1) $$\frac{3}{4} + 1\frac{2}{3} - \frac{7}{9} = \,\frac{3}{4} + \frac{5}{3} - \frac{7}{9} = \frac{{27}}{{36}} + \frac{{60}}{{36}} - \frac{{28}}{{36}} = \frac{{59}}{{36}}$$

2) $$\frac{1}{7} \cdot {\left( {\frac{4}{3}} \right)^2} - \frac{1}{7}:\frac{9}{{11}}$$$$= \frac{1}{7} \cdot \frac{{16}}{9} - \frac{1}{7} \cdot \frac{{11}}{9}$$

$$= \frac{1}{7}\left( {\frac{{16}}{9} - \frac{{11}}{9}} \right) = \frac{1}{7} \cdot \frac{5}{9} = \frac{5}{{63}}$$

3) $$\left( { - \frac{4}{9} + \frac{3}{5}} \right):1\frac{1}{5} + \left( {\frac{1}{5} - \frac{5}{9}} \right):1\frac{1}{5}$$

$$= \left( { - \frac{4}{9} + \frac{3}{5}} \right):\frac{6}{5} + \left( {\frac{1}{5} - \frac{5}{9}} \right):\frac{6}{5}$$

$$= \left( { - \frac{4}{9} + \frac{3}{5}} \right).\frac{5}{6} + \left( {\frac{1}{5} - \frac{5}{9}} \right).\frac{5}{6}$$

$$= \left( { - \frac{4}{9} + \frac{3}{5} + \frac{1}{5} - \frac{5}{9}} \right).\frac{5}{6}$$$$=  - \frac{1}{5}.\frac{5}{6} =  - \frac{1}{6}$$.

4) $$\,0,5.\frac{4}{9} + {\left( {\frac{1}{3} - 1,5} \right)^2} - {\left( {\frac{2}{3}} \right)^9}:{\left( {\frac{2}{3}} \right)^7}$$

$$= \frac{1}{2}.\frac{4}{9} + {\left( {\frac{1}{3} - \frac{3}{2}} \right)^2} - {\left( {\frac{2}{3}} \right)^2}$$$$= \frac{2}{9} + {\left( {\frac{{ - 7}}{6}} \right)^2} - \frac{4}{9}$$

$$= \frac{2}{9} + \frac{{49}}{{36}} - \frac{4}{9} = \frac{{41}}{{36}}$$.

1) $$\frac{5}{2}x - \frac{3}{4} = \frac{1}{4} \Leftrightarrow \,\,\frac{5}{2}x = \frac{1}{4} + \frac{3}{4}$$

$$\Leftrightarrow \,\frac{5}{2}x = 1 \Leftrightarrow \,x = 1:\frac{5}{2} \Leftrightarrow \,x = \frac{2}{5}$$

2) $$\frac{{x + 4}}{{20}} = \frac{5}{{x + 4}} \Leftrightarrow \,\,(x + 4)(x + 4) = 5\,.\,20$$

$$\Leftrightarrow \,{(x + 4)^2} = 100$$

$$\, \Leftrightarrow \,\,\left[ \begin{array}{l}x + 4 = 10\\x + 4 =  - 10\end{array} \right. \Leftrightarrow \,\,\left[ \begin{array}{l}x = 6\\x =  - 14\end{array} \right.$$

Vậy $$x \in \left\{ {6;\,\, - 14} \right\}$$.

GT, KL:

Để làm ý 1,2 HS không cần thiết phải vẽ hình

1) Ta có: $$AB\parallel Mx$$ Þ$\widehat {ABM} + \widehat {{M_1}} = 180^\circ$

Hay $135^\circ  + \widehat {{M_1}} = 180^\circ$Þ$\widehat {{M_1}} = 180^\circ  - 135^\circ  = 45^\circ$

    Vì $\widehat {BMN} = 135^\circ$Þ$\widehat {{M_1}} + \widehat {{M_2}} = 180^\circ$Þ$\widehat {{M_2}} = 135^\circ  - 45^\circ  = 90^\circ$

2) Vì $\widehat {{M_2}} = 90^\circ$Þ $Mx \bot MN$

   Mà $NP \bot MN$ (gt)

   Nên $$Mx\parallel NP$$  (vì cùng vuông góc với $$MN$$)

   Do đó $$AB\parallel NP$$ (vì cùng song song với $$Mx$$).

3) Vì MQ là tia phân giác của $\widehat {MNP}$

Þ$\widehat {MNQ} = \widehat {QNP} = \frac{1}{2}\widehat {MNP}$ Þ$$\widehat {MNQ} = \widehat {QNP} = \frac{1}{2}\,.\,90^\circ  = 45^\circ$$

Vì $$Mx\parallel NP$$ Þ$\widehat {NQM} = \widehat {QNP} = 45^\circ$ (cặp góc so le trong)

Þ$\widehat {NQM} = \widehat {{M_1}}( = 45^\circ )$ Þ $$NQ\parallel MB$$ (đpcm)

Ta có:$$\frac{{3x - 2y}}{4} = \frac{{2z - 4x}}{3} = \frac{{4y - 3z}}{2}$$

$$\Rightarrow \frac{{12x - 8y}}{{16}} = \frac{{6z - 12x}}{9} = \frac{{8y - 6z}}{4}$$

$$= \frac{{12x - 8y + 6z - 12x + 8y - 6z}}{{14 + 9 + 4}} = 0$$

Do đó:

$$\left\{ \begin{array}{l}3x - 2y = 0\\2z - 4x = 0\\4y - 3z = 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}3x = 2y\\2z = 4x\\4y = 3z\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\frac{x}{2} = \frac{{y\,}}{3}\,\,\,\,(1)\\\frac{z}{4} = \frac{x}{2}\\\frac{y}{3} = \frac{z}{4}\,\,\,\,(2)\end{array} \right.$$

Từ (1) và (2) ta có: $$\frac{x}{2} = \frac{{y\,}}{3} = \frac{z}{4}$$

Suy ra : $$\frac{x}{2} = \frac{{y\,}}{3} = \frac{z}{4} = \frac{{x + y - z}}{{2 + 3 - 4}} = \frac{{ - 10}}{1} =  - 10$$

Từ đó ta có: $$x =  - 20;{\rm{ }}y =  - 30;{\rm{ }}z =  - 40$$.