Question

Tìm x a, (x+4)^2-(x+1)(x-1)=19 b, (2x-1)^2+(x+3)^2-5(x+7)(x-7)=0

Answer 1

a) \(\Rightarrow x^2+8x+16-x^2+1=19\)

\(\Rightarrow8x=2\Rightarrow x=\dfrac{1}{4}\)

b) \(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Rightarrow2x=-255\Rightarrow x=-\dfrac{255}{2}\)

Answer 2

a: Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=19\)

\(\Leftrightarrow x^2+8x+16-x^2+1=19\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

b: Ta có: \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Leftrightarrow2x=-255\)

hay \(x=-\dfrac{255}{2}\)