Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(\frac{x^2-5x+1}{2x+1}+2=\frac{x^2-5x+4x+1+2}{2x+1}\)
=\(\frac{x^2-x+3}{2x+1}=\frac{x^2-4x+1}{x+1}\)
=> (x2 - x +3)(x+1)=(x2 - 4x+1)(2x+1)
=>x3 +2x+3=2x3-7x2-2x+1
=>0=x3-7x2-4x-2
Đây là cách làm của mình :
\(\Leftrightarrow\frac{x^2-5x+1}{2x+1}+1+1=\frac{x^2-4x+1}{x+1}\)
\(\Leftrightarrow\frac{x^2-5x+1}{2x+1}+1=\frac{x^2-4x+1}{x+1}-1\)
\(\Leftrightarrow\frac{x^2-3x+2}{2x+1}=\frac{x^2-5x}{x+1}\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{2x+1}=\frac{x^2-5x}{x+1}\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=\left(2x+1\right)\left(x^2-5x\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-1\right)=\left(2x+1\right)\left(x^2-5x\right)\)
Bạn tự nhân phân phối vào nha :
\(\Leftrightarrow x^3-2x^2-x+2=2x^3-9x^2-5x\)
\(\Leftrightarrow x^3-7x^2-4x-2=0\)
Đến đây chỉ có nước bấm máy tính thôi chứ phân tích bình thường không ra được đâu
CASIO fx-570VN PLUS : Mode --> 5 --> 4 : giải pt bậc 3 một ẩn
Kết quả cho là x = 7.563793497...
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 \(x=1;x=2;x=-\frac{2}{3}\)
\(\Leftrightarrow\frac{x^2-4x+1}{x+1}+1=-\frac{x^2-5x+1}{2x+1}-1.DKXD:x\ne-1;x\ne-\frac{1}{2}\)
\(\Leftrightarrow\frac{x^2-3x+2}{x+1}=\frac{-x^2+3x-2}{2x+1}\)
\(\Leftrightarrow\frac{x^2-3x+2}{x+1}+\frac{x^2-3x+2}{2x+1}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{x+1}+\frac{1}{2x+1}\right)=0\)
\(\Leftrightarrow\left(x^2-x-2x+2\right)\left[\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left[\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}\right]=0\)
\(\Leftrightarrow x-1=0\Leftrightarrow x=1\left(n\right)\)
\(hay:x-2=0\Leftrightarrow x=2\left(n\right)\)
\(hay:\frac{3x+2}{\left(x+1\right)\left(2x+1\right)}=0\Leftrightarrow3x+2=0\Leftrightarrow x=-\frac{2}{3}\left(n\right)\)
\(V...S=\left\{1:2:-\frac{2}{3}\right\}\)
Bài làm
a) \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x-4}\)
\(\Leftrightarrow\frac{3x+2}{3x-2}-\frac{6}{3x+2}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Leftrightarrow\frac{(3x+2)\left(3x+2\right)}{(3x-2)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{(3x+2)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)
\(\Rightarrow\left(3x+2\right)^2-\left(18x-12\right)=9x^2\)
\(\Leftrightarrow9x^2+12x+4-18x+12x-9x^2=0\)
\(\Leftrightarrow6x+4=0\)
\(\Leftrightarrow x=-\frac{4}{6}\)
\(\Leftrightarrow x=-\frac{2}{3}\)
Vậy x = -2/3 là nghiệm.
@Tao Ngu :))@ 9x-4 không tách thành (3x+4)(3x-4) được đâu bạn. Chỗ đó phải là: 9x2-4
Bài thiếu đkxđ của x \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)
$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$###############################@@@@@@@@@@@@@@@@@@@@@@@$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$###############################@@@@@@@@@@@@@@@@@@@@@@@
\(x^2-4x+\frac{1}{x+1}+2=-x^2-5x+\frac{1}{2x+1}\left(ĐK:x\ne-1;-\frac{1}{2}\right)\)
\(< =>x^2-4x+\frac{1}{x+1}+2+x^2+5x-\frac{1}{2x+1}=0\)
\(< =>2x^2+x+\frac{2x+3}{x+1}-\frac{1}{2x+1}=0\)
\(< =>2x^2+x=\frac{1}{2x+1}-\frac{2x+3}{x+1}\)
\(< =>2x^2+x=\frac{x+1-\left(2x+1\right)\left(2x+1\right)+4x+2}{\left(x+1\right)\left(x+1\right)+x^2+x}\)
\(< =>2x^2+x=\frac{x+1-4x^2-4x-1+4x+2}{x^2+2x+1+x^2+x}\)
\(< =>2x^2+x=\frac{x-4x^2+2}{2x^2+3x+1}\)
\(< =>\left(2x^2+x\right)^2+\left(2x+1\right)^2x=x-4x^2+2\)
\(< =>4x^4+8x^3+9x^2-2=0\)
nhờ bạn nào đó giải giúp ạ
\(ĐKXĐ:x\ne-1;x\ne-\frac{1}{2}\)
\(PT:\Leftrightarrow\frac{x^2-4x+1}{x+1}+1+\frac{x^2-5x+1}{2x+1}=0\)
\(\Leftrightarrow\frac{x^2-3x+2}{x+1}+\frac{x^2-3x+2}{2x+1}=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(\frac{1}{x+1}+\frac{1}{2x+1}\right)=0\)
\(\Leftrightarrow\left(x^2-3x+2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(3x+2\right)=0\)
\(x-1=0\Leftrightarrow x=1\)
\(x-2=0\Leftrightarrow x=2\)
\(3x+2=0\Leftrightarrow3x=-2\Leftrightarrow x=-\frac{2}{3}\)
\(\Rightarrow\hept{\begin{cases}x=1\\x=2\\x=-\frac{2}{3}\end{cases}}\)
\(\frac{x^2-4x+1}{x+1}+2=-\frac{x^2-5x+1}{2x+1}\)
\(\Leftrightarrow\left(x^2-4x+1\right)\left(x+1\right)+2\left(x+1\right)\left(2x+1\right)=-\left(x^2-5x+1\right)\left(x+1\right)\)
\(\Leftrightarrow2x^3-3x^2+4x+3=-x^3+4x^2+4x-1\)
\(\Leftrightarrow2x^3-3x^2+3+x^2-4x+1=0\)
\(\Leftrightarrow3x^2-7x^2+4=0\)
\(\Leftrightarrow\left(3x^2-4x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x^2+2x-6x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[x\left(3x+2\right)-2\left(3x+2\right)\right]\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}3x+2=0\\x-2=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{2}{3}\\x=2\\x=1\end{cases}}\)
vậy:...