{5})=-2+2\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|+|3-2\sqrt{2}|=2\sqrt{2}-\sqrt{7}+(3-2\sqrt{2})$
$=3-\sqrt{7}$
c.
$=|x-3|=x-3$
d.
$=|1-x|=x-1$
$=\sqrt{(3a^2)^2}=|3a^2|=3a^2$
e.
$=\sqrt{(10a)^2}=|10a|=-10a$
B 3. Tính
a)\(\sqrt{\left(\sqrt{7}-1\right)^2}\) b)\(\sqrt{\left(2-\sqrt{3}\right)^2}\)
c)\(\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{2}\) d)\(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-6\right)^2}\)
a.\(\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)
b.\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
c.\(\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{2}=\left|\sqrt{2}+5\right|-\sqrt{2}=\sqrt{2}+5-\sqrt{2}=5\)
d.\(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-6\right)^2}=\left|3+\sqrt{5}\right|+\left|\sqrt{5}-6\right|=3+\sqrt{5}+6-\sqrt{5}=9\)
a: \(=\sqrt{7}-1\)
b: \(=2-\sqrt{3}\)
c: \(=5+\sqrt{2}-\sqrt{2}=5\)
d: \(=3+\sqrt{5}+6-\sqrt{5}=9\)
a)\(=\sqrt{7}-1\)
b)\(=2-\sqrt{3}\)
c)\(=\sqrt{2}+5-\sqrt{2}=5\)
d)\(=3+\sqrt{5}+\sqrt{5}+6=9\)
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
Thực hiện phép tính:
a) \(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\) b) \(\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}\)
c) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}\) d) \(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
e) \(\sqrt{\left(\sqrt{5-\sqrt{2}}\right)^2}+\sqrt{\left(\sqrt{5+\sqrt{2}}\right)^2}\) f) \(\sqrt{\left(\sqrt{2+1}\right)^2-\sqrt{\left(\sqrt{2-5}\right)^2}}\)
a) \(\sqrt{(3-2\sqrt{2})^2}+\sqrt{(3+2\sqrt{2})^2}=3-2\sqrt{2}+3-2\sqrt{2}=6\)
b\(\sqrt{(5-2\sqrt{6})^2}+\sqrt{(5+2\sqrt{6})^2}=5-2\sqrt{6}+5+2\sqrt{6}=10 \)
các ý còn lại làm tương tự