= 1/1x2 + 1/2x3 + 1/3x4 ...... +1/9x10

= 1-1/2+1/2-1/3+1/3-1/4+........+1/9-1/10

=1-1/10=9/10

đặt A=1/1 x 1/2 + 1/2 x 1/3 + 1/3 + 1/4 + .......... + 1/9 x 1/10

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

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

đặt B=2/1 x 2 + 2/2 x 3 + 2/3 x4 + .............. + 2/98 x 99 + 2/99 x 100

\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\left(1-\frac{1}{100}\right)\)

\(=2\times\frac{99}{100}\)

\(=\frac{99}{50}\)

câu 1 

=> x+1/2+x+1/3+x+1/4-x-1/5-x-1/6=0

=> (x+x+x-x-x)+(1/2+1/3+1/4-1/5-1/6)=0

=> x+43/60=0

=> x = -43/60

câu dưới làm tương tự bạn nhé!

\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}........\frac{2010}{2009}=\frac{3.4.5...2010}{2.3.4....2009}=\frac{2010}{2}=1005\)

\(B=\frac{1.2.3......99}{1.2.3.4.....100}=\frac{1}{100}\)

Ta có\(M=\left[\left(1+\frac{1}{98}\right)+\left(\frac{1}{2}+\frac{1}{97}\right)+...+\left(\frac{1}{49}+\frac{1}{50}\right)\right].2.3...98\)

\(=\left[\frac{99}{1.98}+\frac{99}{2.97}+...+\frac{99}{49.50}\right].2.3...98=99\left(\frac{1}{1.98}+\frac{1}{2.97}+...+\frac{1}{49.50}\right).2.3...98\)

\(=99\left(\frac{k_1+k_2+...+k_{49}}{1.2.3...98}\right).2.3...98\left(k_1,k_2...k_{49}\varepsilonℕ^∗\right)=99\left(k_1+k_2+...+k_{49}\right)⋮99\Rightarrow M⋮99\left(đpcm\right)\)

a: Ta có: \(\left(\dfrac{3}{2}x-\dfrac{1}{5}\right)^2\cdot\left(x^2+\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow x\cdot\dfrac{3}{2}=\dfrac{1}{5}\)

hay \(x=\dfrac{1}{5}:\dfrac{3}{2}=\dfrac{2}{15}\)

b: Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\)

\(\Leftrightarrow x+100=0\)

hay x=-100

Mn nhớ giải chi tiết ra nha

a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )

Vậy x = 1

b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)

=> x + 100 = 0

=> x           = -100

c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)

=> x - 100 = 0

=> x           = 100

Chúc bạn học tốt

có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai