Đổi $25\%$ = $\genfrac{}{}{0ex}{}{1}{4}$.

Ta có $28$m vải còn lại ứng với:

     $1-\genfrac{}{}{0ex}{}{1}{3}=\genfrac{}{}{0ex}{}{2}{3}$ (số mét vải còn lại sau ngày thứ nhất).

Sau ngày thứ nhất người đó bán còn lại số mét vải là:

     $28:\genfrac{}{}{0ex}{}{2}{3}=42$ (m)

Số mét vải ban đầu là:

     $(42+15):(1-\genfrac{}{}{0ex}{}{1}{4})=57:\genfrac{}{}{0ex}{}{3}{4}=76$ (m).

a) $\genfrac{}{}{0ex}{}{3}{8}-\genfrac{}{}{0ex}{}{1}{6}x$ $=\genfrac{}{}{0ex}{}{1}{4}$          

Vậy $x=\genfrac{}{}{0ex}{}{3}{4}$.

b) ${\left(x-1\right)}^2=\genfrac{}{}{0ex}{}{1}{4}$                               

Suy ra [(�−1)2=(12)2(�−1)2=(−12)2​​(x−1)2=(21​)2(x−1)2=(2−1​)2​ hay [�−1=12 �−1=−12 ​​x−1=21​ x−1=2−1​ ​ 

[�=12+1 �=−12+1 ​​x=21​+1 x=2−1​+1 ​ suy ra [�=32 �=12 ​​x=23​ x=21​ ​

Vậy $x\in \left\{\genfrac{}{}{0ex}{}{3}{2};\genfrac{}{}{0ex}{}{1}{2}\right\}$.

c) $\left(x-\genfrac{}{}{0ex}{}{-1}{2}\right).\left(x+\genfrac{}{}{0ex}{}{1}{3}\right)=0$.

Suy ra [�−−12=0�+13=0​​x−2−1​=0x+31​=0​  hay [�=−12 �=−13 ​​x=2−1​ x=3−1​ ​       

Vậy $x\in \left\{\genfrac{}{}{0ex}{}{-1}{2};\genfrac{}{}{0ex}{}{-1}{3}\right\}$.

a) $\genfrac{}{}{0ex}{}{3}{8}-\genfrac{}{}{0ex}{}{1}{6}x$ $=\genfrac{}{}{0ex}{}{1}{4}$          

Vậy $x=\genfrac{}{}{0ex}{}{3}{4}$.

b) ${\left(x-1\right)}^2=\genfrac{}{}{0ex}{}{1}{4}$                               

Suy ra [(�−1)2=(12)2(�−1)2=(−12)2​​(x−1)2=(21​)2(x−1)2=(2−1​)2​ hay [�−1=12 �−1=−12 ​​x−1=21​ x−1=2−1​ ​ 

[�=12+1 �=−12+1 ​​x=21​+1 x=2−1​+1 ​ suy ra [�=32 �=12 ​​x=23​ x=21​ ​

Vậy $x\in \left\{\genfrac{}{}{0ex}{}{3}{2};\genfrac{}{}{0ex}{}{1}{2}\right\}$.

c) $\left(x-\genfrac{}{}{0ex}{}{-1}{2}\right).\left(x+\genfrac{}{}{0ex}{}{1}{3}\right)=0$.

Suy ra [�−−12=0�+13=0​​x−2−1​=0x+31​=0​  hay [�=−12 �=−13 ​​x=2−1​ x=3−1​ ​       

Vậy $x\in \left\{\genfrac{}{}{0ex}{}{-1}{2};\genfrac{}{}{0ex}{}{-1}{3}\right\}$.

a) $\genfrac{}{}{0ex}{}{1}{3}+\genfrac{}{}{0ex}{}{3}{4}-\genfrac{}{}{0ex}{}{5}{6}$

$=\genfrac{}{}{0ex}{}{4}{12}+\genfrac{}{}{0ex}{}{9}{12}-\genfrac{}{}{0ex}{}{10}{12}$

$=\genfrac{}{}{0ex}{}{4+9-10}{12}$

$=\genfrac{}{}{0ex}{}{3}{12}$

$=\genfrac{}{}{0ex}{}{1}{4}$.

b) $\genfrac{}{}{0ex}{}{-2}{3}+\genfrac{}{}{0ex}{}{6}{5}:\genfrac{}{}{0ex}{}{2}{3}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{-2}{3}+\genfrac{}{}{0ex}{}{6}{5}.\genfrac{}{}{0ex}{}{3}{2}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{-2}{3}+\genfrac{}{}{0ex}{}{18}{10}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{-2}{3}+\genfrac{}{}{0ex}{}{9}{5}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{-10}{15}+\genfrac{}{}{0ex}{}{27}{15}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{1}{3}$.

c) $\genfrac{}{}{0ex}{}{-3}{7}+\genfrac{}{}{0ex}{}{5}{13}+\genfrac{}{}{0ex}{}{-4}{7}$

$=\left(\genfrac{}{}{0ex}{}{-3}{7}+\genfrac{}{}{0ex}{}{-4}{7}\right)+\genfrac{}{}{0ex}{}{5}{13}$

$=\genfrac{}{}{0ex}{}{-7}{7}+\genfrac{}{}{0ex}{}{5}{13}$

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

$=\genfrac{}{}{0ex}{}{-8}{13}$.

d) $\genfrac{}{}{0ex}{}{12}{19}+\genfrac{}{}{0ex}{}{-8}{13}-\genfrac{}{}{0ex}{}{12}{19}+\genfrac{}{}{0ex}{}{5}{-13}+2$

$=\genfrac{}{}{0ex}{}{12}{19}+\genfrac{}{}{0ex}{}{-8}{13}-\genfrac{}{}{0ex}{}{12}{19}+\genfrac{}{}{0ex}{}{-5}{13}+2$

$=\left(\genfrac{}{}{0ex}{}{12}{19}-\genfrac{}{}{0ex}{}{12}{19}\right)+\left(\genfrac{}{}{0ex}{}{-8}{13}+\genfrac{}{}{0ex}{}{-5}{13}\right)+2$

$=0+\genfrac{}{}{0ex}{}{-13}{13}+2$

$=1$.

a) $\genfrac{}{}{0ex}{}{1}{3}+\genfrac{}{}{0ex}{}{3}{4}-\genfrac{}{}{0ex}{}{5}{6}$

$=\genfrac{}{}{0ex}{}{4}{12}+\genfrac{}{}{0ex}{}{9}{12}-\genfrac{}{}{0ex}{}{10}{12}$

$=\genfrac{}{}{0ex}{}{4+9-10}{12}$

$=\genfrac{}{}{0ex}{}{3}{12}$

$=\genfrac{}{}{0ex}{}{1}{4}$.

b) $\genfrac{}{}{0ex}{}{-2}{3}+\genfrac{}{}{0ex}{}{6}{5}:\genfrac{}{}{0ex}{}{2}{3}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{-2}{3}+\genfrac{}{}{0ex}{}{6}{5}.\genfrac{}{}{0ex}{}{3}{2}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{-2}{3}+\genfrac{}{}{0ex}{}{18}{10}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{-2}{3}+\genfrac{}{}{0ex}{}{9}{5}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{-10}{15}+\genfrac{}{}{0ex}{}{27}{15}-\genfrac{}{}{0ex}{}{2}{15}$

$=\genfrac{}{}{0ex}{}{1}{3}$.

c) $\genfrac{}{}{0ex}{}{-3}{7}+\genfrac{}{}{0ex}{}{5}{13}+\genfrac{}{}{0ex}{}{-4}{7}$

$=\left(\genfrac{}{}{0ex}{}{-3}{7}+\genfrac{}{}{0ex}{}{-4}{7}\right)+\genfrac{}{}{0ex}{}{5}{13}$

$=\genfrac{}{}{0ex}{}{-7}{7}+\genfrac{}{}{0ex}{}{5}{13}$

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

$=\genfrac{}{}{0ex}{}{-8}{13}$.

d) $\genfrac{}{}{0ex}{}{12}{19}+\genfrac{}{}{0ex}{}{-8}{13}-\genfrac{}{}{0ex}{}{12}{19}+\genfrac{}{}{0ex}{}{5}{-13}+2$

$=\genfrac{}{}{0ex}{}{12}{19}+\genfrac{}{}{0ex}{}{-8}{13}-\genfrac{}{}{0ex}{}{12}{19}+\genfrac{}{}{0ex}{}{-5}{13}+2$

$=\left(\genfrac{}{}{0ex}{}{12}{19}-\genfrac{}{}{0ex}{}{12}{19}\right)+\left(\genfrac{}{}{0ex}{}{-8}{13}+\genfrac{}{}{0ex}{}{-5}{13}\right)+2$

$=0+\genfrac{}{}{0ex}{}{-13}{13}+2$

$=1$.

+ $S=\genfrac{}{}{0ex}{}{1}{31}+\genfrac{}{}{0ex}{}{1}{32}+\genfrac{}{}{0ex}{}{1}{33}+\dots +\genfrac{}{}{0ex}{}{1}{60}$

$S<\left(\genfrac{}{}{0ex}{}{1}{30}+\genfrac{}{}{0ex}{}{1}{30}+\dots +\genfrac{}{}{0ex}{}{1}{30}\right)+\left(\genfrac{}{}{0ex}{}{1}{40}+\genfrac{}{}{0ex}{}{1}{40}+\dots +\genfrac{}{}{0ex}{}{1}{40}\right)+\left(\genfrac{}{}{0ex}{}{1}{50}+\genfrac{}{}{0ex}{}{1}{50}+\dots +\genfrac{}{}{0ex}{}{1}{50}\right)$

$S<\genfrac{}{}{0ex}{}{10}{30}+\genfrac{}{}{0ex}{}{10}{40}+\genfrac{}{}{0ex}{}{10}{50}<\genfrac{}{}{0ex}{}{48}{60}=\genfrac{}{}{0ex}{}{4}{5};$

+ $S>\left(\genfrac{}{}{0ex}{}{1}{40}+\genfrac{}{}{0ex}{}{1}{40}+\dots +\genfrac{}{}{0ex}{}{1}{40}\right)+\left(\genfrac{}{}{0ex}{}{1}{50}+\genfrac{}{}{0ex}{}{1}{50}+\dots +\genfrac{}{}{0ex}{}{1}{50}\right)+\left(\genfrac{}{}{0ex}{}{1}{60}+\genfrac{}{}{0ex}{}{1}{60}+\dots +\genfrac{}{}{0ex}{}{1}{60}\right)$

$S>\genfrac{}{}{0ex}{}{10}{40}+\genfrac{}{}{0ex}{}{10}{50}+\genfrac{}{}{0ex}{}{10}{60}>\genfrac{}{}{0ex}{}{3}{5}.$

a) Số lượng bóng rổ bán được trong tháng 1, tháng 2, tháng 3 lần lượt là:

$15$ quả; $20$ quả; $10$ quả.

b) Cả ba tháng cửa hàng bán được:

$15+20+10=45$ (quả)

c) Tháng 2 cửa hàng bán được nhiều hơn tháng 3:

$20–10=10$ (quả)

d) Tỉ số giữa số lượng bóng bán được trong tháng 1 và tháng 2 là:

$3:4=\genfrac{}{}{0ex}{}{3}{4}$
 

1. a) $O$ thuộc các đoạn thẳng: $AB;CD;OA;OB;OC;OD.$

b) Ta có $O$ nằm giữa hai điểm $A$ và $B$ và $OA=OB=3$ cm nên $O$ là trung điểm của đoạn thẳng $AB.$

2. a) Số đo góc $xOy$ bằng $30^{\circ }$.

b) 

a) $\genfrac{}{}{0ex}{}{5}{17}-\genfrac{}{}{0ex}{}{25}{31}+\genfrac{}{}{0ex}{}{12}{17}+\genfrac{}{}{0ex}{}{-6}{31}$

$=\genfrac{}{}{0ex}{}{5}{17}-\genfrac{}{}{0ex}{}{25}{31}+\genfrac{}{}{0ex}{}{12}{17}+\genfrac{}{}{0ex}{}{-6}{31}$

$=\left(\genfrac{}{}{0ex}{}{5}{17}+\genfrac{}{}{0ex}{}{12}{17}\right)+\left(-\genfrac{}{}{0ex}{}{25}{31}+\genfrac{}{}{0ex}{}{-6}{31}\right)$

$=1+(-1)$

$=0$

b) $\genfrac{}{}{0ex}{}{17}{8}:\left(\genfrac{}{}{0ex}{}{27}{8}+\genfrac{}{}{0ex}{}{11}{4}\right)$

$=\genfrac{}{}{0ex}{}{17}{8}:\left(\genfrac{}{}{0ex}{}{27}{8}+\genfrac{}{}{0ex}{}{22}{8}\right)$

$=\genfrac{}{}{0ex}{}{17}{8}:\genfrac{}{}{0ex}{}{49}{8}$

$=\genfrac{}{}{0ex}{}{17}{49}$.

c) $\genfrac{}{}{0ex}{}{1}{5}\cdot \genfrac{}{}{0ex}{}{11}{16}+\genfrac{}{}{0ex}{}{1}{5}\cdot \genfrac{}{}{0ex}{}{5}{16}+\genfrac{}{}{0ex}{}{4}{5}$

$=\genfrac{}{}{0ex}{}{1}{5}\cdot \left(\genfrac{}{}{0ex}{}{11}{16}+\genfrac{}{}{0ex}{}{5}{16}\right)+\genfrac{}{}{0ex}{}{4}{5}$

$=\genfrac{}{}{0ex}{}{1}{5}\cdot 1+\genfrac{}{}{0ex}{}{4}{5}$

$=\genfrac{}{}{0ex}{}{1}{5}+\genfrac{}{}{0ex}{}{4}{5}=1.$

d) $\genfrac{}{}{0ex}{}{5}{6}:25-2+\genfrac{}{}{0ex}{}{-7}{3}\cdot \genfrac{}{}{0ex}{}{2}{7}$

$=\genfrac{}{}{0ex}{}{5}{6}:25-2+\genfrac{}{}{0ex}{}{-7}{3}\cdot \genfrac{}{}{0ex}{}{2}{7}$

$=\genfrac{}{}{0ex}{}{5}{6}\cdot \genfrac{}{}{0ex}{}{1}{25}-2+\genfrac{}{}{0ex}{}{-2}{3}$

$=\genfrac{}{}{0ex}{}{1}{30}-2+\genfrac{}{}{0ex}{}{-2}{3}$

$=\genfrac{}{}{0ex}{}{1}{30}-\genfrac{}{}{0ex}{}{60}{30}+\genfrac{}{}{0ex}{}{-20}{30}$

$=\genfrac{}{}{0ex}{}{1}{30}-\genfrac{}{}{0ex}{}{60}{30}+\genfrac{}{}{0ex}{}{-20}{30}=\genfrac{}{}{0ex}{}{-79}{30}$
 

I dream of having a robot named Franky. It's a small home robot. Franky can tidy up rooms, make simple meals, and even water my plants. I want Franky because it can help me with daily chores, so I will have more time to relax and play with my friends.