Question

Tìm x : (x > 0)

x^2 + (2x)^2 + (3x)^2 + (4x)^2 + (5x)^2 = 220

Answer 1

\(x^2+2x^2+3x^2+4x^2+5x^2=220\)

\(\Leftrightarrow x^2+2^2.x^2+3^2.x^2+4^2.x^2+5^2.x^2=220\)

\(\Leftrightarrow x^2\left(1+2^2+3^2+4^2+5^2\right)=220\)

\(\Leftrightarrow x^2\left(1+4+9+16+25\right)=220\)

\(\Leftrightarrow x^2.55=220\)

\(\Leftrightarrow x^2=\frac{220}{55}\)

\(\Leftrightarrow x^2=4\)

\(\Leftrightarrow x=2\left(x>0\right)\)

Vậy \(x=2\)

Answer 2

x^2+4x^2+9x^2+16x^2+25x^2=220

55x^2=220

x^2=220:55

x^2=4

x=2

Vậy x có giá trị bằng 2