x+(-1) + (-2) + (-3) +........+(-89)=1000
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(1000 - 13) x (1000 - 23) x (1000 - 33) x ... x (1000 - 103) x ... x (1000 - 503)
= (1000 - 13) x (1000 - 23) x (1000 - 33) x ... x (1000 - 1000) x ... x (1000 - 503)
= (1000 - 13) x (1000 - 23) x (1000 - 33) x ... x 0 x ... x (1000 - 503)
= 0
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}=\dfrac{x+4+96}{96}+\dfrac{x+5+95}{95}\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
=>x+100=0
hay x=-100
a,
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{x.\left(x+2\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\times\frac{x+1}{x+2}\)
\(=\frac{2x+2}{x+2}\)
Hơ hơ =v
Làm đại phần a đúng sai mặc kệ ~~
a,
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{x\left(x+2\right)}\)
\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{x\left(x+2\right)}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{x+2}\right)\)
\(=\frac{1}{2}\cdot\frac{x+1}{x+2}\)
\(=\frac{2x+2}{x+2}\)
b,
x = 1.2 + 2.3 + 3.4 + ....+ 89.90
3x = 1.2.3 + 2.3.3 + 3.4.3 + .... + 89.90
3x = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 89.90.(91 - 88)
3x = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 89.90.91 - 88.89.90
3x = 89.90.91
x = \(\frac{89\cdot90\cdot91}{3}=242970\)
\(Ta có : x - 1 - 2 - 3 - ... - 89 = 1000\)
\(\Rightarrow\)\(x - ( 1 + 2 + 3 + ... + 89 ) = 1000\)
\(\Rightarrow\)\(x - [ ( 89 - 1 ) ÷ 1 + 1 ] . [ ( 89 + 1 ) ÷ 2 ]\)\(= 1000\)
\(\Rightarrow\)\(x - 4005 = 1000 \)
\(\Rightarrow\)\(x = 5005\)
\(Vậy : x = 5005\)