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

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) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)

\(=\sqrt{5}-1\)

b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)

\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)

\(=2\sqrt{2}\)

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

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

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

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

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

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

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=sqrt{(2-sqrt5)^2}+sqrt{(2sqrt2-sqrt5)^2}`

`A=|2-sqrt5|+|2sqrt2-sqrt5|`

`A=\sqrt5-2+2sqrt2-sqrt5`

`A=2sqrt2-2`

`b)B=sqrt{(sqrt7-2sqrt2)^2}+sqrt{(3-2sqrt2)^2}`

`B=|sqrt7-2sqrt2|+|3-2sqrt2|`

`A=2sqrt2-sqrt7+3-2sqrt2`

`A=3-sqrt7`

a,=> A=\(\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\sqrt{2}\right)^2}=2-\sqrt{5}+\sqrt{5}-2\sqrt{2}=2-2\sqrt{2}\)

b tương tự

a) Ta có: \(A=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)

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

\(=\sqrt{2}\)

b) Ta có: \(B=\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)

\(=2\sqrt{2}-\sqrt{7}+3-2\sqrt{2}\)

\(=3-\sqrt{7}\)

a) \(\left(2-\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)

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

\(=4-3=1\)

b) \(\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{3}+\sqrt{5}\right)\)

\(=\left(2\sqrt{3}\right)^2-\left(\sqrt{5}\right)^2\)

\(=12-5=7\)

a) (2 - √3)(2 + √3)

= 2² - (√3)²

= 4 - 3

= 1

b) (2√3 - √5)(2√3 + √5)

= (2√3)² - (√5)²

= 12 - 5

= 7

1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)

2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)

3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)

4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)

a)\(\sqrt{400.0,81}=\sqrt{4.81}=\sqrt{2^2.9^2}=2.9=18\)

b)\(\sqrt{\dfrac{5}{27}.\dfrac{3}{20}}=\sqrt{\dfrac{5}{3^3}.\dfrac{3}{2^2.5}}=\sqrt{\dfrac{1}{3^2.2^2}}=\dfrac{1}{3.2}=\dfrac{1}{6}\)

c)\(\sqrt{\left(-5\right)^2.3^2}=\sqrt{5^2.3^2}=5.3=15\)

d)\(\sqrt{\left(2-\sqrt{5}\right)^2\left(2+\sqrt{5}\right)^2}=\sqrt{\left[2^2-\left(\sqrt{5}\right)^2\right]^2}=\sqrt{\left(-1\right)^2}=1\)

mk chịu !!!!

ai làm đk giúp mik vs ạ