Question

[Vũ Thu Hiền]

Bài 1:Giải phương trình

a, (3x-2)(4+5x) = 0

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

c, \(\frac{10x+3}{2009}\)+ \(\frac{10x-1}{2013}\) = \(\frac{10x+1}{2011}\) - \(\frac{2-10x}{2014}\)

Bài 2: Giải các phương trình sau

a, \(\frac{4x-8}{2x^2+1}\) = 0

b, \(\frac{x^2-x-6}{x-3}\) = 0

c, \(\frac{x+5}{3x-6}\) - \(\frac{1}{2}\) = \(\frac{2x-3}{2x-4}\)

d, \(\frac{12}{1-9x^2}\) = \(\frac{1-3x}{1+3x}\) - \(\frac{1+3x}{1-3x}\)

e, \(\frac{x+5}{x-1}\) = \(\frac{x+1}{x-3}\) - \(\frac{8}{x^2-4x+3}\)

f,\(\frac{x+1}{x-2}\)- \(\frac{5}{x+2}\) = \(\frac{12}{x^2-4}\) +1

Bài 3:Tìm giá trị nhỏ nhất của biểu thức A=\(^{x^2}\)+ x+2012

Answer 1

[Nguyễn Ngọc Lộc]

Bài 3 :

Ta có : \(A=x^2+x+2012\)

=> \(A=x^2+x+\left(\frac{1}{2}\right)^2+\frac{8047}{4}\)

=> \(A=\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\)

- Ta thấy : \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)

=> \(\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\ge\frac{8047}{4}\forall x\)

- Dấu "=" xảy ra <=> \(x+\frac{1}{2}=0\)

<=> \(x=-\frac{1}{2}\)

Vậy Min_A = \(\frac{8047}{4}\) <=> x = \(-\frac{1}{2}\) .

Bài 1 :

a, Ta có : \(\left(3x-2\right)\left(4+5x\right)=0\)

=> \(\left[{}\begin{matrix}3x-2=0\\4+5x=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}3x=2\\5x=-4\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=-\frac{4}{5}\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = \(\frac{2}{3}\), x = \(-\frac{4}{5}\) .

b,- ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

=> \(x\ne\pm1\)

Ta có : \(\frac{x+1}{x-1}-\frac{4}{x+1}=\frac{3-x^2}{1-x^2}\)

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

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

=> \(x^2+2x+1-4x+4=x^2-3\)

=> \(-2x=-3-5\)

=> \(x=4\left(TM\right)\)

Vậy phương trình có nghiệm là x = 4 .

c, Ta có : \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}-\frac{2-10x}{2014}\)

=> \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}+\frac{10x-2}{2014}\)

=> \(\frac{10x+3}{2009}+1+\frac{10x-1}{2013}+1=\frac{10x+1}{2011}+1+\frac{10x-2}{2014}+1\)

=> \(\frac{10x+3}{2009}+\frac{2009}{2009}+\frac{10x-1}{2013}+\frac{2013}{2013}=\frac{10x+1}{2011}+\frac{2011}{2011}+\frac{10x-2}{2014}+\frac{2014}{2014}\)

=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}=\frac{10x+2012}{2011}+\frac{10x+2012}{2014}\)

=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}-\frac{10x+2012}{2011}-\frac{10x+2012}{2014}=0\)

=> \(\left(10x+2012\right)\left(\frac{1}{2009}+\frac{1}{2013}-\frac{1}{2011}-\frac{1}{2014}\right)=0\)

=> \(10x+2012=0\)

=> \(x=-\frac{2012}{10}\)

Vậy phương trình có nghiệm là x = \(-\frac{2012}{10}\) .

Answer 2

[Jeong Soo In]

Bài 3:

Giải:

Ta có : A = x² + x + 2012

= x² + 2.\(\frac{1}{2}\).x + \(\frac{1}{4}\) + \(\frac{8047}{4}\)

= (x + \(\frac{1}{2}\))² + \(\frac{8047}{4}\) ≥ \(\frac{8047}{4}\)

⇒ A_min = \(\frac{8047}{4}\) ⇔ (x + \(\frac{1}{2}\))² = 0 ⇔ x = \(-\frac{1}{2}\)

Vậy A_min = \(\frac{8047}{4}\) tại x = \(-\frac{1}{2}\)

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