\(\text{Δ}=\left[-2\left(a-1\right)\right]^2-4\cdot\left(a+1\right)\left(-a-3\right)\)

\(=4\left(a-1\right)^2+4\left(a+1\right)\left(a+3\right)\)

\(=4\left(a^2-2a+1\right)+4\left(a^2+4a+3\right)\)

\(=4a^2-8a+4+4a^2+16a+12\)

\(=8a^2+8a+16\)

\(=8\left(a^2+a+2\right)\)

\(=8\left(a^2+a+\dfrac{1}{4}+\dfrac{7}{4}\right)\)

\(=8\left[\left(a+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]>=8\cdot\dfrac{7}{4}>0\forall a\) khác -1

=>Phương trình luôn có hai nghiệm phân biệt khi \(a\ne-1\)

\(a.A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2}{x+\sqrt{x}+1}\) ( x ≥ 0 ; x # 1 )

\(b.\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\) \(c.\) \(\dfrac{2}{x+\sqrt{x}+1}\) ≤ \(\dfrac{2}{1}=2\left(x\text{≥ }0\right)\)

⇒ \(A_{Max}=2."="\) ⇔ \(x=0\left(TM\right)\)

giả sử : \(x+y+xy=-1\) \(\Rightarrow x+y+xy+1=0\)

\(\Rightarrow\left(x+1\right)\left(y+1\right)=0\rightarrow x+1=0\) hoặc \(y+1=0\)

\(\Rightarrow x=-1\) hoặc \(y=-1\) ( trái giả thiết )

vậy nếu \(x\ne-1\) và \(y\ne-1\) thì \(x+y+xy\ne-1\)

Áp dụng tỉ dãy số bằng nhau. Ta có:

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Leftrightarrow\frac{1+1+1}{a+b+c}=1\)

\(\Rightarrow a=b=c\)

\(\Rightarrow\frac{a}{b}\Leftrightarrow1-1\Leftrightarrow0\)

\(\Rightarrow PT=\frac{a-c}{c-b}=\frac{\left(a-c\right)^0}{\left(c-b\right)^0}=0\)

Vậy dấu = xảy ra khi a - c = a               , c - b = b

Ta có ĐPCM

Ps: Chả biết đúng hay không nữa

như này mới đúng nè 

ta có\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{c}.2\)

\(\Rightarrow\frac{b}{ab}+\frac{a}{ba}=\frac{2}{c}\)

\(\Rightarrow\frac{b+a}{ab}=\frac{2}{c}\)

\(\Rightarrow\left(b+a\right)c=2ab\)

\(\Rightarrow cb+ca=ab+ab\)

\(\Rightarrow ca-ab=ab-cb\)

\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

\(\Rightarrow\frac{a-c}{c-b}=\frac{a}{b}\)

a: \(=\dfrac{x^3-x^2+x+3\left(x^2-1\right)+x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

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

\(=\dfrac{x^2+x+1}{x^2-x+1}\)

b: \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

\(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

=>A>0 với mọi x<>-1

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

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

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

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