Question

[nimbaoi12xyz]

Ai giúp em với ạ e cảm ơn ai làm được câu nào thì làm ạ

Answer 1

[Quỳnh Anh]

Trả lời:

Bài 3:

1, \(C=\frac{x-1}{2}:\left(\frac{x^2+2}{x^3-1}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\right)\) \(\left(ĐKXĐ:x\ne1\right)\) 

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

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

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

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

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

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

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

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

2, Ta có: \(x^2+x+1=\left(x^2+x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\ne1\)

\(\Rightarrow\frac{x^2+x+1}{2}>0\forall x\ne1\)

Vậy C lớn hơn 0 với mọi x khác 1.

Answer 2

[Nguyễn Minh Quang]

\(1.C=\frac{x-1}{2}:\left(\frac{x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x}{x^2+x+1}+\frac{1}{1-x}\right)\)

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

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

2. ta có : \(C=\frac{x^2+x+\frac{1}{4}}{2}+\frac{3}{8}=\frac{\left(x+\frac{1}{2}\right)^2}{2}+\frac{3}{8}\ge\frac{3}{8}>0\)

3. Vậy GTNN của \(C=\frac{3}{8}\)