Diện tích xung quanh của bể bơi là:

$S=(12+5).2.2,75=34\; .\; 2,75=93,5$ (m2)

Diện tích mặt đáy của bể bơi là: $S=12\; .\; 5=60$ (m2)

Tổng diện tích xung quanh và một mặt đáy của bể bơi là:

$S=93,5+60=153,5$ (m2)

Diện tích một viên gạch men là: $20.25=500$ (cm2)

Đổi: $500$ cm2 $=0,05$ m2.

Số viên gạch men cần dùng là: $153,5:0,05=3070$ (viên)

Vậy người thợ phải dùng $3070$ viên gạch men hình chữ nhật để lát đáy và xung quanh thành bể đó.

a) $x+\genfrac{}{}{0ex}{}{5}{6}=\genfrac{}{}{0ex}{}{4}{3}$

$x=\genfrac{}{}{0ex}{}{4}{3}-\genfrac{}{}{0ex}{}{5}{6}=\genfrac{}{}{0ex}{}{8}{6}-\genfrac{}{}{0ex}{}{5}{6}$

$x=\genfrac{}{}{0ex}{}{3}{6}=\genfrac{}{}{0ex}{}{1}{2}$

b) $x:2^4=8^3$

$x:2^4=(2^3{)}^3$

$x:2^4=2^9$

$x=2^9.2^4=2^{13}$

c) $\genfrac{}{}{0ex}{}{13}{4}.(\genfrac{}{}{0ex}{}{5}{52}-x)=\genfrac{}{}{0ex}{}{1}{4}$

$\genfrac{}{}{0ex}{}{5}{52}-x=\genfrac{}{}{0ex}{}{1}{4}:\genfrac{}{}{0ex}{}{13}{4}$

$\genfrac{}{}{0ex}{}{5}{52}-x=\genfrac{}{}{0ex}{}{1}{13}$

$x=\genfrac{}{}{0ex}{}{5}{52}-\genfrac{}{}{0ex}{}{1}{13}$

$x=\genfrac{}{}{0ex}{}{1}{52}$
 

a) $\genfrac{}{}{0ex}{}{5}{9}-{\left(\genfrac{}{}{0ex}{}{1}{3}\right)}^2=\genfrac{}{}{0ex}{}{5}{9}-\genfrac{}{}{0ex}{}{1}{9}=\genfrac{}{}{0ex}{}{4}{9}$

b) $\genfrac{}{}{0ex}{}{1}{5}.\genfrac{}{}{0ex}{}{-3}{2}+\genfrac{}{}{0ex}{}{-17}{2}.\genfrac{}{}{0ex}{}{1}{5}$

$=\genfrac{}{}{0ex}{}{1}{5}.(\genfrac{}{}{0ex}{}{-3}{2}+\genfrac{}{}{0ex}{}{-17}{2})$

$=\genfrac{}{}{0ex}{}{1}{5}.(-10)=-2$

c) $1+(\genfrac{}{}{0ex}{}{-2}{5}+\genfrac{}{}{0ex}{}{11}{13})-(\genfrac{}{}{0ex}{}{3}{5}-\genfrac{}{}{0ex}{}{2}{13})$

$=1+\genfrac{}{}{0ex}{}{-2}{5}+\genfrac{}{}{0ex}{}{11}{13}-\genfrac{}{}{0ex}{}{3}{5}+\genfrac{}{}{0ex}{}{2}{13}$

$=1+(\genfrac{}{}{0ex}{}{-2}{5}-\genfrac{}{}{0ex}{}{3}{5})+(\genfrac{}{}{0ex}{}{11}{13}+\genfrac{}{}{0ex}{}{2}{13})$

$=1+(-1)+1=1$

Ta có: $1^2+2^2+3^2+\dots +10^2=385$

Suy ra: $\left(1^2+2^2+3^2+\dots +10^2\right).3^2=385.3^2$

$(1.3{)}^2+(2.3{)}^2+(3.3{)}^2+\dots +(10.3{)}^2=385.3^2$

Do đó $A=3^2+6^2+9^2+\dots +30^2=3465$.

Số tiền 3 quyển sách là:

$3.120000=360000$ (đồng)

Số tiền Lan phải trả khi có thẻ thành viên là:

$360000.(100\%-10\%)=324000$ (đồng)

Ta có: $350000-324000=26000$ (đồng).

Do đó Lan được trả lại $26000$ đồng.​

a) Ta có $\widehat{CAx}+\widehat{BAC}=180^{\circ }$ (hai góc kề bù).

Suy ra: $\widehat{CAx}=180^{\circ }-\widehat{BAC}$

$\widehat{CAx}=180^{\circ }-100^{\circ }=80^{\circ }$.

b) Vì $Ay$ là tia phân giác của $\widehat{CAx}$, nên

$\widehat{CAy}=\widehat{xAy}=\genfrac{}{}{0ex}{}{1}{2}.\widehat{CAx}=\genfrac{}{}{0ex}{}{1}{2}.80^{\circ }=40^{\circ }$.

Vậy $\widehat{CAy}=\widehat{ACB}$, mà hai góc này ở vị trí so le trong, do đó $Ay$ // $BC$.

c) Do $Ay$ // $BC$, nên $\widehat{xAy}=\widehat{ABC}$ (hai góc đồng vị).

Suy ra $\widehat{ABC}=40^{\circ }$.

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

$x=\genfrac{}{}{0ex}{}{1}{6}+\genfrac{}{}{0ex}{}{2}{3}$

$x=\genfrac{}{}{0ex}{}{1}{6}+\genfrac{}{}{0ex}{}{4}{6}$

$x=\genfrac{}{}{0ex}{}{5}{6}$.

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

$2x=-\genfrac{}{}{0ex}{}{5}{3}-\genfrac{}{}{0ex}{}{1}{2}$

$2x=-\genfrac{}{}{0ex}{}{13}{6}$

$x=-\genfrac{}{}{0ex}{}{13}{12}$.

c) $3x+\genfrac{}{}{0ex}{}{3}{2}=x-\genfrac{}{}{0ex}{}{5}{3}$

$3x-x=-\genfrac{}{}{0ex}{}{5}{3}-\genfrac{}{}{0ex}{}{3}{2}$

$2x=\genfrac{}{}{0ex}{}{-19}{6}$

$x=\genfrac{}{}{0ex}{}{-19}{12}$.

a) $\genfrac{}{}{0ex}{}{11}{24}-\genfrac{}{}{0ex}{}{5}{41}+\genfrac{}{}{0ex}{}{13}{24}+0,5-\genfrac{}{}{0ex}{}{36}{41}=(\genfrac{}{}{0ex}{}{11}{24}+\genfrac{}{}{0ex}{}{13}{24})-(\genfrac{}{}{0ex}{}{5}{41}+\genfrac{}{}{0ex}{}{36}{41})+0,5=1-1+0,5=0,5$.

b) $\genfrac{}{}{0ex}{}{1}{2}\cdot \genfrac{}{}{0ex}{}{3}{4}+\genfrac{}{}{0ex}{}{1}{2}\cdot \genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{2}=\genfrac{}{}{0ex}{}{1}{2}\cdot (\genfrac{}{}{0ex}{}{3}{4}+\genfrac{}{}{0ex}{}{1}{4}+1)=\genfrac{}{}{0ex}{}{1}{2}\cdot 2=1$.

c) $(\genfrac{}{}{0ex}{}{-3}{4}{)}^2:(\genfrac{}{}{0ex}{}{-1}{4}{)}^2+9\cdot \left(\genfrac{}{}{0ex}{}{-1}{9}\right)+\left(\genfrac{}{}{0ex}{}{-3}{2}\right)=\genfrac{}{}{0ex}{}{9}{16}:\genfrac{}{}{0ex}{}{1}{16}-1-\genfrac{}{}{0ex}{}{3}{2}=9-1-\genfrac{}{}{0ex}{}{3}{2}=\genfrac{}{}{0ex}{}{13}{2}.$

d) $\sqrt{0,25}\cdot (-3{)}^3-\sqrt{\genfrac{}{}{0ex}{}{1}{81}}:(\genfrac{}{}{0ex}{}{-1}{3}{)}^3=0,5\cdot (-27)-\genfrac{}{}{0ex}{}{1}{9}:\genfrac{}{}{0ex}{}{-1}{27}=\genfrac{}{}{0ex}{}{-27}{2}+3=\genfrac{}{}{0ex}{}{-21}{2}$.

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

a) $\mathrm{A}=\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}.\genfrac{}{}{0ex}{}{3}{5}-\genfrac{}{}{0ex}{}{2}{7}.\genfrac{}{}{0ex}{}{3}{5}$
$=\genfrac{}{}{0ex}{}{3}{5}\cdot \left(\genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}-\genfrac{}{}{0ex}{}{2}{7}\right)=\genfrac{}{}{0ex}{}{3}{5}$
b) $\mathrm{B}=\left(-13\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{-2}{9}\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{5}{2}$
$=\left(-13-\genfrac{}{}{0ex}{}{2}{9}+\genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{5}{2}=-13+\left(\genfrac{}{}{0ex}{}{11}{9}-\genfrac{}{}{0ex}{}{2}{9}\right)=-12.$
c) $\mathrm{C}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}+\left(\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{-1}{5}\right)+\left(\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{2}{7}\right)\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=0.$
d) $\mathrm{D}=\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{1}{15}-\genfrac{}{}{0ex}{}{10}{15}\right)+\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{2}{22}-\genfrac{}{}{0ex}{}{5}{22}\right)$
$=\genfrac{}{}{0ex}{}{4}{9}:\genfrac{}{}{0ex}{}{-3}{5}+\genfrac{}{}{0ex}{}{4}{9}:\genfrac{}{}{0ex}{}{-3}{22}=\genfrac{}{}{0ex}{}{4}{9}\cdot \genfrac{}{}{0ex}{}{-5}{3}+\genfrac{}{}{0ex}{}{4}{9}.\genfrac{}{}{0ex}{}{-22}{3}$
$=\genfrac{}{}{0ex}{}{4}{9}\cdot \left(\genfrac{}{}{0ex}{}{-5}{3}+\genfrac{}{}{0ex}{}{-22}{3}\right)=\genfrac{}{}{0ex}{}{4}{9}.\genfrac{}{}{0ex}{}{-27}{3}=-4.$