\(\left(x+2\right)^2=x^2\)

\(\Rightarrow x^2+4x+4=x^2\)

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

\(\Rightarrow4x+4=0\)

\(\Rightarrow4x=\left(-4\right)\)

\(\Rightarrow x=\left(-1\right)\)

\(\Rightarrow x\in\left\{-1\right\}\)

\(\left(x+2\right)^2=x^2\)

=>\(x^2+2.2.x+2^2=x^2\)

=>\(x^2+4x+4=x^2\)

vì \(x^2+4x+4=x^2\)

mà \(x^2=x^2\)

=>\(4x+4=0\)

=>\(x.4=1.4\)

=>\(x=4:4=1\)

=>\(x=1\)

đề bài thực hiện pháp tính nha

2x^3-5x^2-x+1 2x+1 x^2-3x+1 2x^3+x^2 - -6x^2-x+1 -6x^2-3x - 2x+1 2x+1 - 0

Ta có: x² + xy + 5 - 6x - y = (x² - x) + (xy - y) + (5 - 5x)

= x(x - 1) + y(x - 1) - 5(x - 1)

= 9x + y - 5)(x - 1)

https://olm.vn/hoi-dap/detail/245049015319.html?pos=572115847211

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

\(=2\left(x^2-\frac{5}{2}x+\frac{3}{2}\right)\)

\(=2\left(x^2-\frac{5}{2}x+\frac{25}{16}-\frac{1}{16}\right)\)

\(=2\left[\left(x-\frac{5}{4}\right)^2-\frac{1}{16}\right]\)

\(=2\left[\left(x-\frac{5}{4}\right)^2\right]-\frac{1}{32}\ge\frac{-1}{32}\)

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

\(=2\left(x^2-\frac{5}{2}x+\frac{3}{2}\right)\)

\(=2\left(x^2-\frac{5}{4}\cdot2x+\left(\frac{5}{4}\right)^2-\left(\frac{5}{4}\right)^2+\frac{3}{2}\right)\)

\(=2\left[\left(x-\frac{5}{4}\right)^2-\frac{25}{16}+\frac{3}{2}\right]\)

\(=2\left[\left(x-\frac{5}{4}\right)^2-\frac{1}{16}\right]\)

\(=2\left(x-\frac{5}{4}\right)^2-\frac{1}{8}\)

có\(2\left(x-\frac{5}{4}\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{5}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)

\(\Rightarrow GTNNB=-\frac{1}{8}\)

 với \(\left(x-\frac{5}{4}\right)^2=0;x=\frac{5}{4}\)