Có: \(A=x^3-x^2+2\)

\(=x^3+1-x^2+1\)

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

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

A là số dương 

<=>  \(\left(x+1\right)\left(x^2-2x+2\right)>0\)

Vì \(x^2-2x+2=\left(x-1\right)^2+1>0\)

=> \(\left(x+1\right)>0\)

<=> x > - 1

A là số nguyên => x nguyên 

Vậy để A là số nguyên dương  thì x là số nguyên và x > -1.

\(\text{a) Thay a = 4 vào pt ta có:}\)
      \(\frac{x+4}{x+2}+\frac{x-2}{x-4}=2\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+4\right)+\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=2\)
\(\Leftrightarrow\frac{x^2-16+x^2-4}{x^2-4x+2x-8}=2\)
\(\Leftrightarrow\frac{2x^2-20}{x^2-2x-8}=2\)
\(\Leftrightarrow2x^2-20=2.\left(x^2-2x-8\right)\)
\(\Leftrightarrow2x^2-20=2x^2-4x-16\)
\(\Leftrightarrow2x^2-2x^2+4x=-16+20\)
\(\Leftrightarrow4x=4\)
\(\Leftrightarrow x=1\)

\(\text{b) Thay x = -1 vào pt ta có:}\)
     \(\frac{-1+a}{-1+2}+\frac{-1-2}{-1-a}=2\)
\(\Leftrightarrow\frac{a-1}{1}+\frac{-3}{-\left(a+1\right)}=2\)
\(\Leftrightarrow\left(a-1\right)+\frac{3}{a+1}=2\)
\(\Leftrightarrow\frac{\left(a-1\right)\left(a+1\right)+3}{a+1}=2\)
\(\Leftrightarrow\frac{a^2-1+3}{a+1}=2\)
\(\Leftrightarrow a^2+2=2.\left(a+1\right)\)
\(\Leftrightarrow a^2+2=2a+2\)
\(\Leftrightarrow a^2-2a=2-2\)
\(\Leftrightarrow a\left(a-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=0\\a-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=0\\a=2\end{cases}}}\)
Vậy để pt có nghiệm là x = 1 thì a = {0 ; 2}
 

\(a.Thay:a=4\Leftrightarrow\frac{x+4}{x+2}+\frac{x-2}{x-4}=2\)

                    \(\Leftrightarrow\frac{\left(x+4\right)\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x-4\right)\left(x+2\right)}=\frac{2\left(x+2\right)\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}\)

                    \(\Rightarrow\left(x+4\right)\left(x-4\right)+\left(x-2\right)\left(x+2\right)=2\left(x+2\right)\left(x-4\right)\)

                    \(\Leftrightarrow x^2-4x+4x-16+x^2+2x-2x-4=\left(2x+4\right)\left(x-4\right)\)

                    \(\Leftrightarrow2x^2-20=2x^2-8x+4x-16\)

                    \(\Leftrightarrow2x^2-20-2x^2+8x-4x+16=0\)

                    \(\Leftrightarrow4x-4=0\)

                    \(\Leftrightarrow x=1\)

Ta co:

\(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2=\left(\frac{1}{a}-2\right)^2+\left(\frac{1}{b}-2\right)^2+6\left(\frac{1}{a}+\frac{1}{b}\right)-6\ge\frac{24}{a+b}-6=18\)

Dau '=' xay ra khi \(a=b=\frac{1}{2}\)

TH1: \(|x|=x\)\(\Leftrightarrow x\ge0\)\(\Rightarrow x^2-3x-4=0\)

\(\Leftrightarrow x^2-4x+x-4=0\)\(\Leftrightarrow x\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-4\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=4\end{cases}}\)

mà \(x\ge0\)\(\Rightarrow x=4\)

TH2: \(|x|=-x\)\(\Leftrightarrow x< 0\)\(\Rightarrow x^2+3x-4=0\)

\(\Leftrightarrow x^2-x+4x-4=0\)\(\Leftrightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-1\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

mà \(x< 0\)\(\Rightarrow x=-4\)

Vậy \(x=4\)hoặc \(x=-4\)

Mình chỉ biết đáp án nó là -4 và 4, còn cách trình bày thì ko biết :/

Thân.

thek tl lm j???

Ta có:

\(3x-\frac{x^2}{x^2-9}=0\)

\(\Leftrightarrow3x=\frac{x^2}{x^2-9}\)

\(\Leftrightarrow3x\left(x^2-9\right)=x^2\)

\(\Leftrightarrow3x^3-27x^2=x^2\)

\(\Leftrightarrow3x^3=x^2+27x^2\)

\(\Leftrightarrow3x^3=28x^2\)

\(\Leftrightarrow3x=28\)

\(\Leftrightarrow x=\frac{28}{3}\)

Vậy \(x=\frac{28}{3}\)

Cậu ơi ! 

TH x = 0 thì sao ạ ?

Ta có \(3-\frac{8}{x^3}\)

\(=0=3-\frac{8}{x^3}=0=x=\frac{2}{^3\sqrt{3}}=y=\frac{9}{^3\sqrt{3}}=3^3\sqrt{9}\)

Vậy \(min\)của hàm số \(3x^2+\frac{4}{x}=3^3\sqrt{9}\)