từ đề suy ra 7x-7+3x-6=-3

suy ra 10x-13+3=0

suy ra 10x-10=0

suy ra 10x=10

suy ra x=1

\(7\left(x-1\right)+3.\left(x-2\right)=-3\)

\(\Rightarrow7x-7+3x-6=-3\)

\(\Rightarrow\left(7x+3x\right)+\left(-7-6\right)=-3\)

\(\Rightarrow10x-13=-3\)

\(\Rightarrow10x=10\)

\(\Rightarrow x=1\)

Vậy x=1.

_Học tốt_

Lời giải:
Xét các TH để bỏ dấu trị tuyệt đối thôi bạn.

TH1: $3\geq x\geq -2$ thì: $|3-x|=3-x; |x+2|=x+2$
$\Rightarrow (x+2)-(3-x)=7-2(x+1)$

$\Rightarrow 2x-1=7-2(x+1)$

$\Rightarrow x=1,5$ (thỏa mãn)

TH2: $x>3$ thì $|3-x|=x-3; |x+2|=x+2$

$\Rightarrow (x+2)-(x-3)=7-2(x+1)$

$\Rightarrow x=0$ (không thỏa mãn)

TH3: $x<-2$ thì $|x+2|=-(x+2); |3-x|=3-x$

$\Rightarrow -(x+2)-(3-x)=7-2(x+1)$

$\Rightarrow x=5$ (vô lý do $x<-2$)

Vậy.......

Gọi số số hạng của dãy trên là n (n thuộc N*)

Theo công thức, ta có:

[11 + (x - 3)]n : 2 = 0

=> [11 + (x - 3)] = 0

=> 11 + (x - 3) = 0 (Vì n khác 0)

=> x - 3 = -11

=> x = -8

Vậy...

x=8,sai thì cho tui xin lỗi

\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)

\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)

\(\left(2x-1\right)=-\frac{4}{63}\)

2x= -4/63 + 1

2x = 59/63

x = 59/63 : 2

x = 59/126

1/3:(2.x-1)=-5-1/4

1/3:(2.x-1)=-21/4

2.x-1=1/3:-21/4

2.x-1=-4/63

2.x=-4/63+1

2.x=\(3\frac{59}{63}\)

x=\(3\frac{59}{63}\):2

x=\(1\frac{61}{63}\)

\(\frac{1}{2\cdot x}-2021-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{222}=\frac{6}{11}\)

\(\frac{1}{2\cdot x}-2021-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\right)=\frac{6}{11}\)

....

Cái dãy \(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{222}\) nó không có quy luật, không tính được

Sửa đề\(\frac{1}{2x-2021}-\frac{1}{4}-\frac{1}{12}-\frac{1}{24}-...-\frac{1}{220}=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{220}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{10}-\frac{1}{11}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}\left(1-\frac{1}{11}\right)=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{1}{2}.\frac{10}{11}=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}-\frac{5}{11}=\frac{6}{11}\)

=> \(\frac{1}{2x-2021}=1\)

=> 2x - 2021 = 1

=> 2x = 2022

=> x = 1011

Vậy x = 1011

A = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+..+9\right)}{1\times2+2\times3+3\times4+...+19\times20}\)

 \(=\frac{\frac{1\times\left(1+1\right)}{2}+\frac{2\times\left(2+1\right)}{2}+\frac{3\times\left(3+1\right)}{2}...+\frac{9\times\left(9+1\right)}{2}}{1\times2+2\times3+3\times4+...+19\times20}\)

\(=\frac{\frac{1\times2}{2}+\frac{2\times3}{2}+\frac{3\times4}{2}+...+\frac{9\times10}{2}}{1\times2+2\times3+3\times4+...+9\times10}\)

\(=\frac{\frac{1}{2}\times\left(1\times2+2\times3+3\times4+...+9\times10\right)}{1\times2+2\times3+3\times4+...+9\times10}=\frac{\frac{1}{2}}{1}=\frac{1}{2}\)