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a ) \(\dfrac{1}{x-1}-\dfrac{7}{x+2}=\dfrac{3}{x^2+x-2}\) (1)
ĐKXĐ : x\(\ne1;-2.\)
\(\left(1\right)\Leftrightarrow x+2-7x+7=3\)
\(\Leftrightarrow-6x=-6\)
\(\Leftrightarrow x=1\left(loại\right)\)
Vậy pt vô nghiệm .
b ) \(\dfrac{x^2+2x+1}{x^2+2x+2}+\dfrac{x^2+2x+2}{x^2+2x+3}=\dfrac{7}{6}\)
Đặt \(x^2+2x+1=t\) ta được :
\(\dfrac{t}{t+1}+\dfrac{t+1}{t+2}=\dfrac{7}{6}\)
\(\Leftrightarrow6t^2+12t+6t^2+12t+6=7\left(t^2+3t+2\right)\)
\(\Leftrightarrow5t^2+3t-8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{8}{5}\end{matrix}\right.\)
Khi t = 1
\(\Leftrightarrow\left(x+1\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Khi \(t=-\dfrac{8}{5}\)
\(\Leftrightarrow\left(x+1\right)^2=-\dfrac{8}{5}\) ( vô lí )
Vậy ............
a) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\dfrac{x}{x+2}=\dfrac{x^2+4}{x^2-4}\)
\(\Leftrightarrow\dfrac{x}{x+2}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2+4}{\left(x+2\right)\left(x-2\right)}\)
\(\Rightarrow x\left(x-2\right)=x^2+4\)
\(\Leftrightarrow x^2-2x=x^2+4\)
\(\Leftrightarrow-2x=4\Leftrightarrow x=-2\)(KTMĐK)
Vậy phương trình vô nghiệm
b) ĐKXĐ: \(x\ne3;x\ne-1\)
Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}+\dfrac{2x}{\left(x+1\right)\left(3-x\right)}=0\)
\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)-2.2x=0\)
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐK\right)\\x=3\left(KTMĐK\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=0\)
Đặt t=x2-2x+3(t\(\ge\)2)
PTTT: \(\dfrac{1}{t-1}+\dfrac{1}{t}=\dfrac{9}{2\left(t+1\right)}\)
<=>2t2+2t+2t2-2=9t2-9
<=>5t2-2t-7=0
<=>(t+1)(5t-7)=0
Do t\(\ge\)2
=>t+1>0 5t-7>0
Vậy pt vô nghiệm
\(\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2-2x+3}=\dfrac{9}{2\left(x^2-2x+4\right)}\)
Đặt \(t=x^2-2x+2=\left(x-1\right)^2+1\ge1\)
Thì ta có:
\(PT\Leftrightarrow\dfrac{1}{t}+\dfrac{1}{t+1}=\dfrac{9}{2\left(t+2\right)}\)
\(\Leftrightarrow5t^2-t-4=0\)
\(\Leftrightarrow\left(5t^2-5t\right)+\left(4t-4\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(5t+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5t+4=0\\t-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{4}{5}\left(l\right)\\t=1\end{matrix}\right.\)
\(\Rightarrow x^2-2x+2=1\)
\(\Leftrightarrow x=1\)
Vậy PT có 1 nghiệm là x = 1
b: \(\Leftrightarrow\dfrac{2}{\left(x+7\right)\left(x-3\right)}=\dfrac{3x+21}{\left(x-3\right)\left(x+7\right)}\)
=>3x+21=2
=>x=-19/3
d: \(\Leftrightarrow\left(2x+1\right)^2-\left(2x-1\right)^2=8\)
\(\Leftrightarrow4x^2+4x+1-4x^2+4x-1=8\)
=>8x=8
hay x=1
a/
\(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{\left(3-x\right)\left(3+x\right)}\)
\(\Leftrightarrow\dfrac{\left(x^2-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2.\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{-\left(7x^2-3x\right)}{\left(x-3\right)\left(x+3\right)}\) \(\Leftrightarrow x^3-4x^2+3x-x^3-3x^2=7x^2+3x\)
\(\Leftrightarrow x^3-4x^2+3x-x^3-3x^2-7x^2-3x=0\)
\(\Leftrightarrow-14x^2=0\)
\(\Leftrightarrow x=0\)
ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\2-x\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
Pt \(\Leftrightarrow\) \(\dfrac{\left(x-2\right)}{x^2-4}+\dfrac{-5\left(x+2\right)}{x^2-4}=\dfrac{2x-3}{x^2-4}\)
\(\Leftrightarrow x-2-5x-10=2x-3\)
\(\Leftrightarrow x-5x-2x=10+2-3\)
\(\Leftrightarrow-6x=9\)
\(\Leftrightarrow x=\dfrac{-3}{2}\) ( thỏa mãn)
Vậy nghiệm của pt là \(x=\dfrac{-3}{2}\)
a) \(\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)
=>\(\left(x+1\right)^2-\left(x-1\right)^2=16\)
=>\(x^2+2x+1-x^2+2x-1=16\)
=>4x=16=>x=4
b)\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\)
=>\(\dfrac{12}{x^2-4}-\dfrac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\dfrac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\)
=>\(12-\left(x+1\right)\left(x+2\right)+\left(x+7\right)\left(x-2\right)=0\)
=>\(12-x^2-3x-2+x^2+5x-14=0\)
=>2x-4=0=>2x=4=>x=2
c)\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)
=>\(\dfrac{12}{8+x^3}=\dfrac{x^3+8}{x^3+8}+\dfrac{x^2-2x+4}{x^3+8}\)
=>\(12=x^3+8+x^2-2x+4\)
=>\(x^3+x^2-2x=0\)
=>\(x^3-x+x^2-x=0\)
a) Đk : \(x\ne0;\ne1\)
\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)
\(\Rightarrow\dfrac{x^2+3x}{x\left(x+1\right)}+\dfrac{x^2-x-2}{x\left(x+1\right)}-\dfrac{2x^2+2x-2}{x\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{x^2+3x+x^2-x-2-2x^2-2x+2}{x\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{0}{x-1}=0\)
=> Phương trình có vô số nghiệm x
b) Đk : \(x\ne2;x\ne3\)
\(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)
\(\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}\)
=0
\(\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0\)
=> Phương trình vô nghiệm
c)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{-1}{x^4+x^2+1}=0\)
=> PTVN
d) Thôi tự làm đi, câu này dễ :Vvv
e)
\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)\)=40
\(\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40\)
\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt
\(x^2+6x+7=t\)
Phương trình tương đương
\(\left(t-1\right)\left(t+1\right)=40\)
\(t^2=41\)
\(\)\(t=\pm\sqrt{41}\)
Thay vào tìm x.
2) \(\dfrac{x}{2}\)-\(\dfrac{x}{10}\)<\(\dfrac{1}{2}-\dfrac{1}{3}\)
<=>\(\dfrac{x}{2}\)-\(\dfrac{x}{10}\)<\(\dfrac{1}{6}\)
=>15x-3x<5
<=>12x<5
<=>x<\(\dfrac{5}{12}\)
=> S={x|x<\(\dfrac{5}{12}\)}
\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{x^2+3}{x^2-2x}\)
<=> \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{x^2+3}{x\left(x-2\right)}\)
<=> \(\frac{x\left(x+2\right)-x+2}{x\left(x-2\right)}=\frac{x^2+3}{x\left(x-2\right)}\)
=> x2+2x-x+2=x2+3
<=>x=3