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a4 + a3 + a + 1 ≥ 0
<=> a3( a + 1 ) + ( a + 1 ) ≥ 0
<=> ( a + 1 )( a3 + 1 ) ≥ 0
<=> ( a + 1 )2( a2 - a + 1 ) ≥ 0 ( đúng )
Vậy ta có đpcm. Dấu "=" xảy ra <=> a = -1
Ta có: \(a^4+a^3+a+1\)
\(=a^3\left(a+1\right)+\left(a+1\right)\)
\(=\left(a+1\right)\left(a^3+1\right)\)
\(=\left(a+1\right)\left(a+1\right)\left(a^2-a+1\right)\)
\(=\left(a+1\right)^2\left[\left(a^2-a+\frac{1}{4}\right)+\frac{3}{4}\right]\)
\(=\left(a+1\right)^2\left[\left(a-\frac{1}{2}\right)^2+\frac{3}{4}\right]\ge0\left(\forall a\right)\) (luôn đúng)
Dấu "=" xảy ra khi: a = -1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(x-1\right)\left(x-3\right)+2=x^2-4x+3+2=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\ge1>0\forall x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a^2+a+1=a^2+a+\frac{1}{4}+\frac{3}{4}=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(a^2-a+1=a^2-a+\frac{1}{4}+\frac{3}{4}=\left(a-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow\frac{a^2+a+1}{a^2-a+1}>0\forall a\in R\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=x^2-2x+2=\left(x-1\right)^2+1>0\forall x\inℝ\)
b) \(x-x^2-3=-\left(x^2-x+3\right)\)
\(=-\left(x^2-x+\frac{1}{4}+\frac{11}{4}\right)\)
\(=-\left[\left(x-\frac{1}{2}\right)^2+\frac{11}{4}\right]\)
\(=-\left[\left(x-\frac{1}{2}\right)^2\right]-\frac{11}{4}\le\frac{-11}{4}< 0\forall x\inℝ\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có (a+2)3-(a+6)(a2+12)+64=a3+6a2+12a+8-a3-12a-6a2-72+64=0(đpcm)
\(\left(a+2^3\right)-\left(a+6\right).\left(a^2+12\right)+64=0\)
\(\Leftrightarrow\left(a+8\right)-\left(a^3+6a^2+12a+72\right)=-64\)
\(\Leftrightarrow\left(a^3+6a^2+12a+72\right)-\left(a+8\right)=64\)
\(\Leftrightarrow a^3+6a^2+11a+64=64\)
\(\Leftrightarrow a^3+6a^2+11a^2=0\)
\(\Leftrightarrow a.\left(a^2+6a+11\right)=0\)
\(\Leftrightarrow a.\left[\left(a^2+2.a.3+9\right)+2\right]=0\)
\(\Leftrightarrow a.\left[\left(a+3\right)^2+2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\\left(a+3\right)^2+2=0\left(\text{Vô lí}\right)\end{matrix}\right.\)
\(\Rightarrow a=0\)
\(\Rightarrow\) Đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2+x+1\)
\(A=x^2+x+\dfrac{1}{4}-\dfrac{1}{4}+1\)
\(A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
mà \(\left(x+\dfrac{1}{2}\right)^2\ge0\)
\(\Rightarrow A=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>\dfrac{3}{4}>0\) với mọi x
\(\Rightarrow Dpcm\)