Question

bài này là tìm x

(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0

giải hộ mk nha

Answer 1

(2x - 1)^2 + (x + 3)^2 - 5(x + 7)(x - 7) = 0
<=>4x^2-4x+1+x^2+6x+9-5x^2+245=0
<=>2x+255=0
<=>2x=-255
<=>x=-255/2

Có trên google ( ghi nguồn đầy đủ )

Answer 2

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

\(\Leftrightarrow\)\(4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow\)\(5x^2+2x+10-5x^2+245=0\)

\(\Leftrightarrow\)\(2x=-255\)

\(\Leftrightarrow\)\(x=-127,5\)

Vậy...

Answer 3

4x^2 - 2.2x.1 + 1^2 + x^2 + 2.x.3 + 3^2 - 5.x^2 + 5. 7^2 = 0

4x^2 - 4x + 1 + x^2 + 6x + 9 - 5x^2 + 245 = 0

2x + 1 + 9 + 245 = 0

=> x = 127,5

Answer 4

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

\(=4x^2-4x+1+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)\)

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

\(\Rightarrow x=-\frac{255}{2}\left(=-127.5\right)\)