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

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

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

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

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

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

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

\(=2\sqrt{6}\)

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

`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.\(\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)

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

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

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{(5-sqrt3)^2}+sqrt{(2-sqrt3)^2}`

`=5-sqrt3+2-sqrt3`

`=7-2sqrt3`

`B=sqrt{(3-sqrt2)^2}-sqrt{(1-sqrt2)^2}`

`=3-sqrt2-(sqrt2-1)`

`=4-2sqrt2`

`C=sqrt{(3+sqrt7)^2}-sqrt{(2-sqrt7)^2}`

`=3+sqrt7-(sqrt7-2)`

`=5`

`D=sqrt{4-2sqrt3}+sqrt{7+4sqrt3}`

`=sqrt{3-2sqrt3+1}+sqrt{4+2.2.sqrt3+3}`

`=sqrt{(sqrt3-1)^2}+sqrt{(2+sqrt3)^2}`

`=sqrt3-1+2+sqrt3=1+2sqrt3`

\(A=\left|5-\sqrt{3}\right|+\left|2-\sqrt{3}\right|=5-\sqrt{3}+2-\sqrt{3}=7-2\sqrt{3}\)

\(B=\left|3-\sqrt{2}\right|-\left|1-\sqrt{2}\right|=3-\sqrt{2}-\sqrt{2}+1=4-2\sqrt{2}\)

\(C=\left|3+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=3+\sqrt{7}-\sqrt{7}+2=5\)

\(D=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2.2\sqrt{3}+3}\)

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

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

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

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

\(=2\sqrt{2}\)

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: Ta có: \(\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}\)

\(=5-\sqrt{19}-\sqrt{19}+4\)

\(=9-2\sqrt{19}\)

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

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

=0

c.

Căn bậc 2 không xác định do $2-\sqrt{5}< 0$

d.

\(=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}\)

e.

\(=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1\)

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

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

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

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

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

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