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\(B=4x^2+4x+2=\left(2x\right)^2+2.2x+1+1=\left(2x+1\right)^2+1\)

Ta có: \(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2+1\ge1\)

Dấu '' = '' xảy ra khi: \(\left(2x+1\right)^2=0\Rightarrow x=\frac{-1}{2}\)

Vậy giá trị nhỏ nhất của \(B=1\) khi \(x=\frac{-1}{2}\)

\(B=\left(4x^2+4x+1\right)+1=\left(2x+1\right)^2+1=\)

Vì \(\left(2x+1\right)^2\ge0\Leftrightarrow\left(2x+1\right)^2+1\ge1\Leftrightarrow B\ge1\Rightarrow min_B=1\)

Đẳng thức xảy ra \(\Leftrightarrow2x+1=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy \(min_B=1\Leftrightarrow x=-\frac{1}{2}\)

\(x^2-6x+11=\left(x^2-6x+9\right)+2\)\(=\left(x-3\right)^2+2\)

Vì \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2+2\ge2\)

Mặt khác 2 > 0 nên \(\left(x-3\right)^2+2>0\Leftrightarrow x^2-6x+11>0\)\(\forall x\inℝ\)

\(x^2-6x+11\)

\(=x^2-6x+9+2\)

\(=\left(x^2-6x+9\right)+2\)

\(=\left(x-3\right)^2+2\)

Với mọi \(x\) ta có: \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2+2\ge2>0,\forall x\)

Vậy \(x^2-6x+11>0\forall x\)

DÀI VẬY BẠN??

Bài 3:

\(\left(4x^2+12xy+9y^2\right):\left(2x+3y\right)\)

\(=[\left(2x\right)^2+2.2x.3y+\left(3y\right)^2]:\left(2x+3y\right)\)

\(=\left(2x+3y\right)^2:\left(2x+3y\right)\)

\(=2x+3y\)

\(\left(125x^3+27\right):\left(5x+3\right)\)

\(=[\left(5x\right)^3+3^3]:\left(5x+3\right)\)

\(=\left(5x+3\right).[\left(5x\right)^2-3.5x+3^2]:\left(5x+3\right)\)

\(=\left(5x+3\right).\left(25x^2-15x+9\right):\left(5x+3\right)\)

\(=25x^2-15x+9\)

\(2xy+\frac{1}{4}-x^2-y^2\)\(=\left(\frac{1}{2}\right)^2-\left(x^2-2xy+y^2\right)\)\(=\left(\frac{1}{2}\right)^2-\left(x-y\right)^2=\left(\frac{1}{2}+x-y\right)\left(\frac{1}{2}-x+y\right)\)

2xy + 1/4 - x² - y²

=> - (-2xy + x² + y² ) + 1/4

=> - (x - y)² + (1/2)²

=> - (x - y + 1/2)(x - y - 1/2)

ĐÚNG ĐÓ NHA, NHỚ NHÉ!

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\(8y+49x^2-16-y^2=49x^2-\left(y^2-8y+16\right)\)\(=\left(7x\right)^2-\left(y-4\right)^2=\left(7x+y-4\right)\left(7x-y+4\right)\)

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