ax+b^2=a^2+bx,a\ne b

解答

a x + b^{2} = a^{2} + b x, a \neq = b

x = a + b; a \neq = b

求解步骤

a x + b^{2} = a^{2} + b x

将

b^{2}

到右边

a x = a^{2} + b x - b^{2}

将

b x

para o lado esquerdo

a x - b x = a^{2} - b^{2}

分解

a x - b x : x (a - b)

x (a - b) = a^{2} - b^{2}

两边除以

a - b; a \neq = b

x = a + b; a \neq = b

so l v e f or x, ∣ z ∣ i + x + y = 2 i

so l v e f or x, - 5 (x + 2 y) - \frac{3}{2} + x = y

3.9 x - 0.7 = 8.66

so l v e f or x, 2 x + 4 y = - 1

3 (x + 7) - 2 (x + 4) = 8 (x + 10) - 7 (- x - 3)