a) $\frac{{11}}{{24}} - \frac{5}{{41}} + \frac{{13}}{{24}} + 0,5 - \frac{{36}}{{41}}$

$= \left( {\frac{{11}}{{24}} + \frac{{13}}{{24}}} \right) - \left( {\frac{5}{{41}} + \frac{{36}}{{41}}} \right) + 0,5$

$= 1 - 1 + 0,5 = 0,5$.

b) $\frac{1}{2}.\frac{3}{4} + \frac{1}{2}.\frac{1}{4} + \frac{1}{2} = \frac{1}{2}.\left( {{\kern 1pt} \frac{3}{4} + \frac{1}{4} + 1} \right) = \frac{1}{2}.2 = 1$.

c) ${\left( {\frac{{ - 3}}{4}} \right)^2}:{\left( {\frac{{ - 1}}{4}} \right)^2} + 9.\left( {\frac{{ - 1}}{9}} \right) + \left| {\frac{{ - 3}}{2}} \right|$

$= \frac{9}{{16}}:\frac{1}{{16}} - 1 + \frac{3}{2} = 9 - 1 + \frac{3}{2} = \frac{{19}}{2}.$

d) $\sqrt {0,25} .{\left( { - 3} \right)^3} - \sqrt {\frac{1}{{81}}} :{\left( {\frac{{ - 1}}{3}} \right)^3} = 0,5.( - 27) - \frac{1}{9}:\frac{{ - 1}}{{27}}$

$= \frac{{ - 27}}{2} + 3 = \frac{{ - 21}}{2}$.

a) $5x + 7 = 2$

$5x = 2 - 7$

$x=-1$

b) $4+\sqrt{x}=20$

$x = {16^2}$

$x = 256$

c) ${\left( {x + 2} \right)^2} = 16$

${\left( {x + 2} \right)^2} = {4^2}$

TH1: $x + 2 = 4$

               $x = 2$

TH2: $x+2=-4$

               $x=-6$

a) $A = 6{x^3} - 3{x^2} + 2\left| x \right| + 4$

Thay $x = \frac{{ - 2}}{3}$ vào biểu thức $A$, ta được:

$A = 6.{\left( {\frac{{ - 2}}{3}} \right)^3} - 3.{\left( {\frac{{ - 2}}{3}} \right)^2} + 2.\left| {\frac{2}{3}} \right| + 4$

$= 6.\frac{{ - 8}}{{27}} - 3.\frac{4}{9} + 2.\frac{2}{3} + 4$

$= \frac{{ - 16}}{9} - \frac{4}{3} + \frac{4}{3} + 4$

$= \frac{{ - 16}}{9} + 4 = \frac{{20}}{9}$.

b) $B = 2\left| x \right| - 3\left| y \right|$

$= 2\left| {\frac{1}{2}} \right| - 3\left| { - 3} \right|$

$= 2.\frac{1}{2} - 3.3$

$=1-9=-8$.

a) Ta có: $\widehat {DCx} = \widehat {ACm}$ (hai góc đối đỉnh)

Suy ra $\widehat {DCx} = 60^\circ$

b) Ta có $$\widehat {DCx} + \widehat {ACD} = 180^\circ$$ (hai góc kề bù)

$\widehat{ACD}={180}^{\circ }-\widehat{DCx}$

$\widehat{ACD}={180}^{\circ }-{60}^{\circ }={120}^{\circ }$

Nên $\widehat {ACD} = \widehat {CDy} = 120^\circ$

Mà hai góc ở vị trí so le trong

Nên $$Ax\parallel By$$ (dấu hiệu nhận biết)

c) Do $$\left. \begin{array}{l}Ax//By\\Ax \bot AB\end{array} \right\}$$

Suy ra $$By \bot AB$$