Giải Phương Trình : \(\frac{x^2-4x+1}{x+1}+2=\frac{x^2-5x+1}{2x+1}\)
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S
2
2 tháng 8 2017
ĐK \(x\ne\left\{-1;-\frac{1}{2}\right\}\)
Phương trình \(\Leftrightarrow\frac{x^2-4x+1}{x+1}+1=\frac{-x^2+5x-1}{2x+1}-1\)\(\Leftrightarrow\frac{x^2-4x+1+x+1}{x+1}=\frac{-x^2+5x-1-2x-1}{2x+1}\)
\(\Leftrightarrow\frac{x^2-3x+2}{x+1}=\frac{-\left(x^2-3x+2\right)}{2x+1}\Leftrightarrow\left(x^2-3x+2\right)\left[\frac{1}{x+1}+\frac{1}{2x+1}\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-3x+2=0\\\frac{1}{x+1}+\frac{1}{2x+1}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)\left(x-2\right)=0\\\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}=0\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1;x=2\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1;x=2\\x=-\frac{2}{3}\end{cases}}\left(tm\right)}\)
Vậy hệ có 3 nghiệm \(x=1;x=2;x=-\frac{2}{3}\)
3 tháng 3 2019
\(\Leftrightarrow\frac{x^2-4x+1}{x+1}+1=-\frac{x^2-5x+1}{2x+1}-1.DKXD:x\ne-1;x\ne-\frac{1}{2}\)
\(\Leftrightarrow\frac{x^2-3x+2}{x+1}=\frac{-x^2+3x-2}{2x+1}\)
\(\Leftrightarrow\frac{x^2-3x+2}{x+1}+\frac{x^2-3x+2}{2x+1}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{x+1}+\frac{1}{2x+1}\right)=0\)
\(\Leftrightarrow\left(x^2-x-2x+2\right)\left[\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left[\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}\right]=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\left(n\right)\)
\(hay:x-2=0\Leftrightarrow x=2\left(n\right)\)
\(hay:\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}=0\Leftrightarrow3x+2=0\Leftrightarrow x=-\frac{2}{3}\left(n\right)\)
\(V...S=\left\{1:2:-\frac{2}{3}\right\}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 6 2020
b/ ĐKXĐ: ...
\(2x^3-2y^3+5x-5y=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2\right)+5\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(2x^2+2xy+2y^2+5\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left[\left(x+y\right)^2+x^2+y^2+5\right]=0\)
\(\Leftrightarrow x=y\) (ngoặc sau luôn dương)
Thế vào pt dưới:
\(\frac{3x}{x^2+x+1}+\frac{5x}{x^2+3x+1}=2\)
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{3}{x+\frac{1}{x}+1}+\frac{5}{x+\frac{1}{x}+3}=2\)
Đặt \(x+\frac{1}{x}+1=t\)
\(\Rightarrow\frac{3}{t}+\frac{5}{t+2}=2\Leftrightarrow3\left(t+2\right)+5t=2t\left(t+2\right)\)
\(\Leftrightarrow2t^2-4t-6=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}+1=-1\\x+\frac{1}{x}+1=3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=0\\x^2-2x+1=0\end{matrix}\right.\) \(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 6 2020
a/ ĐKXĐ: ...
\(2x-\frac{1}{y}=2y-\frac{1}{x}\Leftrightarrow\frac{2xy-1}{y}=\frac{2xy-1}{x}\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\2xy-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\xy=\frac{1}{2}\end{matrix}\right.\)
TH1: \(x=y\Rightarrow6x^2=7x^2-8\Rightarrow x^2=8\Rightarrow...\)
TH2: \(xy=\frac{1}{2}\Rightarrow y=\frac{1}{2x}\)
\(\Rightarrow2\left(2x^2+\frac{1}{4x^2}\right)+4\left(x-\frac{1}{2x}\right)=\frac{7}{2}-8\)
\(\Leftrightarrow4\left(x^2+\frac{1}{4x^2}\right)+8\left(x-\frac{1}{2x}\right)+9+4x^2=0\)
Đặt \(x-\frac{1}{2x}=t\Rightarrow x^2+\frac{1}{4x^2}=t^2+1\)
\(\Rightarrow4\left(t^2+1\right)+8t+9+4x^2=0\)
\(\Leftrightarrow4\left(t+1\right)^2+4x^2+9=0\)
Vế trái luôn dương nên pt vô nghiệm
NL
2
HN
12 tháng 7 2020
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PN
12 tháng 7 2020
\(x^2-4x+\frac{1}{x+1}+2=-x^2-5x+\frac{1}{2x+1}\left(ĐK:x\ne-1;-\frac{1}{2}\right)\)
\(< =>x^2-4x+\frac{1}{x+1}+2+x^2+5x-\frac{1}{2x+1}=0\)
\(< =>2x^2+x+\frac{2x+3}{x+1}-\frac{1}{2x+1}=0\)
\(< =>2x^2+x=\frac{1}{2x+1}-\frac{2x+3}{x+1}\)
\(< =>2x^2+x=\frac{x+1-\left(2x+1\right)\left(2x+1\right)+4x+2}{\left(x+1\right)\left(x+1\right)+x^2+x}\)
\(< =>2x^2+x=\frac{x+1-4x^2-4x-1+4x+2}{x^2+2x+1+x^2+x}\)
\(< =>2x^2+x=\frac{x-4x^2+2}{2x^2+3x+1}\)
\(< =>\left(2x^2+x\right)^2+\left(2x+1\right)^2x=x-4x^2+2\)
\(< =>4x^4+8x^3+9x^2-2=0\)
nhờ bạn nào đó giải giúp ạ
7 tháng 2 2018
ĐKXĐ : \(\hept{\begin{cases}x^2+x-6\ne0\\x^2+4x+3\ne0\\2x-1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+3\right)\left(x-2\right)\ne0\\\left(x+1\right)\left(x+3\right)\ne0\\x\ne\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x\ne2;-3\\x\ne-1;-3\\x\ne\frac{1}{2}\end{cases}}}}\)
TXĐ : \(x\ne\left\{-3;-1;\frac{1}{2};2\right\}\)
\(pt\Leftrightarrow\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3x+9}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{1}{x^2-x-2}=\frac{1}{1-2x}\)
\(\Leftrightarrow x^2-x-2-1+2x=0\)
\(\Leftrightarrow x^2+x-3=0\)
\(\Leftrightarrow\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{1-\sqrt{13}}{2}\right)\left(x+\frac{1+\sqrt{13}}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{2}\\x=\frac{-\sqrt{13}-1}{2}\end{cases}}\)
BT
7 tháng 2 2018
\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4+3}=-\frac{3}{2x-1}\)
<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3x+9}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{1}{x-2}=-\frac{1}{2x-1}\)
<=> x-2=1-2x <=> 3x=3
=> x=1
Đáp số: x=1
H
5 tháng 5 2019
\(\frac{2x+1}{x^2-5x+4}+\frac{5}{x-1}=\frac{2}{x-4}\)ĐKXĐ : \(x\ne1;4\)
\(\Leftrightarrow\frac{2x+1}{\left(x-1\right)\left(x-4\right)}+\frac{5\left(x-4\right)}{\left(x-1\right)\left(x-4\right)}=\frac{2\left(x-1\right)}{\left(x-1\right)\left(x-4\right)}\)
\(\Leftrightarrow2x+1+5x-20=2x-2\)
\(\Leftrightarrow2x+5x-2x=-1+20-2\)
\(\Leftrightarrow5x=17\)
\(\Leftrightarrow x=\frac{17}{5}\)
KL : Nghiệm của PT là S={ 17/5 }
H
5 tháng 5 2019
\(\frac{7}{8x}-\frac{x-5}{4x^2-8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\) ĐKXĐ : \(x\ne0;2\)
\(\Leftrightarrow\frac{7}{8x}-\frac{x-5}{4x\left(x-2\right)}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8\left(x-2\right)}\)
\(\Leftrightarrow\frac{7\left(x-2\right)}{8x\left(x-2\right)}-\frac{2\left(x-5\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)
\(\Leftrightarrow7x-14-2x+10=4x-4+x\)
\(\Leftrightarrow7x-2x-4x-x=14-10-4\)
\(\Leftrightarrow0x=0\)
=> PT vô số nghiệm
\(\frac{x^2-4x+1}{x+1}+2=\frac{x^2-5x+1}{2x+1}\)
\(\Leftrightarrow\frac{\left(x^2-4x+1\right)\left(2x+1\right)+2\left(x+1\right)\left(2x+1\right)}{\left(x+1\right)\left(2x+1\right)}=\frac{\left(x^2-5x+1\right)\left(x+1\right)}{\left(2x+1\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{2x^3+x^2-8x^2-4x+2x+1+2\left(2x^2+x+2x+1\right)}{\left(x+1\right)\left(2x+1\right)}=\frac{x^3+x^2-5x^2-5x+x+1}{\left(2x+1\right)\left(x+1\right)}\)
\(\Rightarrow2x^3-7x^2-2x+1+4x^2+2x+4x+2=x^3-4x^2-4x+1\)
\(\Leftrightarrow2x^3-3x^2+4x+3-x^3+4x^2+4x-1=0\)
\(\Leftrightarrow x^3+x^2+8x-2=0\)