\(\frac{2}{1^2}.\frac{6}{2^2}.\frac{10}{3^2}.\frac{20}{4^2}.......\frac{110}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)

\(\Rightarrow\frac{1.2}{1.1}.\frac{2.3}{2.2}.\frac{3.4}{3.3}.\frac{4.5}{4.4}......\frac{10.11}{10.10}\left(x-2\right)=-20x-20+60\)

\(\Rightarrow\frac{1.2.3.4.....10}{1.2.3.4.....10}.\frac{2.3.4.5.....11}{1.2.3.4.....10}\left(x-2\right)=-20x+40\)

\(\Rightarrow11\left(x-2\right)=-20x+40\)

\(\Rightarrow11x-22=-20x+40\)

\(\Rightarrow11x+20x=22+40\)

\(\Rightarrow31x=62\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

\(\dfrac{1.2}{1^2}.\dfrac{2.3}{2^2}.\dfrac{3.4}{3^2}...\dfrac{9.10}{9^2}.\dfrac{10.11}{10^2}\left(x-2\right)=-20\left(x+1\right)+60\)

\(\Leftrightarrow\dfrac{1.2^2.3^2.4^2...10^2.11}{1^2.2^2.3^2....10^2}\left(x-2\right)+20\left(x+1\right)=60\)

\(\Leftrightarrow11\left(x-2\right)+20\left(x+1\right)=60\)

\(\Leftrightarrow31x=62\)

\(\Rightarrow x=2\)

\(\Rightarrow10-\left(x-3\right):2=49-1=48\\ \Rightarrow\left(x-3\right):2=-38\\ \Rightarrow x-3=-76\\ \Rightarrow x=-73\)

\(10-\left(x-3\right):2=7^2-1^{10}\)

\(\Rightarrow10-\left(x-3\right):2=48\)

\(\Rightarrow\left(x-3\right):2=-38\)

\(\Rightarrow x-3=-76\Rightarrow x=-73\)

\(10-\left(x-3\right):2=7^2-1^{10}\\ \Rightarrow\left(x-3\right):2=10-49-1\\ \Rightarrow x-3=-38.2\\ \Rightarrow x=-76+3\\ \Rightarrow x=-73\)

\(\dfrac{1.2}{1.1}.\dfrac{2.3}{2.2}.\dfrac{3.4}{3.3}.\dfrac{4.5}{4.4}...\dfrac{10.11}{10.10}\left(x-2\right)=-20x+40\)

\(\Leftrightarrow\dfrac{2.3.4...11}{1.2.3...10}\left(x-2\right)=-20x+40\)

\(\Leftrightarrow11\left(x-2\right)=-20x+40\)

\(\Leftrightarrow11x-22=-20x+40\)

\(\Leftrightarrow31x=62\)

\(\Rightarrow x=2\)

\(=>\dfrac{2\cdot1}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot......\cdot\dfrac{10\cdot11}{10\cdot10}\cdot\left(x-2\right)=-20\left(x+1\right)+60\)=>11*(x-2)=-20*(x+1)+60

=>11x-22=-20x-20+60

=>31x=62

=>x=2

haha,có vẻ cái"người bạn đang nhắc đến"đã quen với việc nói bậy rồi ^^

10+20+30+40+50+60...+110

mình ko đáng cái j linh tinh hết đây là các bài toán mà mình ko giải đc

b. (x-7)^x+1-(x-7)^x+11=0

(x-7)^x+1.[1-(x-7)¹⁰]=0

=> (x-7)^x+1=0 hoặc 1-(x-7)¹⁰=0

• (x-7)^x+1= 0 => x-7=0 => x=7

• 1-(x-7)¹⁰=0=> (x-7)¹⁰=1=>x-7=1 hoặc x-7=-1 => x=8 hoặc x=6

Vậy x thuộc {6;7;8}

lớp mấy đấy

\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+10\right)=110\)

\(\left(x+x+x+...+x\right)+\left(1+2+3+...+10\right)=110\)

\(\left(x+x+x+...+x\right)+55=110\)

\(11x=110-55\)

\(11x=55\)

\(x=55:11\)

\(x=5\)

(x-3)+(x-2)+(x-1)+10+11=110

(x-3)+(x-2)+(x-1)=89

x-3+x-2+x-1=89

3x-6=89

3x=95

x=95/3

Vậy x=95/3

=>x-3+x-2+x-1=110-21=89

=>3x-6=89

=>3x=95

=>x=95/3