a, (x+2)2+(x-3)2=2x(x+7)
x.2+2.2+x.2+(-3).2-2x=8
2x+4+2x-6-2x=8
(2x+2x)+(4-6)=8
4x-2=8
4x=8+2
4x=10
X=10:4
X=5/2
a) (x+2)2+(x-3)2 = 2x(x+7)
⇒x2+4x+4+x2-6x+9=2x2+14x
⇒x2+4x+4+x2-6x+9-2x2-14x=0
⇒ -16x+13=0
⇒ x=\(\dfrac{13}{16}\)
b) (x+3)(x2-3x+9) = x(x2+4)-1
⇒x(x2-3x+9)+3(x2-3x+9)=x3+4x-1
⇒x3-3x2+9x+3x2-9x+27-x3-4x+1=0
⇒-4x+28=0
⇒x=7
a: Ta có: \(\left(x+2\right)^2+\left(x-3\right)^2=2x\left(x+7\right)\)
\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(\Leftrightarrow-16x=-13\)
hay \(x=\dfrac{13}{16}\)
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)=x\left(x^2+4\right)-1\)
\(\Leftrightarrow x^3+27-x^3-4x=-1\)
\(\Leftrightarrow-4x=-28\)
hay x=7