\(A=\frac{6}{3.5}+\frac{9}{5.8}+\frac{12}{8.12}+\frac{15}{12.17}\)

\(A=3.\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{17}\right)< 3.\frac{1}{3}=1\)

=> A < 1

đề sai thì phải

\(A=\frac{10}{2\cdot12}+\frac{2}{3\cdot5}+\frac{3}{5\cdot8}+\frac{1}{2\cdot3}+\frac{5}{12\cdot17}+\frac{6}{17\cdot23}+\frac{7}{23\cdot30}\)

\(A=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{2}-\frac{1}{3}+\frac{1}{12}-\frac{1}{17}+\frac{1}{17}-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)

\(A=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}\)

\(A=\frac{101}{120}\)

\(A=\frac{1}{2.12}+\frac{2}{3.5}+\frac{3}{5.8}+...+\frac{7}{23.30}\)

\(=\frac{1}{2}-\frac{1}{12}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...-\frac{1}{23}+\frac{1}{23}-\frac{1}{30}\)

\(=\frac{1}{2}+\frac{1}{2}-\frac{1}{8}-\frac{1}{30}=1-\frac{19}{120}=\frac{101}{120}\)

Ta có :

\(A=\frac{6}{3.5}+\frac{9}{5.8}+\frac{12}{8.12}+\frac{15}{12.17}\)

\(A=3.\left(\frac{2}{3.5}\right)+3.\left(\frac{3}{5.8}\right)+3.\left(\frac{4}{8.12}\right)+3.\left(\frac{5}{12.17}\right)\)

\(A=3.\left(\frac{2}{3.5}+\frac{3}{5.8}+\frac{4}{8.12}+\frac{5}{12.17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{17}\right)\)

\(A=3.\left(\frac{1}{3}-\frac{1}{17}\right)\)

\(A=3.\frac{14}{51}\)

\(A=\frac{14}{17}< 1\)

Vậy A < 1

_Chúc bạn học tốt_

1/7+1/91+1/247+1/475+1/775+1/1147=? (1)
ta có: (1) <=>: 1/(1.7)+1/(7.13)+1/(13.19)+1/(19.25)+1/(25.31)+1/(31.37)
=1/6.(1-1/7+1/7-1/13+1/13-1/19+1/19-1/25+1/25-1/31+1/31-1/37)
=1/6.(1-1/37)=6/37

a: \(A=x^2-10x+25+1\)

\(=\left(x-5\right)^2+1\)

\(=100^2+1=10001\)

b: \(B=2\left(a^2+a-5a-5\right)-\left(a^2-10a+25\right)+36\)

\(=2a^2-8a-10-a^2+10a-25+36\)

\(=a^2+2a+1\)

\(=\left(a+1\right)^2=100^2=10000\)

c: \(C=a^3+3a^2+3a+1=\left(a+1\right)^3=100^3=1000000\)

d: \(E=a^3+3a^2+3a+1+5\)

\(=\left(a+1\right)^3+5\)

\(=30^3+5=27005\)

Bài 2:

a) 11. x = -55

x = -55 : 11

x = -5

Vậy x = -5

b) -3 . x = -12

x = -12 : (-3)

x = 4

Vậy x = 4

c) 3x - 12 = -48

3x = -48 + 12

3x = -36

x = -36 : 3

x = -12

Vậy x = -12

d) 10 + 23 . (2x - 10) = -36

23 . (2x - 10) = -36 - 10

23 . (2x - 10) = -46

2x - 10 = -46 : 23

2x - 10 = -2

2x = -2 + 10

2x = 8

x = 8: 2

x = 4

Vậy x = 4

f) | 2x - 1| + 3 = 8

|2x - 1| = 8 - 3

|2x - 1| = 5

\(\Rightarrow\left[\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy x = 3 hoặc x = -2

g) |2x² - 3| - 4 = 11

|2x²- 3| = 11 + 4

|2x² - 3| = 15

\(\Rightarrow\left[\begin{matrix}2x^2-3=15\\2x^2-3=-15\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x^2=18\\2x^2=-12\end{matrix}\right.\Rightarrow\left[\begin{matrix}x^2=9\\x^2=-6\end{matrix}\right.\Rightarrow\left[\begin{matrix}x^2=3^2\\\left(loại\right)\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=3\\\end{matrix}\right.\)

Vậy x = 3.

Bài 1:

\((-48) . 72 +36 . (-304) \)

\(=\left[\left(-48\right).\left(36.2\right)\right]+36.\left(-340\right)\)

\(=\left[\left(-48\right).2\right].36+36.\left(-340\right)\)

\(=\left(-96\right).36+36.\left(-304\right)\)

\(=36.\left[\left(-96\right)+\left(-304\right)\right]\)

\(=36.\left(-400\right)\)

\(=-14400\)

a. 37-7.(x+1)=40-8.3

37 - 7(x+1) = 16 

7(x+1) = 21

x+1 = 3

x=2

b. (x+5).3=11=10+(3+4).4

(sao có 2 dấu bằng, dấu nào là dấu cộng do viết nhầm ???)

c. 25-5.3+4.10=100-5.x

50 = 100 - 5x 

5x = 50

x=10

d. 36+2.(x-7)=12+8.(3+5)-36

36 + 2(x-7) = 40

2(x-7) = 4

x-7 =2

x=9

e. 3.(x+7)-9=11.5-5-8

3(x+7) - 9 = 42 

3(x+7) = 51

x+7 = 17 

x=10