Question

Tìm x\(\in\)Z, biết:

a)x+(x-1)+(x-2)+......+(x-100)=110x+8

b)3x-7x+11x-15x+.....+155x-159x=1680

Answer 1

\(a.x+\left(x-1\right)+\left(x-2\right)+...+\left(x-100\right)=110x+8\)

\(\Leftrightarrow101x-\left(1+2+...+100\right)=110x+8\)

\(\Leftrightarrow101x-\frac{100\left(100+1\right)}{2}=110x+8\)

\(\Leftrightarrow101x-5050=110x+8\)

\(\Leftrightarrow101x-110x=8+5050\)

\(\Leftrightarrow-9x=5058\)

\(\Leftrightarrow x=5058\div\left(-9\right)\)

\(\Leftrightarrow x=-562\)

Answer 2

\(b.3x-7x+11x-15x+...+155x-159x=1680\)

\(\Leftrightarrow\left(3x-7x\right)+\left(11x-15x\right)+...+\left(155x-159x\right)=1680\)

\(\Leftrightarrow-4x-4x-...-4x=1680\)

\(\Leftrightarrow-80x=1680\)

\(\Leftrightarrow x=1680\div\left(-80\right)\)

\(\Leftrightarrow x=-21\)