Phân thức đại số - Hỏi đáp

4.

\((x-2)(3x+5)=(2x-4)(x+1)\)

\(\Leftrightarrow (x-2)(3x+5)-(2x-4)(x+1)=0\)

\(\Leftrightarrow (x-2)(3x+5)-2(x-2)(x+1)=0\)

\(\Leftrightarrow (x-2)[(3x+5)-2(x+1)]=0\)

\(\Leftrightarrow (x-2)(x+3)=0\Rightarrow \left[\begin{matrix} x-2=0\\ x+3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)

5.

\((2x+5)(x-4)=(x-5)(4-x)\)

\(\Leftrightarrow (2x+5)(x-4)-(x-5)(4-x)=0\)

\(\Leftrightarrow (2x+5)(x-4)+(x-5)(x-4)=0\)

\(\Leftrightarrow (x-4)[(2x+5)+(x-5)]=0\)

\(\Leftrightarrow (x-4).3x=0\)

\(\Rightarrow \left[\begin{matrix} x-4=0\\ 3x=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=4\\ x=0\end{matrix}\right.\)

1.

\((2x+1)(x^2+2)=0\Rightarrow \left[\begin{matrix} 2x+1=0\\ x^2+2=0\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=\frac{-1}{2}\\ x^2=-2< 0(\text{vô lý})\end{matrix}\right.\)

Vậy \(x=-\frac{1}{2}\)

2.

\((x^2+4)(7x-3)=0\Rightarrow \left[\begin{matrix} x^2+4=0\\ 7x-3=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x^2=-4< 0(\text{vô lý})\\ x=\frac{3}{7}\end{matrix}\right.\)

Vậy \(x=\frac{3}{7}\)

3.

\((x-5)(3-2x)(3x+4)=0\)

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

a) \(\left(x-3\right)\left(2x+1\right)\left(4-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)

Vậy ..................

b) \(2x^3-5x^2+3x=0\)

\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)

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

\(\Leftrightarrow x\left[2x\left(x-1\right)-3\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy .................

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

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

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

\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)

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

Vậy .......................

d) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Vậy ...................

Sửa lại câu d, như @Gia Hân Ngô là đúng rồi đó.Tôi nhầm dấu.

@Nguyễn Việt Lâm

@Khôi Bùi

\(\dfrac{x-3}{2013}+\dfrac{x-2}{2014}=\dfrac{x-2014}{2}+\dfrac{x-2013}{3}\)

\(\Leftrightarrow(\dfrac{x-3}{2013}-1)+(\dfrac{x-2}{2014}-1)=(\dfrac{x-2014}{2}-1)+(\dfrac{x-2013}{3}-1)\)

\(\Leftrightarrow\dfrac{x-2016}{2013}+\dfrac{x-2016}{2014}=\dfrac{x-2016}{2}+\dfrac{x-2016}{3}\)

\(\Leftrightarrow\left(x-2016\right)\left(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x-2016=0\) (Vì \(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\))

\(\Leftrightarrow x=2016\)

Vậy \(S=\left\{2016\right\}\)

\(\dfrac{x-3}{2013}+\dfrac{x-2}{2014}=\dfrac{x-2014}{2}+\dfrac{x-2013}{3}\)

\(\Leftrightarrow\dfrac{x-3}{2013}-1+\dfrac{x-2}{2014}-1=\dfrac{x-2014}{2}-1+\dfrac{x-2013}{3}-1\)

\(\Leftrightarrow\dfrac{x-2016}{2013}+\dfrac{x-2016}{2014}=\dfrac{x-2016}{2}+\dfrac{x-2016}{3}\)

\(\Leftrightarrow\left(x-2016\right)\left(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x-2016=0\) (do \(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\))

\(\Rightarrow x=2016\)

A = \(\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+5\right)^2}{x^2+5x+5}=x^2+5x+5\)B = \(\dfrac{\left|x-1\right|+\left|x\right|+x}{3x^2-4x+1}\)với x < 0

Với x < 0 thì |x-1| = 1-x, |x| = -x, ta có:

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

Mysterious PersonNguyễn Đình DũngNguyễn Thanh Hằng๖ۣۜĐặng♥๖ۣۜQuý

giúp!! 2h30 đi rồi

ĐK: \(x\ne0;\pm1\)

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

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

b/ Để \(A\in Z\Rightarrow3⋮\left(x+1\right)\Rightarrow x+1=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(x+1=-3\Rightarrow x=-4\)

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

\(x+1=1\Rightarrow x=0\left(l\right)\)

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

c/ \(A< 0\Leftrightarrow\dfrac{3}{x+1}< 0\Leftrightarrow x+1< 0\Rightarrow x< -1\)

ta có: \(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)

<=>\(\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right)\left(a+b+c\right)=a+b+c\)

<=>\(\dfrac{a^2}{b+c}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}+\dfrac{b^2}{a+c}+\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{c^2}{a+b}+\dfrac{ac}{a+b}+\dfrac{cb}{a+b}=a+c+b\)

<=>\(\dfrac{a^2}{b+c}+\dfrac{ab+ac}{b+c}+\dfrac{b^2}{a+c}+\dfrac{ab+bc}{a+c}+\dfrac{c^2}{a+b}+\dfrac{ac+cb}{a+b}=a+c+b\)

<=>\(\dfrac{a^2}{b+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{b^2}{a+c}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c^2}{a+b}+\dfrac{c\left(a+b\right)}{a+b}=a+c+b\)

<=>\(\dfrac{a^2}{b+c}+a+\dfrac{b^2}{a+c}+b+\dfrac{c^2}{a+b}+c=a+c+b\)

<=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=a+c+b-a-c-b=0\) (đpcm)

chúc bạn học tốt ^ ^

1)\(-\dfrac{4x-3}{x-5}=\dfrac{29}{3}\Leftrightarrow\dfrac{3-4x}{x-5}=\dfrac{29}{3}\)

\(\Leftrightarrow3\left(3-4x\right)=29\left(x-5\right)\Leftrightarrow9-12x=29x-145\)

\(\Leftrightarrow29x+12x=9+145\Leftrightarrow41x=154\Leftrightarrow x=\dfrac{154}{41}\)

2)\(\dfrac{2x-1}{5-3x}=2\Leftrightarrow2\left(2x-1\right)=5-3x\)

\(\Leftrightarrow4x-2=5-3x\)

\(\Leftrightarrow4x+3x=5+2\Leftrightarrow7x=7\Leftrightarrow x=1\)

3)\(\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)

\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\)

\(\Rightarrow4x-5=2x-2+x\)

\(\Leftrightarrow4x-2x-x=-2+5\)

\(\Leftrightarrow x=3\)

\(1)-\dfrac{4x-3}{x-5}=\dfrac{29}{3} (x \neq 5) \\\Leftrightarrow\dfrac{3-4x}{x-5}=\dfrac{29}{3}\) \(\Leftrightarrow3\left(3-4x\right)=29\left(x-5\right)\\\Leftrightarrow9-12x=29x-145\) \(\Leftrightarrow29x+12x=9+145\\\Leftrightarrow41x=154\\\Leftrightarrow x=\dfrac{154}{41}(TM)\)

Vậy \(S=\left\{\dfrac{154}{41}\right\}\)

\(2)\dfrac{2x-1}{5-3x}=2 (x \neq \dfrac{5}{3}) \)

\(\Leftrightarrow2x-1=2\left(5-3x\right)\\ \Leftrightarrow2x-1=10-6x\\ \Leftrightarrow2x+6x=10+1\\ \Leftrightarrow8x=11\\ \Leftrightarrow x=\dfrac{11}{8}\left(TM\right)\)

Vậy \(S=\left\{\dfrac{11}{8}\right\}\)

\(3)\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1} (x \neq 1) \\\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2\left(x-1\right)}{x-1}+\dfrac{x}{x-1}\) \(\Leftrightarrow4x-5=2x-2+x\) \(\Leftrightarrow4x-2x-x=-2+5\) \(\Leftrightarrow x=3(TM)\)

Vậy \(S=\left\{3\right\}\)

1) \(\left(5x-4\right)\left(4x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)

2) \(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)