a) $\Delta ABC$ vuông cân nên $\hat{B}=\hat{C}={45}^{\circ }.$

$\Delta BHE$ vuông tại $H$ có $\widehat{BEH}+\hat{B}={90}^{\circ }$

Suy ra $\widehat{BEH}={90}^{\circ }-{45}^{\circ }={45}^{\circ }$ nên $\hat{B}=\widehat{BEH}={45}^{\circ }$.

Vậy $\Delta BEH$ vuông cân tại $H.$

b) Chứng minh tương tự câu a ta được $\Delta CFG$ vuông cân tại $G$ nên $GF=GC$ và $HB=HE$

Mặt khác $BH=HG=GC$ suy ra $EH=HG=GF$ và $EH$ // $FG$ (cùng vuông góc với $BC)$

Tứ giác $EFGH$ có $EH$ // $FG,EH=FG$ nên là hình bình hành.

Hình bình hành $EFGH$ có một góc vuông $\hat{H}$ nên là hình chữ nhật

Hình chữ nhật $EFGH$ có hai cạnh kề bằng nhau $EH=HG$ nên là hình vuông.

Tứ giác $OBAC$ có ba góc vuông $\hat{B}=\hat{C}=\widehat{BOC}={90}^{\circ }$

Nên $OBAC$ là hình chữ nhật.

Mà $A$ nằm trên tia phân giác $OM$ suy ra $AB=AC$.

Khi đó $OBAC$ là hình vuông.

Dọc theo chiều dài, ta trồng được:

$5.5:\genfrac{}{}{0ex}{}{1}{4}=22$ (khóm hoa)

Dọc theo chiều rộng, ta trồng được:

$3,75:\genfrac{}{}{0ex}{}{1}{4}=15$ (khóm hoa)

Như vậy, số khóm hoa trồng được dọc theo hai cạnh của mảnh vườn là:

$[(22+15).2]-4=70$ (khóm hoa)

a) $\genfrac{}{}{0ex}{}{1}{5}+\genfrac{}{}{0ex}{}{4}{5}:x=0,75$

$\genfrac{}{}{0ex}{}{1}{5}+\genfrac{}{}{0ex}{}{4}{5}:x=\genfrac{}{}{0ex}{}{3}{4}$

$\genfrac{}{}{0ex}{}{4}{5}:x=\genfrac{}{}{0ex}{}{3}{4}-\genfrac{}{}{0ex}{}{1}{5}$

$\genfrac{}{}{0ex}{}{4}{5}:x=\genfrac{}{}{0ex}{}{11}{20}$

$x=\genfrac{}{}{0ex}{}{16}{11}$;

b) $x+\genfrac{}{}{0ex}{}{1}{2}=1-x$

 $2x=1-\genfrac{}{}{0ex}{}{1}{2}$

 $2x=\genfrac{}{}{0ex}{}{1}{2}$

 $x=\genfrac{}{}{0ex}{}{1}{4}$.

a) $\genfrac{}{}{0ex}{}{2}{3}\cdot \genfrac{}{}{0ex}{}{5}{4}-\genfrac{}{}{0ex}{}{3}{4}\cdot \genfrac{}{}{0ex}{}{2}{3}=\genfrac{}{}{0ex}{}{2}{3}\cdot \left(\genfrac{}{}{0ex}{}{5}{4}-\genfrac{}{}{0ex}{}{3}{4}\right)=\genfrac{}{}{0ex}{}{2}{3}\cdot \genfrac{}{}{0ex}{}{1}{2}=\genfrac{}{}{0ex}{}{1}{3}$;
b) $2\cdot {\left(\genfrac{}{}{0ex}{}{-3}{2}\right)}^2-\genfrac{}{}{0ex}{}{7}{2}=2\cdot \genfrac{}{}{0ex}{}{9}{4}-\genfrac{}{}{0ex}{}{7}{2}=\genfrac{}{}{0ex}{}{9}{2}-\genfrac{}{}{0ex}{}{7}{2}=1$;
c) $-\genfrac{}{}{0ex}{}{3}{4}\cdot 5\genfrac{}{}{0ex}{}{3}{13}-0,75\cdot \genfrac{}{}{0ex}{}{36}{13}=-\genfrac{}{}{0ex}{}{3}{4}\cdot 5\genfrac{}{}{0ex}{}{3}{13}-\genfrac{}{}{0ex}{}{3}{4}\cdot \genfrac{}{}{0ex}{}{36}{13}$
$=-\genfrac{}{}{0ex}{}{3}{4}\left(5\genfrac{}{}{0ex}{}{3}{13}+\genfrac{}{}{0ex}{}{36}{13}\right)$
$=-\genfrac{}{}{0ex}{}{3}{4}\cdot 8=-6$.

a)$\left(\genfrac{}{}{0ex}{}{1}{2}+1,5\right)\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$2\cdot x=\genfrac{}{}{0ex}{}{1}{5}$

$x=\genfrac{}{}{0ex}{}{1}{5}:2$

$x=\genfrac{}{}{0ex}{}{1}{10}$
b) $\left(-1\genfrac{}{}{0ex}{}{3}{5}+x\right):\genfrac{}{}{0ex}{}{12}{13}=2\genfrac{}{}{0ex}{}{1}{6}$

$-1\genfrac{}{}{0ex}{}{3}{5}+x=\genfrac{}{}{0ex}{}{13}{6}\cdot \genfrac{}{}{0ex}{}{12}{13}$
$x=2+1\genfrac{}{}{0ex}{}{3}{5}$

$x=3\genfrac{}{}{0ex}{}{3}{5}$
c) $\left(x:2\genfrac{}{}{0ex}{}{1}{3}\right)\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x\cdot \genfrac{}{}{0ex}{}{3}{7}\cdot \genfrac{}{}{0ex}{}{1}{7}=\genfrac{}{}{0ex}{}{-3}{8}$

$x=\genfrac{}{}{0ex}{}{-3}{8}:\genfrac{}{}{0ex}{}{3}{49}$
$x=\genfrac{}{}{0ex}{}{-49}{8}=-6\genfrac{}{}{0ex}{}{1}{8}$
d) $\genfrac{}{}{0ex}{}{-4}{7}\cdot x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}:\left(-1\genfrac{}{}{0ex}{}{2}{3}\right)$

$\genfrac{}{}{0ex}{}{-4}{7}x+\genfrac{}{}{0ex}{}{7}{5}=\genfrac{}{}{0ex}{}{1}{8}\cdot \genfrac{}{}{0ex}{}{-3}{5}$
$-\genfrac{}{}{0ex}{}{4}{7}x=\genfrac{}{}{0ex}{}{-3}{40}-\genfrac{}{}{0ex}{}{7}{5}x=\genfrac{}{}{0ex}{}{-59}{40}:\genfrac{}{}{0ex}{}{-4}{7}=\genfrac{}{}{0ex}{}{413}{160}=2\genfrac{}{}{0ex}{}{93}{160}$
 

a) $x+\genfrac{}{}{0ex}{}{1}{3}=\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{5}{6}$

$x=-\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{3}$

$x=-\genfrac{}{}{0ex}{}{2}{3}$;
b) $\genfrac{}{}{0ex}{}{3}{8}+\genfrac{}{}{0ex}{}{5}{8}-\genfrac{}{}{0ex}{}{1}{5}-x=\genfrac{}{}{0ex}{}{1}{5}$

$x=1-\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{5}$

$x=\genfrac{}{}{0ex}{}{3}{5}$.

a) $x=7-\genfrac{}{}{0ex}{}{2}{5}+1,62=8,22$
b) $x=4\genfrac{}{}{0ex}{}{3}{5}+\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{2}=4\genfrac{}{}{0ex}{}{3}{10}$
c) $2x-x=\genfrac{}{}{0ex}{}{3}{5}+\genfrac{}{}{0ex}{}{4}{7}$
$x=\genfrac{}{}{0ex}{}{41}{35}$
d) $x=3\genfrac{}{}{0ex}{}{1}{2}-\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{1}{13}-0.25$
$x=2\genfrac{}{}{0ex}{}{223}{364}$

 

$\mathrm{A}=\left[\genfrac{}{}{0ex}{}{2}{7}\left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{3}\right)\right]:\left[\genfrac{}{}{0ex}{}{2}{7}\left(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{2}{5}\right)\right]=\left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{3}\right):\left(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{2}{5}\right)=1\genfrac{}{}{0ex}{}{1}{4}$.
 $\mathrm{B}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{3}{4}\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{2}{7}-\genfrac{}{}{0ex}{}{1}{3}+\genfrac{}{}{0ex}{}{2}{7}\right)}{\genfrac{}{}{0ex}{}{1}{5}\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)-\genfrac{}{}{0ex}{}{1}{3}\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{3}{4}\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{3}\right)}{\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{3}\right)\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)}=1\genfrac{}{}{0ex}{}{11}{52}$.
 

a) $\mathrm{A}=\genfrac{}{}{0ex}{}{3}{5}\cdot \genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}:\genfrac{}{}{0ex}{}{5}{3}-\genfrac{}{}{0ex}{}{2}{7}:1\genfrac{}{}{0ex}{}{2}{3}$;
b) $\mathrm{B}=\left(-13\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{-2}{9}:2\genfrac{}{}{0ex}{}{1}{2}+\genfrac{}{}{0ex}{}{2}{5}.\genfrac{}{}{0ex}{}{11}{9}\right)\cdot 2\genfrac{}{}{0ex}{}{1}{2}$;
c) $\mathrm{C}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}\right):\genfrac{}{}{0ex}{}{2}{3}+\left(\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right):\genfrac{}{}{0ex}{}{2}{3}$;
d) $\mathrm{D}=\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{1}{15}-\genfrac{}{}{0ex}{}{2}{3}\right)+\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{1}{11}-\genfrac{}{}{0ex}{}{5}{22}\right)$.