\(A=\frac{2}{1+2}+\frac{2+3}{1+2+3}+...+\frac{2+3+...+20}{1+2+3+...+20}\)

\(A=\frac{2}{3}+\frac{5}{6}+...+\frac{209}{210}\)

\(A=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{6}\right)+...+\left(1-\frac{1}{210}\right)\)

\(A=\left(1+1+....+1\right)\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{210}\right)\)

\(A=19-\left(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{420}\right)\)

\(A=19-\left(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\right)\)

\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\right)\)

\(A=19-2\cdot\left(\frac{1}{2}-\frac{1}{21}\right)\)

\(A=19-2\cdot\frac{19}{42}=19-\frac{19}{21}=\frac{380}{21}\)

Vậy A= \(\frac{380}{21}\)

\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2005}\right)\left(1-\frac{1}{2006}\right)\)

\(B=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2004}{2005}\cdot\frac{2005}{2006}\)

\(B=\frac{1\cdot2\cdot...\cdot2004\cdot2005}{2\cdot3\cdot...\cdot2005\cdot2006}\)

\(B=\frac{1}{2006}\)

Vậy \(B=\frac{1}{2006}\)

 dễ ơt mik ko bik thì mới hỏi nên mik ko bik hhihi! ^^ ^^

ai giúp mik đi mà !

a) \(\left(x+5\right)\left(2x-1\right)=\left(2x-3\right)\left(x+1\right)\)

\(\Leftrightarrow x\left(2x-1\right)+5\left(2x-1\right)=x\left(2x+3\right)+\left(2x+3\right)\)

\(\Leftrightarrow2x^2-x+10x-5=2x^2+3x+2x+3\)

\(\Leftrightarrow9x-5=5x+3\)

\(\Leftrightarrow9x-5x=3+5\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=8\div4\)

\(\Leftrightarrow x=2\)

Vậy nghiệm duy nhất của phương trình là 2.

b) \(\left(x+1\right)\left(x+9\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow x\left(x+1\right)+9\left(x+1\right)=x\left(x+3\right)+5\left(x+3\right)\)

\(\Leftrightarrow x^2+x+9x+9=x^2+3x+5x+15\)

\(\Leftrightarrow10x+9=8x+15\)

\(\Leftrightarrow10x-8x=15-9\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=6\div3\)

\(\Leftrightarrow x=2\)

Vậy nghiệm duy nhất của phương trình là 2

\(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x.\left(1+5^2\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

Vậy x=2

\(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x.1+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+5^2\right)=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=650\div26\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

Vậy x = 2

3,01;6,02;9,03

Học tốt

bạn giai ra đi

\(N=1115;2105\)

giải ra bạn

\(\left(3x-1\right)^4=81\)

\(\left(3x-1\right)^4=3^4\)

\(\Rightarrow3x-1=3\)

\(3x=3+1\)

\(3x=4\)

\(x=\frac{4}{3}\)

Vậy ....

(3x - 1)⁴ = 81

<=> 3x - 1 = 3⁴ 

<=> 3x = 3 + 1

<=> 3x = 4

<=> x = \(\frac{4}{3}\)

Bình phương 2 vế ta có:

\(a+1+2\sqrt{a}>a+1\)

\(\Leftrightarrow2\sqrt{a}>0\left(true\right)\)

\(\Rightarrow Q.E.D\)

Bình phương 2 vế ta có :

\(a-1-2\sqrt{a}>a-1\)

\(\Leftrightarrow2\sqrt{a}>0\)(đúng với \(\forall\)\(a\))

Vậy \(\sqrt{a}+1>\sqrt{a+1}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}\)

\(\frac{A}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)

\(\frac{A}{2}=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

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

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

\(\frac{A}{2}=\frac{3}{8}\)

\(A=\frac{3}{4}\)