Lời giải:
a.

$21\times 23-3\times 7\times (-17)$

$=21\times 23-21\times (-17)$

$=21\times 23+21\times 17=21\times (23+17)=21\times 40$

$=840$

b.

$42\times 3-7\times [(-34)+18]$

$=7\times 18-7\times (-16)$

$=7\times 18+7\times 16=7\times (18+16)=7\times 34=238$

c.

$=142\times 32+32\times (-7)-13\times 32$

$=32\times (142-7-32)=32\times 103=3296$

d.

$=13\times [(23-17)-(23+17)]$

$=13\times (23-17-23-17)=13\times (-34)=-442$

\(\sqrt{17}-3\sqrt{32}-\sqrt{17}+3\sqrt{32}\\ =\left(\sqrt{17}-\sqrt{17}\right)+\left(3\sqrt{32}-3\sqrt{32}\right)\\ =0+0=0\)

\(=\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{17-2\cdot3\cdot2\sqrt{2}}-\sqrt{17+2\cdot3\cdot2\sqrt{2}}\)

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

=3-2căn 2-3-2căn 2

=-4căn 2

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}=3-2\sqrt{2}+3+2\sqrt{2}=6\)

`\sqrt(17-3\sqrt32)+\sqrt(17+3\sqrt32)`

`=\sqrt(17-12\sqrt2)+\sqrt(17+12\sqrt2)`

`=\sqrt(9-12\sqrt2+8)+\sqrt(9+12\sqrt2+8)`

`=\sqrt(3^2-2.3.2\sqrt2 +(2\sqrt2)^2)+\sqrt(3^2+2.3.2\sqrt2+(2\sqrt2)^2)`

`=\sqrt((3-2\sqrt2)^2)+\sqrt((3+2\sqrt2)^2)`

`=3-2\sqrt2+3+2\sqrt2`

`=6`

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

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

=6

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)

\(=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

\(=\sqrt{9+2.3.\sqrt{8}+8}+\sqrt{9-2.3.\sqrt{8}+8}\)

\(=\sqrt{\left(3+\sqrt{8}\right)^2}+\sqrt{\left(3-\sqrt{8}\right)^2}=\left|3+\sqrt{8}\right|+\left|3-\sqrt{8}\right|\)

\(=3+\sqrt{8}+3-\sqrt{8}\)   (do \(3>\sqrt{8}\))

\(=6\)

\(=\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)

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

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

\(=-4\sqrt{2}\)

\(\sqrt{17-3\sqrt{32}}+\sqrt{17-3\sqrt{32}}=2\sqrt{17-3\sqrt{32}}\)

\(=\sqrt{4\left(17-3\sqrt{32}\right)}=\sqrt{68-12\sqrt{32}}=\sqrt{36-12\sqrt{32}+32}\)

\(=\sqrt{6^2-2.6.\sqrt{32}+\left(\sqrt{32}\right)^2}=\sqrt{\left(6-\sqrt{32}\right)^2}=\left|6-\sqrt{32}\right|\)

\(=6-\sqrt{32}=6-4\sqrt{2}\)

Cần gấp thì bạn cũng nên viết đầy đủ đề bài nhé.

** Bài toán rút gọn**

Lời giải:

\(\sqrt{17-12\sqrt{2}}=\sqrt{17-2\sqrt{72}}=\sqrt{9-2\sqrt{8.9}+8}=\sqrt{(\sqrt{9}-\sqrt{8})^2}\)

\(=\sqrt{9}-\sqrt{8}=3-2\sqrt{2}\)

\(\sqrt{24-8\sqrt{8}}=\sqrt{24-2\sqrt{128}}=\sqrt{16-2\sqrt{16.8}+8}=\sqrt{(\sqrt{16}-\sqrt{8})^2}\)

\(=\sqrt{16}-\sqrt{8}=4-2\sqrt{2}\)

\(\Rightarrow \sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=(3-2\sqrt{2})-(4-2\sqrt{2})=-1\)

--------------------

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{8-2\sqrt{8.9}+9}+\sqrt{8+2\sqrt{8.9}+9}\)

\(=\sqrt{(\sqrt{8}-\sqrt{9})^2}+\sqrt{(\sqrt{8}+\sqrt{9})^2}\)

\(=|\sqrt{8}-\sqrt{9}|+|\sqrt{8}+\sqrt{9}|=3-2\sqrt{2}+3+2\sqrt{2}=6\)

----------------------

\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2\sqrt{9.2}+2}-\sqrt{9-2\sqrt{9.2}+2}\)

\(=\sqrt{(\sqrt{9}+\sqrt{2})^2}-\sqrt{(\sqrt{9}-\sqrt{2})^2}\)

\(=|\sqrt{9}+\sqrt{2}|-|\sqrt{9}-\sqrt{2}|=3+\sqrt{2}-(3-\sqrt{2})=2\sqrt{2}\)

\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)

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

\(=-1\)

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)

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

\(=6\)

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

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

\(=2\sqrt{2}\)

c) Ta có: \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

\(=2\sqrt{2}\)