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Answer:

Có \(a+b+c=3\)

\(\Rightarrow\left(a+b+c\right)^2=3^2\)

      \(a^2+b^2+c^2+2ab+2bc+2ac=9\left(\text{*}\right)\) 

Mà đề ra \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\frac{bc}{abc}+\frac{ac}{abc}+\frac{ab}{abc}=0\)

\(\Leftrightarrow\frac{bc+ac+ab}{abc}=0\)

\(\Leftrightarrow bc+ac+ab=0\left(\text{*}\text{*}\right)\)

Thay (**) vào (*), được \(a^2+b^2+c^2=9\)

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với x >= 0 ; x khác 1 

\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)

\(=\frac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}=\frac{x-\sqrt{x}}{x\sqrt{x}-1}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

Answer:

Với \(x\ne1;x\ge0\) có: 

\(\frac{x+2}{x\sqrt{x}-1}\)\(+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\)

\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)

\(=\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)

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O-O ???

\(3x+2>-2x+12\Leftrightarrow5x>10\Leftrightarrow x>2\)

Vậy bpt có nghiệm S = { x | x > 2 } 

bạn tự biểu diễn trên trục số nhé 

Answer:

\(\frac{x^2-x-2}{x^2-x+2}\)\(\inℤ\)

\(\Rightarrow\frac{x^2-x+2-4}{x^2-x+2}\inℤ\)

\(\Rightarrow1-\frac{4}{x^2-x+2}\inℤ\)

\(\Rightarrow\frac{4}{x^2-x+2}\inℤ\)

\(x\inℤ;\frac{4}{x^2-x+2}\inℤ\)

\(\Rightarrow4⋮\left(x^2-x+2\right)\RightarrowƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)