\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\\ =\left|2-\sqrt{5}\right|+\left|2\sqrt{2}-\sqrt{5}\right|\\ =\sqrt{5}-2+2\sqrt{2}-\sqrt{5}\\ =-2+\sqrt{2}\)

\(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)}\\ =\left|\sqrt{7}-2\sqrt{2}\right|+\left|3-2\sqrt{2}\right|\\ =2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\\ =3-\sqrt{7}\)

\(\sqrt{\left(x-3\right)^2}\\ =\left|x-3\right|\\ =x-3\left(vì.x>3\right)\)

\(\sqrt{\left(1-x\right)^2}\\ =\left|1-x\right|\\ =x-1\left(vì.x>1\right)\)

\(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}\\ =\left|3a^2\right|\\ =3a^2\)

\(\sqrt{100a^2}\\ =\sqrt{\left(10a\right)^2}\\ =\left|10a\right|\\ =-10a\left(vì.a< 0\right)\)

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.

$=\sqrt{(10a)^2}=|10a|=-10a$

\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)

\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)

\(C=\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)

\(A=\left|2-\sqrt{3}\right|+\left|1+\sqrt{3}\right|=2-\sqrt{3}+1+\sqrt{3}=3\)

\(B=\left|4-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=4-\sqrt{5}-\sqrt{5}+2=6-2\sqrt{5}\)

C=\(\left|1-\sqrt{5}\right|-\left|2-\sqrt{5}\right|=\sqrt{5}-1-\sqrt{5}+2=1\)

\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)

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) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)

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: \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}\)

\(=2\sqrt{5}\)

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

\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)

\(=54+8-32=30\)

\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)

\(=5-2\sqrt{2}\)

\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)

\(=2-2\sqrt{3}\)

\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)

\(=2\sqrt{6}\)

\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)

`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`

a: Ta có: \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)

\(=x-3\sqrt{x}-5\sqrt{x}-15\)

\(=x-8\sqrt{x}-15\)

b: Ta có: \(3\left(\sqrt{x}+2\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)

\(=3\sqrt{x}+6+2\sqrt{x}-x+6-3\sqrt{x}\)

\(=-x+2\sqrt{x}+12\)

c: Ta có: \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)

\(=x-9-5\sqrt{x}+5\)

\(=x-5\sqrt{x}-4\)

d: Ta có: \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(=3\sqrt{x}-6-x+1\)

\(=-x+3\sqrt{x}-5\)

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)\(=3+\sqrt{5}+\sqrt{5}+6=9\)