Nếu mua hai gói thì giá của mỗi gói là:

\(90000:2=45000\left(đồng\right)\)

Nếu mua ba gói thì giá của mỗi gói là:

\(130000:3=43333\left(đồng\right)\)

Mà \(50000>45000>43333\) nên mẹ khuyên Mai nên mua 3 gói là rẻ nhất là đúng.

Nếu mua hai gói thì giá của mỗi gói là: 90000/2=45000
 đồng

Nếu mua ba gói thì giá của mỗi gói là: 130000/3 đồng

Vì 130000/3< 45 000 < 50 000 đồng nên mua ba gói là rẻ nhất

Bài 1:

$A=1.2+2.3+3.4+...+201.202$

$3A=1.2.3+2.3(4-1)+3.4(5-2)+....+201.202(203-200)$

$=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+201.202.203-200.201.202$

$=(1.2.3+2.3.4+3.4.5+...+201.202.203)-(1.2.3+2.3.4+....+200.201.202)$

$=201.202.203$

$\Rightarrow A=\frac{201.202.203}{3}=2747402$

Bài 2:

$S=4.5+5.6+6.7+....+100.101$

$3S=4.5(6-3)+5.6.(7-4)+6.7.(8-5)+....+100.101(102-99)$

$=4.5.6-3.4.5+5.6.7-4.5.6+6.7.8-5.6.7+....+100.101.102-99.100.101$

$=(4.5.6+5.6.7+6.7.8+...+100.101.102)-(3.4.5+4.5.6+5.6.7+...+99.100.101)$

$=100.101.102-3.4.5$

$\Rightarrow S=\frac{100.101.102-3.4.5}{3}=343380$

\(a.\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\\ =\dfrac{2}{3}+\dfrac{2}{7}=\dfrac{20}{21}\\ b.\dfrac{7}{12}-\dfrac{27}{7}\cdot\dfrac{1}{18}\\ =\dfrac{7}{12}-\dfrac{3}{14}=\dfrac{31}{84}\\ c.\left(\dfrac{23}{11}-\dfrac{15}{82}\right)\cdot\dfrac{41}{25}\\ =\dfrac{1721}{902}\cdot\dfrac{41}{25}=\dfrac{1721}{550}\\ d.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)\cdot\left(\dfrac{3}{13}-\dfrac{8}{13}\right)\\ =\dfrac{13}{10}\cdot\left(\dfrac{-5}{13}\right)=\dfrac{-1}{2}\)

Số hàng trong kho là: 1600 : \(\dfrac{2}{3}\) =  2400 (sản phẩm)

Số hàng xuất bán hôm nay là: 2400  x \(\dfrac{1}{4}\) = 600 (sản phẩm)

Kết luận:...

5/2 + 18/5 + 25/6 = 616/60

Tôi hơi sợ sai

Chu vi tam giác là:

\(\dfrac{5}{2}+\dfrac{18}{5}+\dfrac{25}{6}=\dfrac{75}{30}+\dfrac{108}{30}+\dfrac{125}{30}=\dfrac{308}{30}=\dfrac{154}{15}\left(cm\right)\)

Đáp số: \(\dfrac{154}{15}cm\)

\(\left(2-\dfrac{1}{2}\right)\cdot\left(\dfrac{-3}{4}+\dfrac{1}{2}\right)\)

\(=\left(\dfrac{4}{2}-\dfrac{1}{2}\right)\cdot\left(\dfrac{-3}{4}+\dfrac{2}{4}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{-1}{4}\)

\(=\dfrac{-3}{8}\)

M< 10 * 1/50 = 1/5
M>1/59*10 = 10/59 >10/60=1/6

=> 1/6<M<1/5

\(\dfrac{21}{4}\cdot\dfrac{-11}{4}+\dfrac{21}{4}\cdot\dfrac{1}{2}+\dfrac{21}{4}\cdot\dfrac{-1}{2}\)

\(=\dfrac{21}{4}\cdot\dfrac{-11}{4}\cdot\dfrac{21}{4}+\dfrac{2}{4}+\dfrac{21}{4}\cdot\dfrac{-2}{4}\)

\(=\dfrac{21}{4}\cdot\left(\dfrac{-11}{4}+\dfrac{2}{4}+\dfrac{-2}{4}\right)\)

\(=\dfrac{21}{4}\cdot\dfrac{-11}{4}\)

\(=\dfrac{-231}{16}\)

Câu 1:

$B=\frac{10}{1.3}+\frac{10}{3.5}+\frac{10}{5.7}+...+\frac{10}{101.103}$

$B=5(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{101.103})$

$=5(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{103-101}{101.103})$

$=5(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103})$

$=5(1-\frac{1}{103})=5.\frac{102}{103}=\frac{510}{103}$

Câu 2:

\(C=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+....+\frac{1}{2022.2024}\\ =\frac{1}{2}\left[\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+....+\frac{2}{2022.2024}\right]\)

\(=\frac{1}{2}\left[\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+....+\frac{2024-2022}{2022.2024}\right]\)

\(=\frac{1}{2}(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2022}-\frac{1}{2024})\\ =\frac{1}{2}(\frac{1}{2}-\frac{1}{2024})=\frac{1011}{4048}\)