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NH
1
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 3 2019
\(\frac{1}{x^2-2x+2}-1+\frac{2}{x^2-2x+3}-1+2-\frac{6}{x^2-2x+4}=0\)
\(\Leftrightarrow\frac{-x^2+2x-1}{x^2-2x+2}+\frac{-x^2+2x-1}{x^2-2x+3}+\frac{2\left(x^2-2x+1\right)}{x^2-2x+4}=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(\frac{2}{x^2-2x+4}-\frac{1}{x^2-2x+2}-\frac{1}{x^2-2x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x+1=0\Rightarrow x=1\\\frac{2}{x^2-2x+4}-\frac{1}{x^2-2x+2}-\frac{1}{x^2-2x+3}=0\left(1\right)\end{matrix}\right.\)
Xét (1), đặt \(a=x^2-2x+3\) pt trở thành:
\(\frac{2}{a+1}-\frac{1}{a-1}-\frac{1}{a}=0\Leftrightarrow\frac{2\left(a-1\right)-\left(a+1\right)}{\left(a^2-1\right)}-\frac{1}{a}=0\)
\(\Leftrightarrow\frac{a-3}{a^2-1}=\frac{1}{a}\Leftrightarrow a^2-3a=a^2-1\Leftrightarrow3a=1\Rightarrow a=\frac{1}{3}\)
\(\Rightarrow x^2-2x+3=\frac{1}{3}\Leftrightarrow x^2-2x+1+\frac{5}{3}=0\)
\(\Leftrightarrow\left(x-1\right)^2+\frac{5}{3}=0\) (vô nghiệm)
Vậy \(x=1\)
AV
12 tháng 3 2019
\(\left(x-1\right)^2+\frac{5}{3}=0\) (ko thỏa đk )
ms đúng. chứ vẫn có no mà!!
TH
1
24 tháng 3 2020
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x};x\ne2;x\ne0\)
\(\Leftrightarrow\frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x^2-2x}=0\)
\(\Leftrightarrow\frac{x+2}{x-2}-\frac{1}{x}-\frac{2}{x\times\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x\times\left(x+2\right)-\left(x-2\right)-2}{x\times\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x\times\left(x+2\right)-x+2-2}{x\times\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x^2+2x-x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x^2+x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x+1}{x-2}=0\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
1 tháng 6 2020
\(\frac{3x+2}{x+4}+\frac{2x+1}{x-2}=5-\frac{x-32}{x^2+2x-8}\)
\(\Leftrightarrow\) \(\frac{\left(3x+2\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}+\frac{\left(2x+1\right)\left(x+4\right)}{\left(x+4\right)\left(x-2\right)}=\frac{5\left(x+4\right)\left(x-2\right)}{\left(x+4\right)\left(x-2\right)}-\frac{x-32}{\left(x+4\right)\left(x-2\right)}\)
\(\Rightarrow\) (3x + 2)(x - 2) + (2x + 1)(x + 4) = 5(x + 4)(x - 2) - x + 32
\(\Leftrightarrow\) 3x2 - 6x + 2x - 4 + 2x2 + 8x + x + 4 = 5x2 - 10x + 20x - 40 - x + 32
\(\Leftrightarrow\) 5x2 + 5x = 5x2 + 9x - 8
\(\Leftrightarrow\) 5x2 + 5x - 5x2 - 9x + 8 = 0
\(\Leftrightarrow\) -4x + 8 = 0
\(\Leftrightarrow\) x - 2 = 0
\(\Leftrightarrow\) x = 2
Vậy S = {2}
\(\frac{x+2m}{x+3}+\frac{x-m}{x-3}=\frac{mx\left(x+1\right)}{x^2-9}\) (đkxđ: x \(\ne\) \(\pm\) 3)
\(\Leftrightarrow\) \(\frac{\left(x+2m\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-m\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{mx\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}\)
\(\Rightarrow\) (x + 2m)(x - 3) + (x - m)(x + 3) = mx(x + 1)
\(\Leftrightarrow\) x2 - 3x + 2mx - 6m + x2 + 3x - mx - 3m - mx2 - mx = 0
\(\Leftrightarrow\) (2 - m)x2 - 9m = 0
Thay m = 1 ta được:
(2 - 1)x2 - 9 . 1 = 0
\(\Leftrightarrow\) x2 - 9 = 0
\(\Leftrightarrow\) (x - 3)(x + 3) = 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(KTM\right)\\x=-3\left(KTM\right)\end{matrix}\right.\)
Vậy S = \(\varnothing\)
Thay m = 2 ta được:
(2 - 2)x2 - 9 . 2 = 0
\(\Leftrightarrow\) -18 = 0
\(\Rightarrow\) Pt vô nghiệm
Vậy S = \(\varnothing\)
Chúc bn học tốt!!
DQ
0
6 tháng 2 2017
1) Nhìn cái pt hết ham, nhưng bấm nghiệm đẹp v~`~
\(\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)=2x\sqrt{2}-\sqrt{2}\)
\(\Leftrightarrow\left(\sqrt{2}+2\right)\left(x\sqrt{2}-1\right)-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-\sqrt{2}+2x\sqrt{2}-2-2x\sqrt{2}+\sqrt{2}=0\)
\(\Leftrightarrow2x-2=0\Leftrightarrow2x=2\Rightarrow x=1\)
MM
1
DH
14 tháng 2 2020
Đề phải vậy chứ nhỉ?
\(\frac{1}{x-1}+\frac{3x^2}{1-x^3}=\frac{2x}{x^2+x+1}\left(Đkxđ:x\ne1\right)\)
\(\Leftrightarrow x^2+x+1-3x^2=2x\left(x-1\right)\)
\(\Leftrightarrow x^2+x+1-3x^2=2x^2-2x\)
\(\Leftrightarrow4x^2-3x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktmđk\right)\\x=-\frac{1}{4}\left(tmđk\right)\end{matrix}\right.\)
Vậy ...........
28 tháng 2 2020
a) ĐKXĐ: x≠0
Ta có: \(\frac{9}{x}+2=-6\)
⇔\(\frac{9}{x}+2+6=0\)
⇔\(\frac{9}{x}+8=0\)
⇔\(\frac{9}{x}+\frac{8x}{x}=0\)
⇔9+8x=0
⇔8x=-9
hay \(x=-\frac{9}{8}\)
Vậy: \(x=-\frac{9}{8}\)
b) ĐKXĐ: x≠0;x≠-1;x≠-3
Ta có: \(\frac{7}{x+1}+\frac{-18x}{x\left(x^2+4x+3\right)}=\frac{-4}{x+3}\)
⇔\(\frac{7}{x+1}+\frac{-18x}{x\left(x+1\right)\left(x+3\right)}-\frac{-4}{x+3}=0\)
⇔\(\frac{7x\left(x+3\right)}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}+\frac{-18x}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}-\frac{-4x\left(x+1\right)}{\left(x+3\right)\cdot x\cdot\left(x+1\right)}=0\)
⇔\(7x^2+21x-18x+4x\left(x+1\right)=0\)
\(\Leftrightarrow7x^2+21x-18x+4x^2+4x=0\)
⇔\(11x^2+7x=0\)
\(\Leftrightarrow x\left(11x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\11x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\11x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\frac{-7}{11}\end{matrix}\right.\)
Vậy: \(x=\frac{-7}{11}\)
c) ĐKXĐ: x≠1; x≠-3
Ta có: \(\frac{3x-1}{x-1}-1=\frac{2x+5}{x+3}+\frac{4}{x^2-2x+3}\)
⇔\(\frac{3x-1}{x-1}-1-\frac{2x+5}{x+3}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
⇔\(\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)
⇔\(\left(3x-1\right)\left(x+3\right)-\left(x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-4=0\)
\(\Leftrightarrow3x^2+9x-x-3-\left(x^2+3x-x-3\right)-\left(2x^2-2x+5x-5\right)-4=0\)
\(\Leftrightarrow3x^2+8x-3-\left(x^2+2x-3\right)-\left(2x^2+3x-5\right)-4=0\)
\(\Leftrightarrow3x^2+8x-3-x^2-2x+3-2x^2-3x+5-4=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow3x=-1\)
hay \(x=\frac{-1}{3}\)
Vậy: \(x=\frac{-1}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 5 2019
\(x\ne\left\{-4;-3;-2;-1\right\}\)
\(\Leftrightarrow\frac{x^2+x+1}{x+1}-1+\frac{x^2+2x+2}{x+2}-1=\frac{x^2+3x+3}{x+3}-1+\frac{x^2+4x+4}{x+4}-1\)
\(\Leftrightarrow\frac{x^2}{x+1}+\frac{x^2+x}{x+2}-\frac{x^2+2x}{x+3}-\frac{x^2+3x}{x+4}=0\)
\(\Leftrightarrow x\left(\frac{x}{x+1}+\frac{x+1}{x+2}-\frac{x+2}{x+3}-\frac{x+3}{x+4}\right)=0\)
\(\Leftrightarrow x\left(1-\frac{1}{x+1}+1-\frac{1}{x+2}+\frac{1}{x+3}-1+\frac{1}{x+4}-1\right)=0\)
\(\Leftrightarrow x\left(\frac{1}{x+3}+\frac{1}{x+4}-\frac{1}{x+1}-\frac{1}{x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\frac{1}{x+3}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+4}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\frac{-2}{\left(x+1\right)\left(x+3\right)}=\frac{2}{\left(x+2\right)\left(x+4\right)}\)
\(\Leftrightarrow\left(x+2\right)\left(x+4\right)+\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow2x^2+10x+11=0\Rightarrow x=\frac{-5\pm\sqrt{3}}{2}\)
\(\frac{x+2}{x-2}\)-\(\frac{1}{x}\)=\(\frac{2}{x^2-2x}\)
\(\frac{x+2}{x-2}\)-\(\frac{1}{x}\)=\(\frac{2}{x\left(x-2\right)}\) \(\frac{\left(x+2\right)x}{\left(x-2\right)x}\)-\(\frac{x-2}{x\left(x-2\right)}\)=\(\frac{2}{x\left(x-2\right)}\) x(x+2)-x+2=2 x2+2x-x+2=2 x2+2x-x=2-2 x2+x=0 x(x+1)=0 x=0 hoặc x+1=0 x=0 hoặc x=-1\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\left(x\ne0;x\ne2\right)\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(kot/m\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)
Vậy pt có nghiệm x =-1