Người mua thùng sữa cần phải trả số tiền là:

\(250000\times\left(100\%-20\%\right)=200000\) (đồng)

Đáp số: \(200000\) đồng

(100%- 20%). 250000= 200000 (đồng)

Câu 2 :

\(a,\dfrac{3}{7}+\dfrac{-10}{7}=1\)  b) ko rõ

\(c,\dfrac{-4}{2}:\dfrac{4}{5}=\dfrac{-4}{2}\times\dfrac{5}{4}=\dfrac{-1}{2}\times5=\dfrac{-5}{2}\)

\(d,\dfrac{3}{9}-\dfrac{3}{5}=\dfrac{15}{45}-\dfrac{27}{45}=\left(-\dfrac{8}{45}\right)\)

Câu 3:

\(x-\dfrac{2}{3}=\dfrac{7}{3}\)        \(\dfrac{3}{5}+3x=1,25\)

\(x=\dfrac{7}{3}+\dfrac{2}{3}\)                \(3x=\dfrac{5}{4}-\dfrac{3}{5}\)

\(x=3\)                         \(3x=\dfrac{25}{20}-\dfrac{12}{20}\) 

                                    \(3x=\dfrac{13}{20}\)

                                      \(x=\dfrac{13}{20}:3\)

                                      \(x=\dfrac{13}{60}\)

Câu 3 : 

\(A=\dfrac{12}{9}+\dfrac{-2}{11}+\dfrac{2}{9}+\dfrac{-9}{11}+\dfrac{4}{9}\)

\(A=\left(\dfrac{12}{9}+\dfrac{2}{9}+\dfrac{4}{9}\right)+\left(\dfrac{-2}{11}+\dfrac{-9}{11}\right)\)

\(A=\dfrac{16}{9}+-1\)

\(A=\dfrac{7}{9}\)

\(B=\dfrac{5}{8}.\dfrac{2}{11}+\dfrac{5}{8}.\dfrac{9}{11}+\dfrac{1}{4}\)

\(B=\dfrac{5}{8}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{1}{4}\)

\(B=\dfrac{5}{8}+\dfrac{1}{4}\)

\(B=\dfrac{5}{8}+\dfrac{2}{8}\)

\(B=\dfrac{7}{8}\)

Đề không hiển thị đầy đủ. Bạn xem lại nhé. 

Câu 2 : D

Câu 3: C

Câu 4 : D

Câu 5 : C

Câu 6 : A

Câu 7 : A

Câu 8 : C

Phần II:

Bài 1 :

\(a,A=\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{3}{7}\)                      

\(A=\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{4}\)

\(A=1+\dfrac{3}{4}\)

\(A=\dfrac{7}{4}\)

\(b,B=\dfrac{4}{12}+\dfrac{-18}{45}+\dfrac{6}{9}+\dfrac{21}{35}+\dfrac{-6}{30}\)

\(B=\dfrac{1}{3}+\dfrac{-2}{5}+\dfrac{2}{3}+\dfrac{3}{5}+\dfrac{-1}{5}\)

\(B=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{5}+\dfrac{3}{5}+\dfrac{-1}{5}\right)\)

\(B=1+0\)

\(B=1\)

Bài 2 :

\(x-1\dfrac{2}{3}=0,2\)

\(x-\dfrac{5}{3}=0,2\)

\(x-\dfrac{5}{3}=\dfrac{2}{10}\)

\(x=\dfrac{2}{10}+\dfrac{5}{3}\)

\(x=\dfrac{6}{30}+\dfrac{50}{30}\)

\(x=\dfrac{56}{30}=\dfrac{28}{15}\)

\(\dfrac{7}{5}+\dfrac{5}{6}:x=\dfrac{5}{6}\)

\(\dfrac{5}{6}:x=\dfrac{5}{6}+\dfrac{7}{5}\)

\(\dfrac{5}{6}:x=\dfrac{25}{30}+\dfrac{42}{30}\)

\(\dfrac{5}{6}:x=\dfrac{67}{30}\)

\(x=\dfrac{5}{6}:\dfrac{67}{30}\)

\(x=\dfrac{5}{6}\times\dfrac{30}{67}\)

\(x=\dfrac{5}{1}\times\dfrac{5}{67}\)

\(x=\dfrac{25}{67}\)

Câu 2 : D

Câu 3: C

Câu 4 : D

Câu 5 : C

Câu 6 : A

Câu 7 : A

Câu 8 : C

Phần II:

Bài 1 :

\(a,A=\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{3}{7}\)                      

\(A=\left(\dfrac{4}{7}+\dfrac{3}{7}\right)+\dfrac{3}{4}\)

\(A=1+\dfrac{3}{4}\)

\(A=\dfrac{7}{4}\)

\(b,B=\dfrac{4}{12}+\dfrac{-18}{45}+\dfrac{6}{9}+\dfrac{21}{35}+\dfrac{-6}{30}\)

\(B=\dfrac{1}{3}+\dfrac{-2}{5}+\dfrac{2}{3}+\dfrac{3}{5}+\dfrac{-1}{5}\)

\(B=\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\left(\dfrac{-2}{5}+\dfrac{3}{5}+\dfrac{-1}{5}\right)\)

\(B=1+0\)

\(B=1\)

Bài 2 :

\(x-1\dfrac{2}{3}=0,2\)

\(x-\dfrac{5}{3}=0,2\)

\(x-\dfrac{5}{3}=\dfrac{2}{10}\)

\(x=\dfrac{2}{10}+\dfrac{5}{3}\)

\(x=\dfrac{6}{30}+\dfrac{50}{30}\)

\(x=\dfrac{56}{30}=\dfrac{28}{15}\)

\(\dfrac{7}{5}+\dfrac{5}{6}:x=\dfrac{5}{6}\)

\(\dfrac{5}{6}:x=\dfrac{5}{6}+\dfrac{7}{5}\)

\(\dfrac{5}{6}:x=\dfrac{25}{30}+\dfrac{42}{30}\)

\(\dfrac{5}{6}:x=\dfrac{67}{30}\)

\(x=\dfrac{5}{6}:\dfrac{67}{30}\)

\(x=\dfrac{5}{6}\times\dfrac{30}{67}\)

\(x=\dfrac{5}{1}\times\dfrac{5}{67}\)

\(x=\dfrac{25}{67}\)

Lời giải:

Gọi giá tiền mỗi quyển vở ban đầu là $a$ (đồng) 

Số tiền bạn An mang theo: $10a$ (đồng)

Giá tiền một quyển vở sau khi giảm giá: $10a:12=\frac{5}{6}a$ (đồng)

Cửa hàng giảm giá số phần trăm là:

$(a-\frac{5}{6}a):a.100=16,7$ (%)

dạ em cảm ơn thầy cô ạ:)))

1: \(\left(\dfrac{3}{4}\right)^2\cdot4^2-\left(\dfrac{1}{2}\right)^2:2-2\dfrac{3}{4}\)

\(=\left(\dfrac{3}{4}\cdot4\right)^2-\dfrac{1}{8}-\dfrac{11}{4}\)

\(=9-\dfrac{1}{8}-\dfrac{22}{8}=9-\dfrac{23}{8}=\dfrac{72-23}{8}=\dfrac{49}{8}\)

2: \(\left(\dfrac{3}{5}\right)^2\cdot5^2-\left(2\dfrac{1}{4}\right)^3:\left(\dfrac{3}{4}\right)^3-3\)

\(=\left(\dfrac{3}{5}\cdot5\right)^2-\left(\dfrac{9}{4}:\dfrac{3}{4}\right)^3-3\)

\(=3^2-3^3-3=9-27-3=9-30=-21\)

3: \(25\cdot\left(-\dfrac{1}{5}\right)^3+\dfrac{1}{5}-2\cdot\left(-\dfrac{1}{2}\right)^2-\dfrac{1}{2}\)

\(=25\cdot\dfrac{-1}{125}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)

\(=-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{2}-\dfrac{1}{2}=-1\)

4: \(4\cdot\left(\dfrac{1}{2}\right)^3+3\cdot\left(\dfrac{1}{2}\right)^2-2\cdot\left(\dfrac{1}{2}\right)^0\)

\(=4\cdot\dfrac{1}{8}+3\cdot\dfrac{1}{4}-2\)

\(=\dfrac{1}{2}+\dfrac{3}{4}-2=\dfrac{5}{4}-2=-\dfrac{3}{4}\)

5: \(\left(-\dfrac{1}{2}\right)^2\cdot4+\dfrac{1}{3}\cdot3^2+\left(\dfrac{1}{2020}\right)^0\)

\(=\dfrac{1}{4}\cdot4+\dfrac{1}{3}\cdot9+1\)

=1+3+1

=5

6: \(5\cdot\left(-\dfrac{2}{5}\right)^2+2\cdot\dfrac{-2}{5}+4\cdot\left(-\dfrac{2}{5}\right)^0\)

\(=5\cdot\dfrac{4}{25}-\dfrac{4}{5}+4\)

\(=\dfrac{4}{5}-\dfrac{4}{5}+4=4\)

a: \(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)

\(10B=\dfrac{10^{2024}+10}{10^{2024}+1}=1+\dfrac{9}{10^{2024}+1}\)

Ta có: \(10^{2023}+1< 10^{2024}+1\)

=>\(\dfrac{9}{10^{2023}+1}>\dfrac{9}{10^{2024}+1}\)

=>\(1+\dfrac{9}{10^{2023}+1}>1+\dfrac{9}{10^{2024}+1}\)

=>10A>10B

=>A>B

b: Số số hạng trong dãy số 5,0;5;2;...;9,8 là:

\(\left(9,8-5,0\right):0,2+1=4,8:0,2+1=25\left(số\right)\)

Tổng của dãy số là \(\left(9,8+5,0\right)\times25:2=14,8\times12,5=185\)

=>\(S=185\cdot0,1=18,5\)

Bạn chắn đây là câu trả lời đúng ko

\(B=\dfrac{1}{99}+\dfrac{2}{98}+...+\dfrac{99}{1}\)

\(=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)

\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)

\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)=100\cdot A\)

=>\(\dfrac{A}{B}=\dfrac{1}{100}\)

\(B=\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)

\(B+99=\dfrac{1}{99}+1+\dfrac{2}{98}+1+\dfrac{3}{97}+1+...+\dfrac{99}{1}+1\)

\(B=\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+1\)

\(B=100\times\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{100}\right)\)

Mà \(A=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{1}{100}\)

Chúc bạn thi tốt.