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8 tháng 1 2017
\(\frac{x-1}{x^2-9x+20}+\frac{2x-2}{x^2-6x+8}+\frac{3x-3}{x^2-x-2}+\frac{4x-4}{x^2+6x+5}=0\)
\(\Leftrightarrow\frac{x-1}{\left(x-5\right)\left(x-4\right)}+\frac{2\left(x-1\right)}{\left(x-4\right)\left(x-2\right)}+\frac{3\left(x-1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{10}{x^2-25}\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
PS: Điều kiện xác đinh bạn tự làm nhé
7 tháng 1 2017
từ đề\(\Leftrightarrow\frac{x-1}{x\left(x-4\right)-5\left(x-4\right)}+\frac{2x-2}{x\left(x-2\right)-4\left(x-2\right)}+\frac{3x-3}{x\left(x+1\right)-2\left(x+1\right)}+\frac{4x-4}{x\left(x+1\right)+5\left(x+5\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{2}{\left(x-2\right)\left(x-4\right)}+\frac{3}{\left(x-2\right)\left(x+1\right)}+\frac{4}{\left(x+1\right)\left(x+5\right)}=0\right)\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{x-4}-\frac{1}{x-5}+\frac{1}{x-2}-\frac{1}{x-4}+\frac{1}{x-2}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x-5}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{2}{x-2}-\frac{2}{x-5}\right)=0\) vì \(\frac{2}{x-2}-\frac{2}{x-5}\)luôn khác 0 nên x-1=0 nên x=1.
Điều kiện xác định : x khác 4,5,2,-1. Do đó x=1 thỏa mãn. Vậy x=1
LA
17 tháng 3 2019
Đặt
6x+7 = 7 , ta có
\(\left(t+1\right)\left(t-1\right)t^2=72\Rightarrow\left(t^2-1\right)t^2=72\)
\(\Rightarrow t^4-t^2-72=0\)
Lại đặt \(t^2=a\) (a \(\ge0\) )
\(\Rightarrow a^2-a-72=0\Rightarrow\left(a+8\right)\left(a-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-8\left(ktm\right)\\a=9\left(tm\right)\end{matrix}\right.\)
a = 9 => \(\left[{}\begin{matrix}t=3\\t=-3\end{matrix}\right.\)
Với t = 3
=> 6x + 7 =3
=> 6x = -4
=> x= \(-\frac{2}{3}\)
Với t = -3
=> 6x + 7 = -3
=> 6x = -10
=> x = \(-\frac{5}{3}\)
Vậy.....
b)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x-4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\Rightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{\left(x+7\right)\left(x+4\right)}=\frac{1}{18}\Rightarrow x^2+11x+28-54=0\Rightarrow x^2+11x-26=0\)
\(\Rightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
17 tháng 3 2019
a) Ta có:
(6x+8)(6x+6)(6x+7)2 = 72
Đặt \(6x+7=a\)
\(\Rightarrow\left(a+1\right)\left(a-1\right)a^2=72\)
\(\Leftrightarrow a^4-a^2-72=0\)
\(\Leftrightarrow\left(a^4+8a^2\right)+\left(-9a^2-72\right)=0\)
\(\Leftrightarrow\left(a^2+8\right)\left(a^2-9\right)=0\)
Đễ thấy \(a^2+8>0\)
\(\Rightarrow a^2-9=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x+7=3\\6x+7=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)
b)
AT
23 tháng 8 2018
c/ đk: x khác 1; x khác -3
\(\dfrac{3x-1}{x-1}+\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
\(\Rightarrow\left(3x+1\right)\left(x+3\right)+\left(2x+5\right)\left(x-1\right)+4=x^2+2x-3\)
\(\Leftrightarrow3x^2+10x+3+2x^2+3x-5+4=x^2+2x-3\)
\(\Leftrightarrow4x^2+11x+5=0\)
\(\Leftrightarrow\left(4x^2+2\cdot2x\cdot\dfrac{11}{4}+\dfrac{121}{16}\right)-\dfrac{41}{16}=0\)
\(\Leftrightarrow\left(2x+\dfrac{11}{4}\right)^2=\dfrac{41}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{11}{4}=\dfrac{\sqrt{41}}{4}\\2x+\dfrac{11}{4}=-\dfrac{\sqrt{41}}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{41}}{8}\\x=\dfrac{-11-\sqrt{41}}{8}\end{matrix}\right.\)
Vậy.........
d/ \(\dfrac{12x+1}{6x-2}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
đk: \(x\ne\pm\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{12x+1}{2\left(3x-1\right)}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Rightarrow2\left(12x+1\right)\left(3x+1\right)-4\left(9x-5\right)\left(3x-1\right)=108x-36x^2-9\)
\(\Leftrightarrow72x^2+24x+6x+2-108x^2+36x-60x-20-108x+36x^2+9=0\)
\(\Leftrightarrow-102x-9=0\)
\(\Leftrightarrow-102x=9\Leftrightarrow x=-\dfrac{3}{34}\)(TM)
Vậy.........
AT
23 tháng 8 2018
a/ \(\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\)
\(\Leftrightarrow\left(x+1\right)^2\left(x+2+x-2\right)=-24\)
\(\Leftrightarrow2x\left(x^2+2x+1\right)=-24\)
\(\Leftrightarrow2x^3+4x^2+2x+24=0\)
\(\Leftrightarrow2x^3-2x^2+8x+6x^2-6x+24=0\)
\(\Leftrightarrow x\left(2x^2-2x+8\right)+3\left(2x^2-2x+8\right)=0\)
\(\Leftrightarrow\left(2x^2-2x+8\right)\left(x+3\right)=0\)
\(\Leftrightarrow2\left(x^2-x+4\right)\left(x+3\right)=0\)
Ta thấy: \(x^2-x+4=\left(x^2-2x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)
=> x+ 3 = 0 <=> x= -3
Vậy......
b/ \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+x^2+5x+2x^2+x+5=0\)
\(\Leftrightarrow x\left(2x^2+x+5\right)+\left(2x^2+x+5\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
Ta thấy: \(2x^2+x+5=\left(\sqrt{2}x+2\cdot\sqrt{2}x\cdot\dfrac{\sqrt{2}}{4}+\dfrac{1}{8}\right)+\dfrac{39}{8}=\left(\sqrt{2}x+\dfrac{\sqrt{2}}{4}\right)^2+\dfrac{39}{8}>0\)
=> x + 1 = 0 <=> x = -1
Vậy....
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 3 2019
a/ Đặt \(6x+7=a\Rightarrow\left\{{}\begin{matrix}6x+8=a+1\\6x+6=a-1\end{matrix}\right.\)
\(\Rightarrow\left(a-1\right)\left(a+1\right)a^2-72=0\)
\(\Leftrightarrow\left(a^2-1\right)a^2-72=0\)
\(\Leftrightarrow a^4-a^2-72=0\)
\(\Leftrightarrow\left(a^2-9\right)\left(a^2+8\right)=0\)
\(\Leftrightarrow a^2=9\) (do \(a^2+8>0\))
\(\Rightarrow\left[{}\begin{matrix}a=3\\a=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}6x+7=3\\6x+7=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ne-4;-5;-6;-7\)
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow x^2+11x-26=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
đề sai
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