Муодиларо ҳал кунед: \((x-5)^2+(x-4)^3+(x-3)^4=2\)

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

\((x-5)^2+(x-4)^3+(x-3)^4-2=0\)

\((x-5)^2+(x-4)^3+(x-3)^4-1-1=0\)

\((x-5)^2+(x-4)^3+(x-3)^4-1^2-1^4=0\)

\(((x-5)^2-1^2)+(x-4)^3+((x-3)^4-1^4)=0\)

\((x-4)(x-6)+(x-4)(x-4)^2+((x-3)^2-1)((x-3)^2+1)=0\)

\((x-4)(x-6)+(x-4)(x^2-8x+16)+(x^2-6x+8)(x^2-6x+10)=0\)

\(x^2-6x+8 = x\cdot x-2x-4x+(-2)\cdot(-4)=\)

\(=(x-4)(x-2)\)

\((x-4)(x-6+x^2-8x+16)+(x-4)(x-2)(x^2-6x+10)=0\)

\((x-4)(x^2-7x+10)+(x-4)(x-2)(x^2-6x+10)=0\)

\(x^2-7x+10=x\cdot x-2x-5x+(-2)\cdot(-5)=\)

\(=(x-2)(x-5)\)

\((x-4)(x-2)(x-5)+(x-4)(x-2)(x^2-6x+10)=0\)

\((x-4)(x-2)(x-5+x^2-6x+10)=0\)

\((x-4)(x-2)(x^2-5x+5)=0\)

\(x-4=0\), аз инҷо

мебарояд, ки

\(x_1=4\).

\(x-2=0\), аз инҷо

мебарояд, ки

\(x_2=2\).

\(x^2-5x+5=0\)

\(D=25-20=5\)

\(x_3,x_4=\frac{5\pm\sqrt{5}}{2}\)

Ҷавоб: \(x_1=4\); \(x_2=2\); \(x_3,x_4=\frac{5\pm\sqrt{5}}{2}\).