Giải:

a)  \(\dfrac{7}{x}< \dfrac{x}{4}< \dfrac{10}{x}\) 

\(\Rightarrow7< \dfrac{x^2}{4}< 10\) 

\(\Rightarrow\dfrac{28}{4}< \dfrac{x^2}{4}< \dfrac{40}{4}\) 

\(\Rightarrow x^2=36\) 

\(\Rightarrow x=6\) 

b) \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}\) 

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}< \dfrac{1}{3.4}\) 

\(...\) 

\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\) 

\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\) 

\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\) 

\(\Rightarrow A< \dfrac{1}{1}-\dfrac{1}{9}\) 

\(\Rightarrow A< \dfrac{8}{9}\left(1\right)\) 

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\) 

 \(...\) 

\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\) 

\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\) 

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\) 

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\) 

\(\Rightarrow A>\dfrac{2}{5}\left(2\right)\) 

Từ (1) và (2), ta có:

\(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\left(đpcm\right)\)

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

...

\(\frac{1}{8^2}=\frac{1}{8\cdot8}< \frac{1}{7\cdot8}\)

Cộng vế theo vế 

\(\Rightarrow B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{7\cdot8}\)

\(\Rightarrow B< \frac{1}{1}-\frac{1}{8}=\frac{7}{8}\)

Lại có \(\frac{7}{8}< 1\)

Theo tính chất bắc cầu => \(B< \frac{7}{8}< 1\)

\(\Rightarrow B< 1\left(đpcm\right)\)

A=1+(2-3-3+5)+(6-7-8+9)+....+(98-99-100+101)+102

=1+0+0+....+102=103

b) |1-2x|>7

=> 1-2x>7 hoặc 1-2x<-7

=> 2x<-6 hoặc 2x>8

=> x<-3 hoặc x>4

\(\frac{A}{B}=\frac{\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+\frac{6}{4}+\frac{5}{5}+\frac{4}{6}+\frac{3}{7}+\frac{2}{8}+\frac{2}{9}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{\frac{10}{2}+\frac{10}{3}+\frac{10}{4}+...+\frac{10}{9}+\frac{10}{10}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)

\(\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}\)

\(\frac{A}{B}=10\)

\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{2}{8}+\frac{1}{9}\)

Tách 9=1+1+...+1 ( có 9 số 1)

\(\Rightarrow A=1+\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{2}{8}+1\right)+\left(\frac{1}{9}+1\right)\)

\(A=\frac{10}{10}+\frac{10}{2}+\frac{10}{3}+...+\frac{10}{8}+\frac{10}{9}\)

\(A=10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

\(\Rightarrow A:B=\frac{10.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}}=10\) ( vì \(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\ne0\) )

Vậy \(A:B=10\)

trong câu hỏi tương tự

Bạn ghi sai đề chỗ 3/11 là sai mà phải 2/11 với là chỗ 2/7 là sai mà là 2/9

\(A=\frac{\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}-\frac{7}{11}}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)

\(=\frac{2}{7}:\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{10}\right)}=\frac{2}{7}:\frac{2}{7}=1\)

b,\(A.x+\frac{5}{6}=-\frac{3}{4}\)

<=>\(1.x=-\frac{3}{5}-\frac{5}{6}\)

<=>x=-43/30

Ai thấy mình làm đúng thì tích nha.Ai tích mình mình tích lại

a,A=1863/623:2/7

A=1863/178

b,

ta có:

1863/178.x+5/6=-3/4

1863/178.x=-19/12

=>x=-1691/11178

1. A = 75(4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 25

    A = 25 . [3 . (4²⁰⁰⁴ + 4²⁰⁰³ +...+ 4² + 4 + 1) + 1]

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 3 + 1)

    A = 25 . (3 . 4²⁰⁰⁴ + 3 . 4²⁰⁰³ +...+ 3 . 4² + 3 . 4 + 4)

    A = 25 . 4 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1)

    A =100 . (3 . 4²⁰⁰³ + 3 . 4²⁰⁰² +...+ 3 . 4 + 3 + 1) \(⋮\) 100

3a) |x| = 1/2 

=> x = 1/2 hoặc x = -1/2

với x = 1/2:

A = \(3.\left(\frac{1}{2}\right)^2-2.\frac{1}{2}+1\)

\(A=\frac{3}{4}-1+1=\frac{3}{4}\)

với x = -1/2

A = \(3.\left(-\frac{1}{2}\right)^2-2\left(-\frac{1}{2}\right)+1\)

\(A=\frac{3}{4}+1+1=\frac{3}{4}+2=\frac{11}{4}\)

2.

A=\(\frac{\left(1+2+3+.....+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+......+99-100}\)\

A=0