a: \(20-\left[30-\left(5-1\right)^2\right]\)

\(=20-\left[30-4^2\right]\)

\(=20-14=6\)

b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)

\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)

\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)

c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)

\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)

\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)

d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)

\(=411-\left[\dfrac{110}{5}-4\right]\)

=410-22+4

=410-18

=392

e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)

\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)

\(=450-5\cdot\left[9\cdot8-12\right]+18\)

=468-5*60

=468-300

=168

f:

\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)

\(=102-150:\left[18-2\cdot4\right]+1\)

\(=103-\dfrac{150}{18-8}=103-15=88\)

\(\sqrt{50}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)

\(=5\sqrt{2}-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{12}\)

\(=-15\sqrt{2}\)

\(B=50-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)

\(=50-3.\sqrt{7^2.2}+2\sqrt{2^2.2}+3\sqrt{4^2.2}-5\sqrt{3^2.2}\)

\(=50-3.7\sqrt{2}+2.2\sqrt{2}+3.4\sqrt{2}-5.3\sqrt{2}\)

\(=50-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{2}\)

\(=50+\sqrt{2}.\left(-21+4+12-15\right)\)

\(=50+\sqrt{2}.\left(-20\right)\)

\(=50-20\sqrt{2}\)

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

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

\(=\sqrt{3}^2+2.\sqrt{3}.\sqrt{5}+\sqrt{5}^2-7\)

\(=2\sqrt{15}+3+5-7\)

\(=2\sqrt{15}+1\)

6: \(=3\cdot2\sqrt{3}-4\cdot3\sqrt{3}+5\cdot4\sqrt{3}=14\sqrt{3}\)

7: \(=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)

8: \(=2\cdot4\sqrt{2}+4\cdot2\sqrt{2}-5\cdot3\sqrt{2}=\sqrt{2}\)

9: \(=3\cdot2\sqrt{5}-2\cdot3\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)

10: \(=2\cdot2\sqrt{6}-2\cdot3\sqrt{6}+3\sqrt{6}-5\sqrt{6}=-4\sqrt{6}\)

a) 2√18 - 4√50 + 3√32

= 6√2 - 20√2 + 12√2

= -2√2

b) √(√8 - 4)² + √8

= 4 - √8 + √8

= 4

c) √(14 - 6√5) + √(6 + 2√5)

= √(3 - √5)² + √(√5 + 1)²

= 3 - √5 + √5 + 1

= 4

\(a,2\sqrt{18}-4\sqrt{50}+3\sqrt{32}\\ =6\sqrt{2}-20\sqrt{2}+12\sqrt{2}=-2\sqrt{2}\\ b,\sqrt{\left(\sqrt{8}-4\right)^2}+\sqrt{8}\\ =4-\sqrt{8}+\sqrt{8}\\ =4\\ c,\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}\\ =\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}=3+\sqrt{5}+\sqrt{5}+1\\ =4+2\sqrt{5}\)

a) \(\sqrt{1-8x+16x^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{1^2-2\cdot4x\cdot1+\left(4x\right)^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{\left(4x-1\right)^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|4x-1\right|=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1=\dfrac{1}{3}\left(ĐK:x\ge\dfrac{1}{4}\right)\\4x-1=\dfrac{1}{3}\left(ĐK:x< \dfrac{1}{4}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=\dfrac{1}{6}\left(tm\right)\end{matrix}\right.\)

b) \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\) (ĐK: \(x\ge2\)) 

\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)

\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)

\(\Leftrightarrow6\sqrt{x-2}=18\)

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

\(\Leftrightarrow x-2=9\)

\(\Leftrightarrow x=9+2\)

\(\Leftrightarrow x=11\left(tm\right)\)

\(a.\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\left(\sqrt{3}>\sqrt{2}\right)=\sqrt{3}+2\sqrt{2}\)\(b.3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)