←Vrati se na zadatke4. ZadatakAko je x1:7=x2:3=x3:2=x4:5x_1:7=x_2:3=x_3:2=x_4:5x1:7=x2:3=x3:2=x4:5, pokaži da vrijedi jednakost x12+x22+x32+x42=(7x1+3x2+2x3+5x4)287.x_1^2+x_2^2+x_3^2+x_4^2=\frac{(7x_1+3x_2+2x_3+5x_4)^2}{87}.x12+x22+x32+x42=87(7x1+3x2+2x3+5x4)2.RješenjeNeka je k=x1:7=x2:3=x3:2=x4:5k=x_1:7=x_2:3=x_3:2=x_4:5k=x1:7=x2:3=x3:2=x4:5. Tada je x1=7kx_1=7kx1=7k, x2=3kx_2=3kx2=3k, x3=2kx_3=2kx3=2k, x4=5kx_4=5kx4=5k. Vrijedi x12+x22+x32+x42=49k2+9k2+4k2+25k2=87k2.x_1^2+x_2^2+x_3^2+x_4^2=49k^2+9k^2+4k^2+25k^2=87k^2.x12+x22+x32+x42=49k2+9k2+4k2+25k2=87k2. Dalje je (7x1+3x2+2x3+5x4)287=(49k+9k+4k+25k)287=(87k)287=872k287=87k2.\frac{(7x_1+3x_2+2x_3+5x_4)^2}{87}=\frac{(49k+9k+4k+25k)^2}{87}=\frac{(87k)^2}{87}=\frac{87^2k^2}{87}=87k^2.87(7x1+3x2+2x3+5x4)2=87(49k+9k+4k+25k)2=87(87k)2=87872k2=87k2. Time je tvrdnja dokazana.
