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\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
\(\Rightarrow\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)\(\Rightarrow\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)
\(\Rightarrow\left(x-2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=\left(x-2004\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
Với \(x-2004\ne0\)
\(\Rightarrow\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\left(KTM\right)\)
Với \(x-2004=0\)
\(\Rightarrow x=2004\)
\(\frac{x+10}{2000}+\frac{x+20}{1990}+\frac{x+30}{1980}+\frac{x+40}{1970}=-4\)
\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)
\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)
Vì \(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}>0\)
\(\Rightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)
\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)
mà\(\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)\ne0\Rightarrow\left(x+2010\right)=0\\ \Rightarrow x=-2010\)
Cac ban giup minh voi
1) Giai cac phuong trinh
a) 2010.(4x-3)-4x2+3=0
b)( x2-\(\frac{25}{4}\))2= 10x +1
\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)
\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)
\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{6\left(x-1\right)}{2\left(x^2+4x+3\right)}\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{3\left(x-1\right)}{x^2+4x+3}\)
\(\Leftrightarrow-x^2-3=3x-3\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\)
Vậy x = 0
\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)
\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow-6x+3=8\)
\(\Leftrightarrow x=-\frac{5}{6}\)
Vậy ...
ĐKXĐ: \(x\ne-1;\) \(x\ne-3;\)\(x\ne-5;\)\(x\ne-7\)
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)
\(\Leftrightarrow\)\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)
\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}\right)=\frac{3}{16}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}-\frac{1}{x+7}=\frac{3}{8}\)
\(\Leftrightarrow\)\(\frac{6}{\left(x+1\right)\left(x+7\right)}=\frac{3}{8}\)
\(\Rightarrow\)\(3\left(x+1\right)\left(x+7\right)=48\)
\(\Leftrightarrow\)\(x^2+8x+7=16\)
\(\Leftrightarrow\)\(x^2+8x-9=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x-9\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=9\left(TMĐKXĐ\right)\end{cases}}\)
Vậy...
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)
\(\Leftrightarrow\frac{1}{x^2+x+3x+3}+\frac{1}{x^2+3x+5x+15}+\frac{1}{x^2+5x+7x+35}=\frac{3}{16}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)
\(\Leftrightarrow\frac{\left(x+5\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}+\frac{\left(x+1\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}\)
\(=\frac{3\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}\)
Mẫu của mỗi phân thức bằng nhau nên => tử của mỗi phân thức cũng phải bằng nhau
=> Đến đây thì dễ rồi, bạn giải ra tìm x
\(\frac{2}{x^2-4x+3}+\frac{2}{x^2-8x+15}+\frac{2}{x^2-12x+35}=-\frac{1}{2}\)(x khác 1;3;5;7)
<=>\(\frac{2}{x^2-3x-x+3}+\frac{2}{x^2-5x-3x+15}+\frac{2}{x^2-5x-7x+35}=-\frac{1}{2}\)
<=>\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{2}{\left(x-3\right)\left(x-5\right)}+\frac{2}{\left(x-5\right)\left(x-7\right)}=-\frac{1}{2}\)
<=>\(\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-5}-\frac{1}{x-3}+\frac{1}{x-7}-\frac{1}{x-5}=-\frac{1}{2}\)
<=>\(\frac{1}{x-7}-\frac{1}{x-1}=-\frac{1}{2}\)
<=>\(2x-2-2x+14=-x^2+8x-7\)
<=>\(x^2-8x+19=0\)
<=>(x-4)2+3=0(vô lí)
Vậy PT vô nghiệm
xin lỗi nha, bài đó bằng có một cái 1/5 thôi, tại viết sai
ĐK : \(X\ne-1;-3;-7;-9\)
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}+\frac{1}{x^2+16x+63}=\frac{1}{5}\)
\(\frac{1}{\left(x+2\right)^2-1}+\frac{1}{\left(x+4\right)^2-1}+\frac{1}{\left(x+6\right)^2-1}+\frac{1}{\left(x-8\right)^2-1}=\frac{1}{5}\)
\(\frac{1}{\left(x+2-1\right)\left(x+2+1\right)}+\frac{1}{\left(x+4-1 \right)\left(x+4+1\right)}+\frac{1}{\left(x+6-1\right)\left(x+6+1\right)}+\frac{1}{\left(x+8-1\right)\left(x+8+1\right)}=\frac{1}{5}\)
\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}=\frac{1}{5}\)
\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+....-\frac{1}{x+9}\right)=\frac{1}{5}\)
\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+9}\right)=\frac{1}{5}\)
\(\frac{1}{x+1}-\frac{1}{x+9}=\frac{1}{5}:\frac{1}{2}=\frac{2}{5}\)
\(\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)
\(2\left(x+1\right)\left(x+9\right)=40\)
\(2x^2+20x+18=40\Leftrightarrow x^2+10x+9=20\)
\(\Leftrightarrow x^2+10x-11=0\Leftrightarrow x^2+10x-10-1=0\)
\(\Leftrightarrow\left(x^2-1\right)+\left(10x-10\right)=0\Leftrightarrow\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+11\right)=0\)
\(\orbr{\begin{cases}x-1=0\\x++11=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}}\)( Thõa mãn )
Vậy ...............
a, DKXD: \(x\ne\pm3\)
\(A=\left(\frac{x}{x+3}+\frac{x-1}{x-3}+\frac{2x^2+x-3}{9-x^2}\right):\frac{-2}{x-3}\)
\(=\left(\frac{x\left(x+3\right)+\left(x-1\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{-2x^2-x+3}{x^2-9}\right):\frac{-2}{x-3}\)
\(=\left(\frac{2x^2+5x-3}{x^2-9}+\frac{-2x^2-x+3}{x^2-9}\right):\frac{-2}{x-3}\)
\(=\frac{4x}{x^2-9}:\frac{-2}{x-3}=\frac{4x}{\left(x-3\right)\left(x+3\right)}\cdot\frac{x-3}{-2}=\frac{4x}{-2\left(x+3\right)}=\frac{-2x}{x+3}\)
b, \(x^2-2x-3=0\Leftrightarrow x^2-3x+x-3=0\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)
Thay x=-1 =>\(A=\frac{-2.\left(-1\right)}{-1+3}=1\)
thay x=3 =>\(A=\frac{-2.3}{3+3}=-1\)
c, De \(A\in Z\Leftrightarrow x+3\in U\left(-2\right)=\left\{1;-1;2;-2\right\}\)
<=>x thuoc {-2;-4;-1;-5}
ĐK: \(x\ne\pm3\)
\(A=\left(\frac{x}{x+3}+\frac{x-1}{x-3}+\frac{2x^2+x-3}{9-x^2}\right):\frac{-2}{x-3}\)
\(=\left(\frac{x\left(x-3\right)+\left(x+3\right)\left(x-1\right)}{\left(x+3\right)\left(x-3\right)}+\frac{-2x^2-x+3}{x^2-9}\right).\frac{x-3}{-2}\)
\(=\left(\frac{x^2-3x+x^2+2x-3}{\left(x-3\right)\left(x+3\right)}+\frac{-2x^2-x+3}{\left(x-3\right)\left(x+3\right)}\right).\frac{x-3}{-2}\)
\(=\frac{-2x}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{-2}=\frac{x}{x+3}\)
b, \(x^2-2x-3=0\Rightarrow x\left(x-3\right)+\left(x-3\right)=0\Rightarrow\left(x-3\right)\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
x = 3 không thỏa mãn ĐKXĐ
Với x = -1 (thỏa mãn ĐKXĐ) thì \(A=\frac{x}{x+3}=\frac{-1}{-1+3}=-\frac{1}{2}\)
c, \(A\in Z\Rightarrow\frac{x}{x+3}\in Z\Rightarrow x⋮\left(x+3\right)\)
\(\Rightarrow\left(x+3\right)-3⋮\left(x+3\right)\Rightarrow-3⋮\left(x+3\right)\Rightarrow x+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x\in\left\{-6;-4;-2;0\right\}\) (thỏa mãn điều kiện)
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\(\frac{2}{\left(x+3\right)\left(x+1\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}=\frac{2}{9}\)
\(\Rightarrow\frac{2}{x+1}-\frac{2}{x+3}+\frac{2}{x+3}-\frac{2}{x+5}+\frac{2}{x+5}-\frac{2}{x+7}=\frac{2}{9}\)
\(\frac{2}{x+1}-\frac{2}{x+7}=\frac{2}{9}\\ \Rightarrow\frac{2x+14-2x-2}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}\\ \Rightarrow\frac{12}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}=\frac{12}{54}\)
\(\Rightarrow\left(x+1\right)\left(x+7\right)=54\\ \Rightarrow x^2+8x-54=0\Rightarrow x=-4\pm\sqrt{70}\)