Question

CMR: biểu thức sau luôn nhan gia trị dương:

a)\(x^2-5x+10\)

b)\(2x^2+8x+10\)

c)\(\left(1-2x\right)\left(x-1\right)-5\)

d)\(\left(x+5\right)\left(x-3\right)+20\)

Answer 1

a) \(x^2-5x+10\)

\(=x^2-2.\frac{5}{2}x+\frac{25}{4}+\frac{15}{4}\)

\(=\left(x-\frac{5}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)

b) \(2x^2+8x+10\)

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

\(=2\left[\left(x+2\right)^2+1\right]\ge2>0\)

Answer 2

c) thay x = 1 vào thì đề sai

d) \(\left(x+5\right)\left(x-3\right)+20=x^2+2x+5\)

\(=\left(x+1\right)^2+4\ge0>0\)

Answer 3

\(x^2-2\cdot\frac{5}{2}x+\frac{25}{4}+\frac{15}{4}=\left(x-\frac{5}{2}\right)^2+\frac{15}{4}>0\left(\forall x\right)\)

\(2x^2+8x+10=2\left(x^2+4x+4\right)+2=2\left(x+2\right)^2+2>0\left(\forall x\right)\)

\(\left(1-2x\right)\left(x-1\right)-5=-2x^2+3x-6=-2\left(x^2-\frac{3}{2}x+\frac{9}{4}\right)-\frac{3}{2}=-2\left(x-\frac{3}{2}\right)^2-\frac{3}{2}< 0\)

\(\left(x+5\right)\left(x-3\right)+20=x^2+2x+5=\left(x+1\right)^2+4>0\left(\forall x\right)\)

Câu c có lẽ mk sai mong bạn thông cảm nha