3 nhân 11 nhân 9 nhân 25

\(\frac{1234.1235+1236.66-66}{1235.1236-1235.1106}\)

\(=\frac{1234.1235+\left(1236-1\right).66}{1235.\left(1236-1106\right)}\)

\(=\frac{1234.1235+1235.66}{1235.130}\)

\(=\frac{1235.\left(1234+66\right)}{1235.130}\)

\(=\frac{1300}{130}=10\)

a) Ta có: \(\dfrac{15}{7}>1\) (tử lớn hơn mẫu)

\(\dfrac{9}{14}< 1\) (tử nhỏ hơn mẫu)

Vậy: \(\dfrac{15}{7}>\dfrac{9}{14}\)

b) Ta có: 

\(\dfrac{899}{900}=1-\dfrac{1}{900}\)

\(\dfrac{1235}{1236}=1-\dfrac{1}{1236}\)

Mà: \(\dfrac{1}{900}>\dfrac{1}{1236}\)

Vậy: \(\dfrac{1235}{1236}>\dfrac{899}{900}\)

c) Ta có:

\(\dfrac{77}{75}=1+\dfrac{2}{75}\)

\(\dfrac{37}{35}=1+\dfrac{2}{35}\)

Mà: \(\dfrac{2}{75}< \dfrac{2}{35}\)

Vậy: \(\dfrac{37}{35}>\dfrac{77}{75}\)

\(\left\{{}\begin{matrix}\dfrac{15}{7}=\dfrac{30}{14}\\\dfrac{9}{14}< \dfrac{30}{14}\end{matrix}\right.\Rightarrow\dfrac{15}{7}>\dfrac{9}{14}\)

\(\left\{{}\begin{matrix}\dfrac{899}{900}=\dfrac{899.1236}{900.1236}=\dfrac{\text{1111164}}{900.1236}\\\dfrac{1235}{1236}=\dfrac{1235.900}{900.1236}=\dfrac{\text{1111500}}{900.1236}>\dfrac{\text{1111164}}{900.1236}\end{matrix}\right.\Rightarrow\dfrac{1235}{1236}>\dfrac{899}{900}\)

\(\left\{{}\begin{matrix}\dfrac{77}{75}=\dfrac{539}{525}\\\dfrac{37}{35}=\dfrac{555}{525}>\dfrac{539}{525}\end{matrix}\right.\Rightarrow\dfrac{77}{73}< \dfrac{37}{35}\)

a) 5182

b) 7064

c) 1814

a,123*3+1234*4-123

=369+4936-123

=5182

b,1457*4+1236*2-1236

=5828+2472-1236

=7064

c,214*4+456*2+46

=856+912+46

=1814

tick nha !!!

a. quy luật: tổng 2 số trừ cho một nửa số thứ 2.

b. (?)=1235+1238=1854

dễ mà phải ko?

a) Quy luật là: Lấy tổng cộng số hạng thứ 2:2

b) (?)= 3092

86+56+5641+546585+56485+58648+6548+674874+456476+74674+9874+748749674+7869774+8968 = 758518363

758 518 363.

1235.(108-8)

1235.100

123500

Đúng thì tích nha