b)

\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)

\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)

\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)

\(x-2=8\)

=> x = 10

a) 

\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)

\(A=\frac{1}{2016}\)

A = ( 1 - 1/2) . ( 1 - 1/3 ) . (1-1/4) ....(1-1/2015) . (1-1/2016)

A= 1/2 . 2/3 . 3/4...2014/2015 . 2015/2016

A = 1 . 2 . 3 . 4 ... 2014 . 2015/ 2 . 3 . 4 ... 2015 . 2016

A = 1/ 2016

Câu hỏi của Tăng Thiện Đạt - Toán lớp 8 - Học toán với OnlineMath

a. Ta có:

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c+a-b\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(c-a\right)\left(c-b\right)\)

và \(ab^2-ac^2-b^3+bc^2=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)=\left(a-b\right)\left(b-c\right)\left(b+c\right)\)

Vậy, \(A=\frac{\left(a-b\right)\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(b-c\right)\left(b+c\right)}=\frac{c-a}{-c-b}=\frac{a-c}{c+b}\)

\(A=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot...\cdot\left(\frac{1}{2014^2}-1\right)\)

\(A=\frac{-3}{2^2}\cdot\frac{-8}{3^2}\cdot\frac{-15}{4^2}\cdot...\cdot\frac{-2014^2+1}{2014^2}\)

\(A=\frac{1\cdot\left(-3\right)}{2^2}\cdot\frac{2\cdot\left(-4\right)}{3^2}\cdot\frac{3\cdot\left(-5\right)}{4^2}\cdot...\cdot\frac{2013\cdot\left(-2015\right)}{2014^2}\)

\(A=\frac{1\cdot2\cdot3\cdot...\cdot2013}{2\cdot3\cdot4\cdot...\cdot2014}\cdot\frac{\left(-3\right)\cdot\left(-4\right)\cdot\left(-5\right)\cdot...\cdot\left(-2015\right)}{2\cdot3\cdot4\cdot...\cdot2014}\)

\(A=\frac{1}{2014}\cdot\frac{-2015}{2}\)

\(A=\frac{-2015}{4028}\)

a) 

( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0

\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)

P/s: đợi xíu làm câu b

b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{-1}{x+3}=\frac{1}{2015}\)

\(\Leftrightarrow x+3=-2015\)

\(\Leftrightarrow x=-2018\)

Vậy,.........

A/ Ta có số nào nhân với 0 cx = 0 

Vậy từ đó suy ra 2 trường hợp 

TH1\(4x-9=0\)

\(=>x=\frac{9}{4}\)

TH2 \(2,5+-\frac{7}{3}x=0\)

 \(=>x=\frac{15}{14}\)