a) x² - 7x + 16
= (x² - 2x\(\frac{7}{2}\)+ \(\frac{49}{4}\)) + \(\frac{15}{4}\)
= (x - \(\frac{7}{2}\))² + \(\frac{15}{4}\)> 0
b) 3x² - 3x + 1
= [\(\left(\sqrt{3x^2}\right)^2\)- 2.\(\sqrt{3x^2}\).\(\frac{\sqrt{3}}{2}\)+ \(\frac{3}{4}\)] + \(\frac{1}{4}\)
= (\(\sqrt{3x^2}\)- \(\frac{\sqrt{3}}{2}\))² + \(\frac{1}{4}\)> 0
c) -x² + 3x - 5
= -(x² - 3x + 5)
= -(x² - 2x\(\frac{3}{2}\)+ \(\frac{9}{4}\)+\(\frac{11}{4}\))
= -[(x - \(\frac{3}{2}\))² + \(\frac{11}{4}\)] < 0
d) Câu này sai đề rồi bạn ơi

a)Ta có: \(a^2+2a+b^2+1=a^2+2a+1+b^2\)

                                                 \(=\left(a+1\right)^2+b^2\)

                         Vì \(\left(a+1\right)^2\ge0;b^2\ge0\)

                  \(\left(a+1\right)^2+b^2\ge0\)

b)\(x^2+y^2+2xy+4=\left(x+y\right)^2+4\)

                 Vì \(\left(x+y\right)^2\ge0\Rightarrow< 0\left(x+y\right)^2+4\left(đpcm\right)\)

c)Ta có:\(\left(x-3\right)\left(x-5\right)+2=x^2-8x+15+2\)

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

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

                    Vì \(\left(x-4\right)^2\ge0\)

                              \(\Rightarrow\left(x-4\right)^2+1\ge1\)

Vậy (x-3)(x-5) + 2 > 0 ∀ x R

Đáp án A

1: A={x∈R|x<=2}

=>A=(-∞;2]

B={x∈R|x>5}

=>B=(5;+∞)

A\(\cap\) B=(-∞;2]\(\cap\) (5;+∞)

=∅

A\(\cup\) B=(-∞;2]\(\cup\) (5;+∞)

A\B=(-∞;2]\(5;+∞)

=(-∞;2]

B\A=(5;+∞)\(-∞;2]

=(5;+∞)

\(C_{R}A\) =R\A=R\(-∞;2]

=(2;+∞)

\(C_{R}B=\) R\B=R\(5;+∞)

=(-∞;5]

2: A={x∈R|x<0 hoặc x>=2}

=>A=(-∞;0)\(\cup\) [2;+∞)

B={x∈R|-4<=x<3}

=>B=[-4;3)

A\(\cap\) B=((-∞;0)\(\cup\) [2;+∞))\(\cap\) [-4;3)

=[-4;0)\(\cup\) [2;3)

A\(\cup\) B=((-∞;0)\(\cup\) [2;+∞))\(\cup\) [-4;3)

=(-∞;+∞)

A\B=((-∞;0)\(\cup\) [2;+∞))\[-4;3)

=(-∞;-4)\(\cup\) [3;+∞)

B\A=[-4;3)\((-∞;0)\(\cup\) [2;+∞))

=[0;2)

\(C_{R}A\) =R\A=[0;2)

\(C_{R}B\) =R\B=R\[-4;3)

=(-∞;-4)\(\cup\) [3;+∞)

3: |x-1|<2

=>-2<x-1<2

=>-1<x<3

=>A=(-1;3)

|x+1|<3

=>-3<x+1<3

=>-4<x<2

=>B=(-4;2)

A\(\cap\) B=(-1;3)\(\cap\) (-4;2)

=(-1;2)

A\(\cup\) B=(-1;3)\(\cup\) (-4;2)

=(-4;3)

A\B=(-1;3)\(-4;2)

=[2;3)

B\A=(-4;2)\(-1;3)

=(-4;-1]

\(C_{R}A\) =R\A=R\(-1;3)

=(-∞;-1]\(\cup\) [3;+∞)

\(C_{R}B\) =R\B=R\(-4;2)

=(-∞;-4]\(\cup\) [2;+∞)