1)\(\frac{x-b-c}{a}+\frac{x-c-a}{b}+\frac{x-a-b}{c}=3\)

=>\(\frac{x-b-c}{a}-1+\frac{x-c-a}{b}-1+\frac{x-a-b}{c}-1=0\)

=>\(\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}+\frac{x-a-b-c}{c}=0\)

=>\(\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

Nếu x - a -b -c = 0 => phương trình có nghiệm duy nhất x = a + b + c

Nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)=> Phương trình có vô số nghiệm x thuộc R

Bùi Vũ Kim Thư nếu mà bn ko hiểu thì cứ hỏi nhae

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

ĐKXĐ: x + 2 \(\ne\) 0 và x - 2 \(\ne\) 0

\(\Rightarrow\) x \(\ne\) \(\pm\) 2

b, \(\frac{x-1}{1-2x}=1\)

ĐKXĐ: 1 - 2x \(\ne\) 0

\(\Leftrightarrow\) x \(\ne\) \(\frac{1}{2}\)

Bài 2:

a, \(\frac{x+2}{x}=\frac{2x+3}{x-2}\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) 2)

\(\Leftrightarrow\) \(\frac{\left(x+2\right)\left(x-2\right)}{x\left(x-2\right)}=\frac{x\left(2x+3\right)}{x\left(x-2\right)}\)

\(\Rightarrow\) (x + 2)(x - 2) = x(2x + 3)

\(\Leftrightarrow\) x² - 4 = 2x² + 3x

\(\Leftrightarrow\) x² - 2x² - 3x = 4

\(\Leftrightarrow\) -x² - 3x = 4

\(\Leftrightarrow\) -x² - 3x - 4 = 0

\(\Leftrightarrow\) -(x² + 3x + 4) = 0

\(\Leftrightarrow\) x² + 3x + 4 = 0

\(\Leftrightarrow\) x² + 3x + \(\frac{9}{4}\) + \(\frac{7}{4}\) = 0

\(\Leftrightarrow\) (x + \(\frac{3}{2}\))² + \(\frac{7}{4}\) = 0

Vì (x + \(\frac{3}{2}\))² + \(\frac{7}{4}\) > 0 với mọi x

\(\Rightarrow\) Pt vô nghiệm

Vậy S = \(\varnothing\)

b, \(\frac{2x+5}{2x}-\frac{x}{x+5}=0\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) -5)

\(\Leftrightarrow\) \(\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{2x^2}{2x\left(x+5\right)}=0\)

\(\Rightarrow\) (2x + 5)(x + 5) - 2x² = 0

\(\Leftrightarrow\) 2x² + 10x + 5x + 25 - 2x² = 0

\(\Leftrightarrow\) 15x + 25 = 0

\(\Leftrightarrow\) x = \(\frac{-5}{3}\) (TMĐKXĐ)

Vậy S = {\(\frac{-5}{3}\)}

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

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

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

\(\Leftrightarrow\) x + 1 - 2(3 - x) = 0

\(\Leftrightarrow\) x + 1 - 6 + 2x = 0

\(\Leftrightarrow\) 3x - 5 = 0

\(\Leftrightarrow\) x = \(\frac{5}{3}\) (TMĐKXĐ)

Vậy S = {\(\frac{5}{3}\)}

d, \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\) (ĐKXĐ: x \(\ne\) \(\pm\) 1)

\(\Leftrightarrow\) \(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{16}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow\) (x + 1)² - (x - 1)² = 16

\(\Leftrightarrow\) (x + 1 - x + 1)(x + 1 + x - 1) = 16

\(\Leftrightarrow\) 4x = 16

\(\Leftrightarrow\) x = 4 (TMĐKXĐ)

Vậy S = {4}

Chúc bn học tốt!!

kcj

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 = \(-\frac{2012}{10}\) .

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 = \(-\frac{1}{2}\)

Chúc bạn học tốt@@

Điều kiện a; b ; c khác 0

\(\Rightarrow x.\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=2.\left(\frac{bc+ac+ab}{abc}\right)+\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\)

\(\Rightarrow x.\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=\frac{2bc+2ac+2ab}{abc}+\frac{a^2}{abc}+\frac{b^2}{abc}+\frac{c^2}{abc}\)

\(\Rightarrow x.\left(\frac{a+b+c}{abc}\right)=\frac{\left(a+b+c\right)^2}{abc}\)

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

Nếu a+ b+ c khác 0 => phương trình có nghiệm duy nhất là \(\Rightarrow x=a+b+c\)

Nếu a+ b + c = 0 => x. 0 = 0 =>  pt có vô số nghiêm

x=a+b+c

c)Ta có: \(x^4+3x^3+4x^2+3x+1=0\)

\(\Leftrightarrow x\left(x^3+2x^2+2x+1\right)+1\left(x^3+2x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+2x^2+2x+1\right)=0\)

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

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

Suy ra x + 1 =0 hay x = -1

X=0 hoặc -1