cmr: gt biểu thức P=a^4-4a^3+5a^2-4a+5>0 với mọi a
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a) Ta có: \(x^2-20x+101=x^2-2.x.10+10^2+1=\left(x-10\right)^2+1\)
Vì \(\left(x-10\right)^2\ge0\left(\forall x\in Z\right)\)
\(\Rightarrow\left(x-10\right)^2+1>1>0\)
Vậy x2-20x+101 >0 với mọi x
b) \(4a^2+4a+2=\left(2a\right)^2+2.2a.1+1+1=\left(2a+1\right)^2+1\)
Vì \(\left(2a+1\right)^2\ge0\left(\forall a\in Z\right)\)
\(\Rightarrow\left(2a+1\right)^2+1>1>0\)
Vậy 4a2+4a+2 > 0 với mọi a
c) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)
\(=\left(x^2+10x+16\right)^2+8\left(x^2+10x+16\right)+16\)
\(=\left(x^2+10x+20\right)^2\) \(\ge0\left(\forall x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$5a^2+2b^2=11ab$
$\Leftrightarrow 5a^2+2b^2-11ab=0$
$\Leftrightarrow (5a^2-10ab)-(ab-2b^2)=0$
$\Leftrightarrow 5a(a-2b)-b(a-2b)=0$
$\Leftrightarrow (a-2b)(5a-b)=0$
Do $a>2b>0$ nên $a-2b>0$. Do dó $5a-b=0$
$\Leftrightarrow b=5a$. Khi đó:
$A=\frac{4a^2-5b^2}{a^2+2ab}=\frac{4a^2-5(5a)^2}{a^2+2a.5a}=\frac{-121a^2}{11a^2}=-11$
![](https://rs.olm.vn/images/avt/0.png?1311)
= (4a^2 -4a + 1) + (b^2 + 2b+ 1) + 1/2
= (2a-1)^2 + (b+1)^2 + 1/2 >0 với mọi a, b
\(\left\{{}\begin{matrix}\left(a^2+a\right)^2\ge0\\\left(a-2\right)^2\ge0\end{matrix}\right.\) \(\forall a\)
\(\Rightarrow\left(a^2+a\right)^2+\left(a-2\right)^2+1\ge1>0\) \(\forall a\)
\(P=\left(a^4-4a^3+4a^2\right)+\left(a^2-4a+4\right)+1\)
\(P=\left(a^2+a\right)^2+\left(a-2\right)^2+1>0\) \(\forall a\)