a) Dễ thấy \(x^2\)luôn dương vậy để A dương thì \(4x\ge0\)

\(\Leftrightarrow x\ge0\)

b) \(B=\left(x-3\right)\left(x+7\right)\)dương khi :

TH1: \(\hept{\begin{cases}x-3>0\\x+7>0\end{cases}\Rightarrow\hept{\begin{cases}x>3\\x>-7\end{cases}\Rightarrow}x>3}\)

TH2: \(\hept{\begin{cases}x-3< 0\\x+7< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 3\\x< -7\end{cases}\Rightarrow}x< -7}\)

c) Tương tự câu b)

a) Ta có ; \(x^2\ge0\forall x\in R\)

Nên A dương khi 4x \(\ge0\forall x\in R\) 

=> \(x\ge0\)

Vậy A dương khi \(x\ge0\)

Để a dương \(< =>\left(x-1\right)\left(x-2\right)-\left(x-3\right)>0\)

\(< =>x^2-2x-x+2-x+3>0\)

\(< =>x^2-4x+5>0\)

\(< =>x\left(x-4\right)>5\)

\(< =>x>6\)

Vậy để a dương thì x > 6

Quân , a lm cái j vậy ?

\(A=\frac{\left(x-1\right)\left(x-2\right)}{x-3}\)

Để A dương => A > 0 

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-2\right)}{x-3}>0\)

\(\Leftrightarrow\frac{x^2-3x+2}{x-3}>0\)

\(\Leftrightarrow\frac{x^2-3x+2}{x-3}>\frac{0}{x-3}\)

\(\Leftrightarrow x^2-3x+2>0\Leftrightarrow1< x< 2\)

\(\Leftrightarrow x-3>0\Leftrightarrow3>x\)

Nhầm lẫn r xD

a,Ta có:

\(\left|4x-\frac{7}{3}\right|\ge0\Rightarrow\left|4x-\frac{7}{3}\right|+2004\ge2004\)

Dấu "=" xảy ra \(\Leftrightarrow\left|4x-\frac{7}{3}\right|=0\Leftrightarrow4x-\frac{7}{3}=0\Leftrightarrow4x=\frac{7}{3}\Leftrightarrow x=\frac{7}{12}\)

b,Ta có:

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

Dấu "=" xảy ra \(\Leftrightarrow\)\(\begin{cases}x-1\ge0\\x-2\ge0\\3-x\ge0\\4-x\ge0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge1\\x\ge2\\x\le3\\x\le4\end{cases}\)\(\Leftrightarrow2\le x\le3\)

Câu C sai đề

A=\(\left|4x-\frac{7}{3}\right|+2004\ge2004\)

Dấu "=" xảy ra khi: x=7/12

Vậy GTNN của A là 2004 tại x=7/12

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x-99|

a.

\(ĐKXĐ:x\ne\pm1;\)

Ta có:

\(P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{x-1}{x+1}+\frac{x+1}{x-1}\right)\cdot\frac{x\left(x+1\right)-\left(1+x\right)}{x^3-1}\)

\(\Rightarrow P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{\left(x+1\right)\left(x-1\right)}{x^3-1}\)

\(\Rightarrow P=\left(\frac{x^4+x^2-4x+1}{x^2-1}-\frac{x^2-2x+1}{x^2-1}+\frac{x^2+2x+1}{x^2-1}\right)\cdot\frac{x^2-1}{x^3-1}\)

\(\Rightarrow P=\frac{x^4+x^2+1}{x^2-1}\cdot\frac{x^2-1}{x^3-1}\)

\(\Rightarrow P=\frac{x^4+x^2+1}{x^3-1}\)

b.

Để P là số nguyên thì  \(x^4+x^2+1⋮x^3-1\)

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

\(\Rightarrow x\left(x^3-1\right)+\left(x^2+x+1\right)⋮\left(x-1\right)\left(x^2+x+1\right)\)

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

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

\(\Rightarrow x^2-x+1⋮x-1\)

\(\Rightarrow x\left(x-1\right)+1⋮x-1\)

\(\Rightarrow1⋮x-1\)

\(\Rightarrow x-1\in\left\{1;-1\right\}\)

\(\Rightarrow x=1\left(KTMĐK\right);x=0\)

Vậy x=0.

P/S:Không chắc chắn lắm đâu nha mn,nếu có j sai thì ib vs em ah.