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\(x^2-4x+\frac{1}{x+1}+2=-x^2-5x+\frac{1}{2x+1}\left(ĐK:x\ne-1;-\frac{1}{2}\right)\)
\(< =>x^2-4x+\frac{1}{x+1}+2+x^2+5x-\frac{1}{2x+1}=0\)
\(< =>2x^2+x+\frac{2x+3}{x+1}-\frac{1}{2x+1}=0\)
\(< =>2x^2+x=\frac{1}{2x+1}-\frac{2x+3}{x+1}\)
\(< =>2x^2+x=\frac{x+1-\left(2x+1\right)\left(2x+1\right)+4x+2}{\left(x+1\right)\left(x+1\right)+x^2+x}\)
\(< =>2x^2+x=\frac{x+1-4x^2-4x-1+4x+2}{x^2+2x+1+x^2+x}\)
\(< =>2x^2+x=\frac{x-4x^2+2}{2x^2+3x+1}\)
\(< =>\left(2x^2+x\right)^2+\left(2x+1\right)^2x=x-4x^2+2\)
\(< =>4x^4+8x^3+9x^2-2=0\)
nhờ bạn nào đó giải giúp ạ
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 \(\hept{\begin{cases}3x-2\ne0\\2+3x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}3x\ne2\\3x\ne-2\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{2}{3}\\x\ne\frac{-2}{3}\end{cases}\Leftrightarrow}x\ne\pm\frac{2}{3}}\)
\(\frac{2x+1}{x^2-5x+4}+\frac{5}{x-1}=\frac{2}{x-4}\)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 }
\(\frac{7}{8x}-\frac{x-5}{4x^2-8x}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8x-16}\) 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
\(\frac{x^2-4x+1}{x+1}+2=\frac{x^2-5x+1}{2x+1}\)
\(\Leftrightarrow\frac{\left(x^2-4x+1\right)\left(2x+1\right)+2\left(x+1\right)\left(2x+1\right)}{\left(x+1\right)\left(2x+1\right)}=\frac{\left(x^2-5x+1\right)\left(x+1\right)}{\left(2x+1\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{2x^3+x^2-8x^2-4x+2x+1+2\left(2x^2+x+2x+1\right)}{\left(x+1\right)\left(2x+1\right)}=\frac{x^3+x^2-5x^2-5x+x+1}{\left(2x+1\right)\left(x+1\right)}\)
\(\Rightarrow2x^3-7x^2-2x+1+4x^2+2x+4x+2=x^3-4x^2-4x+1\)
\(\Leftrightarrow2x^3-3x^2+4x+3-x^3+4x^2+4x-1=0\)
\(\Leftrightarrow x^3+x^2+8x-2=0\)
Cái bài đầu giải BPT bn ghi cái dj ak ,mik cx k hỉu nữa
V mik giải bài 2 nghen, sửa lại đề bài đầu rồi mik giải cho
\(3x-3=|2x+1|\)
Điều kiện: \(3x-3\ge0\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=3x-3\\2x+1=-3x+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1-3\\2x+3x=-1+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-3\\5x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\left(n\right)\\x=\frac{2}{5}\left(l\right)\end{cases}}}\)
Vậy S={3}
Cài đề câu b ,bn xem lại nhé!
\(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}>\frac{x^2}{7}-\frac{2x-3}{5}\)
\(\Leftrightarrow\frac{2x-3}{35}+\frac{5x\left(x-2\right)}{35}-\frac{5x^2}{35}+\frac{7\left(2x-3\right)}{35}>0\)
\(\Leftrightarrow2x-3+5x\left(x-2\right)-5x^2+7\left(2x-3\right)>0\)
\(\Leftrightarrow2x-3+5x^2-10x-5x^2+14x-21>0\)
\(\Leftrightarrow6x-24>0\)
\(\Leftrightarrow x>4\)
VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG TRÌNH LÀ : S = { \(x\text{\x}>4\)}
\(\frac{6x+1}{18}+\frac{x+3}{12}\le\frac{5x+3}{6}+\frac{12-5x}{9}\)
\(\Leftrightarrow\frac{6\left(6x+1\right)}{108}+\frac{9\left(x+3\right)}{108}\le\frac{18\left(5x+3\right)}{108}+\frac{12\left(12-5x\right)}{108}\)
\(\Leftrightarrow36x+6+9x+27\le90x+54+144-60x\)
\(\Leftrightarrow36x+6+9x+27-90x-54-144+60x\le0\)
\(\Leftrightarrow15x-165\le0\)
\(\Leftrightarrow x\le11\)
VẬY TẬP NGHIỆM CỦA BẤT PHƯƠNG trình ..........
tk mk nka !!! chúc bạn học tốt !!!
Sửa đề +2 thành +1
\(\Leftrightarrow\frac{x^2-5x+1}{2x+1}=\frac{x^2-4x+1}{x+1}-1\)