( x - 2 )( x - 4 ) + 3

<=> x² - 6x + 8 + 3

<=> ( x² - 6x + 9 ) + 2

<=> ( x - 3 )² + 2 ≥ 2 > 0 ∀ x ( đpcm )

\(a,P=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

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

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

\(=\left(x^2+5x+5\right)^2\ge0\forall x\)

Vậy \(P\ge0\forall x\)

\(b,P=\left(x^2+5x+5\right)^2\left(cmt\right)\)

Thay \(x=\frac{\sqrt{7}-5}{2}\)vào P ta được

\(P=\left(\left(\frac{\sqrt{7}-5}{2}\right)^2+5.\frac{\sqrt{7}-5}{2}+5\right)^2\)

\(=\left(\frac{7-10\sqrt{7}+25}{4}+\frac{10\sqrt{7}-50}{4}+\frac{20}{4}\right)^2\)

\(=\left(\frac{32-10\sqrt{7}+10\sqrt{7}-50+20}{4}\right)^2\)

\(=\left(\frac{2}{4}\right)^2\)

\(=\frac{1}{4}\)

a,

P=(x+1)(x+2)(x+3)(x+4)+1

P=[(x+1).(x+4)].[(x+2).(x+3)]+1

P=(x^2+5x+4)(x^2+5x+6)+1

P=[(x^2+5x+5)-1].[(x^2+5x+5)+1]+1

P=(x^2+5x+5)^2-1+1

P=\(\left(x^2+5x+5\right)^2\) \(\ge\)0 với mọi x

Câu b thì thay x vào rồi bấm máy ra ra kết quả

\(P=x^2-4x+2x-8+9,5=x^2-2x+1-9+9,5=\)

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

Answer:

\(f\left(x\right)+2f\left(\frac{1}{x}\right)=x^2\)

Thay x = 2 vào, ta được:

 \(f\left(2\right)+2f\left(\frac{1}{2}\right)=2^2\Rightarrow f\left(2\right)=2f\left(\frac{1}{2}\right)=4\left(\text{*}\right)\)

Thay \(x=\frac{1}{2}\) vào, ta được: 

\(f\left(\frac{1}{2}\right)+2\left(\frac{1}{\frac{1}{2}}\right)=\left(\frac{1}{2}\right)^2\Rightarrow f\left(\frac{1}{2}\right)+2f\left(2\right)=\frac{1}{4}\Rightarrow2f\left(\frac{1}{2}\right)+4f\left(2\right)=\frac{1}{2}\left(\text{*}\text{*}\right)\)

Từ (*) và (**) \(\Rightarrow f\left(2\right)+2f\left(\frac{1}{2}\right)-\left(2f\left(\frac{1}{2}\right)+4f\left(2\right)\right)=4-\frac{1}{2}\)

\(\Rightarrow f\left(2\right)+2f\left(\frac{1}{2}\right)-2f\left(\frac{1}{2}\right)-4f\left(2\right)=\frac{7}{2}\)

\(\Rightarrow-3f\left(2\right)=\frac{7}{2}\)

\(\Rightarrow f\left(2\right)=\frac{7}{2}.\left(-3\right)=\frac{-7}{6}\)

các bạn giải giúp mk đi ạ !!!!!

Với a, b, c là các số thực ta có: \(\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)=\dfrac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\).

Chọn \(a=x^2;b=y^2;c=z^2\) ta có \(x^4+y^4+z^4\ge x^2y^2+y^2z^2+z^2x^2\). (1)

Chọn \(a=xy;b=yz;c=zx\) ta có \(x^2y^2+y^2z^2+z^2x^2=\left(xy\right)^2+\left(yz\right)^2+\left(zx\right)^2\ge xy.yz+yz.zx+zx.xy=xyz\left(x+y+z\right)\). (2)

Từ (1), (2) ta có đpcm.