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

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