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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x^{2\:}-2x+2=t\)
Được phương trình: \(\frac{t}{t+1}+\frac{t-1}{t}=\frac{1}{6}\)
Quy đồng và khử mẫu được: \(12t^2-6=t^2+t\)
<=> \(11t^2-t=6\)
r á. đến đó thỳ hk lm đk n~. pn xem lại đề đy na @@
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\left(x\ne-4;-5;-6;-7;-8\right)\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{x}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)
vậy x=2; x=-13
Bài làm:
đkxđ: \(x\ne\left\{-4;-5;-6;-7\right\}\)
Ta có: \(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow x^2+11x+28=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-13\end{cases}}\)
Vậy tập nghiệm của PT \(S=\left\{-13;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\left(x\ne1\right)\)
\(\Leftrightarrow\frac{1}{x-1}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4}{x^2+x+1}=0\)
\(\Leftrightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{4x-4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow\frac{3x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Rightarrow3x=0\)
=> x=0 (tmđk)
Vậy x=0
giải bất phương trình sau :
a , \(\frac{x+1}{x+7}\ge0\)
b , \(\frac{2x-1}{3x+2}< 0\)
cảm ơn trước nhé !
![](https://rs.olm.vn/images/avt/0.png?1311)
a,TH1:\(\hept{\begin{cases}x+1\ge0\\x+7\ge0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x\ge-1\\x\ge-7\end{cases}}\)\(\Rightarrow x\ge-1\)
TH2:\(\hept{\begin{cases}x+1\le0\\x+7\le0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x\le-1\\x\le-7\end{cases}}\)\(\Rightarrow x\le-7\)
Tập nghiệm của BPT là ...
b,TH1:\(\hept{\begin{cases}2x-1< 0\\3x+2>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x< 1\\3x>-2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< \frac{1}{2}\\x>-\frac{2}{3}\end{cases}}\)\(\Rightarrow-\frac{2}{3}< x< \frac{1}{2}\)
TH2:\(\hept{\begin{cases}2x-1>0\\3x+2< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x>1\\3x< -2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>\frac{1}{2}\\x< -\frac{2}{3}\end{cases}}\)(loại)
Tập nghiệm của BPT....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\frac{6x^2+3}{24}-\frac{10x-4}{24}=\frac{6x^2-6}{24}-\frac{4x-12}{24}\)
\(\Leftrightarrow\frac{6x^2+3-10x+4}{24}=\frac{6x^2-6-4x+12}{24}\)
\(\Leftrightarrow6x^2-10x+7=6x^2-4x+6\)
\(\Leftrightarrow-6x+1=0\)
\(\Rightarrow-6x=-1\)
\(\Leftrightarrow x=\frac{1}{6}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2x-1}{x}+\frac{3-x}{4}=2\)
\(ĐKXĐ:x\ne0\)
\(MTC:4x\)
\(\frac{4\left(2x-1\right)}{4x}+\frac{x\left(3-x\right)}{4x}=\frac{8x}{4x}\)
\(\Rightarrow4\left(2x-1\right)+x\left(x-3\right)=8x\)
\(\Leftrightarrow8x-4+x^2-3x=8x\)
\(\Leftrightarrow8x-4+x^2-3x-8x=0\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Leftrightarrow x^2-4x+x-4=0\)
\(\Leftrightarrow\left(x^2-4x\right)+\left(x-4\right)=0\)
\(\Leftrightarrow x\left(x-4\right)+\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)
Hoặc\(\hept{\begin{cases}x-4=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(N\right)\\x=-1\left(N\right)\end{cases}}}\)
Vậy tập nghiệp của pt là \(S=\left\{-1;4\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{2x}{x-1}+\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}\)ĐKXĐ : \(x\ne1;-3\)
\(\Leftrightarrow\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}+\frac{4}{\left(x-1\right)\left(x+3\right)}=\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{2x^2+6x+4}{\left(x-1\right)\left(x+3\right)}=\frac{2x^2-7x+5}{\left(x-1\right)\left(x+3\right)}\)
\(\Rightarrow2x^2+6x+4=2x^2-7x+5\)
\(\Leftrightarrow2x^2+5x+4-2x^2+7x-5=0\)
\(\Leftrightarrow12x-1=0\)
\(\Leftrightarrow x=\frac{1}{12}\)( thỏa mãn ĐKXĐ )
b) c) tương tự
ĐK: x khác 1 và - 1
\(\frac{x+1}{2\left(x-1\right)}-\frac{x-1}{2\left(x+1\right)}=\frac{2}{\left(x-1\right)\left(x+1\right)}\)
<=> \(\frac{\left(x+1\right)^2-\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\frac{4}{2\left(x-1\right)\left(x+1\right)}\)
<=> \(\left(x+1\right)^2-\left(x-1\right)^2=4\)
<=> \(\left(x+1-x+1\right)\left(x+1+x-1\right)=4\)
<=> 2.2x = 4
<=> x = 1 loại
Vậy phương trình vô nghiệm
e trả lời sau đc ko ạ ? ):
\(\frac{x+1}{2x-2}-\frac{x-1}{2x+2}=\frac{2}{x^2-1}\) ĐKXĐ : \(x\ne\pm1\)
\(\frac{x+1}{2\left(x-1\right)}-\frac{x-1}{2\left(x+1\right)}=\frac{2}{\left(x-1\right)\left(x+1\right)}\)
\(\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{2\left(x+1\right)\left(x-1\right)}=\frac{4}{2\left(x+1\right)\left(x-1\right)}\)
Khử mẫu ta đc : \(\left(x+1\right)^2-\left(x-1\right)^2=4\)
\(4x=4\Leftrightarrow x=1\)Theo ĐKXĐ : ktm
Vậy pt vô nghiệm.