13⁵.17³.a⁹.(13.17².a³)

= 13⁵.17³.13.17².a⁹.a³

= (13⁵.13).(17³.17²).(a⁹.a³)

= 13⁶.17⁵.a¹²

\(5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17=5\left(x^2-4\right)-\left(4x^2-12x+9\right)-x^2+17=12x-12\)

5(x+2)(x−2)−(2x−3)2−x2+17(x2−4)−(4x2−12x+9)−x2+17=12x−12

\(5\left(x+2\right)\left(x-2\right)-\left(2x-3\right)^2-x^2+17\)

\(=5x^2-20-4x^2+12x-9-x^2+17\)

=12x-12

\(16.17-\dfrac{5}{16}\times16+11\)

\(=272-5+11\)

\(=278\)

\(16\cdot17-\dfrac{5}{16}\cdot16+11\\ =16\cdot\left(17-\dfrac{5}{16}\right)+11\\ =16\cdot\dfrac{267}{16}+11\\ =267+11=278\)

\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)

\(A=\frac{2.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3.\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}\)

\(A=\frac{2}{3}\)

a.\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2\)

b.\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2\)

c.\(\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3\)

d.\(\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)

-158/17

=-158/17

-158/17

\(=\dfrac{40x-7-17}{5x-3}=\dfrac{40x-24}{5x-3}=\dfrac{8\left(5x-3\right)}{5x-3}=8\)