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\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow x\left(x+8\right)=105\)
\(\Leftrightarrow x^2+8x-105=0\)
\(\Leftrightarrow x^2-7x+15x-105=0\)
\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)
Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x^2-2x}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-2=x^2-2x\)
\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)
Cho mình sửa lại nhé:
\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x-2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-2=x-2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(\dfrac{1}{x+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)
\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+8\right)}=\dfrac{4}{105}\)
\(\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)
\(\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}=\dfrac{8}{105}\)
\(\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)
\(\dfrac{x+8-x}{x\left(x+8\right)}=\dfrac{8}{105}\)
\(\dfrac{8}{x.\left(x+8\right)}=\dfrac{8}{105}\)
\(\Rightarrow x\left(x+8\right)=105\)
\(x^2+8x-105=0\)
\(\left(x-7\right)\left(x+15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)
ĐKXĐ:\(x\ne-2\)
\(\dfrac{1}{x+2}-1=\dfrac{5x+7}{x+2}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{5x+7}{x+2}=1\\ \Leftrightarrow\dfrac{1-5x-7}{x+2}=1\\ \Leftrightarrow-5x-6=x+2\\ \Leftrightarrow x+2+5x+6=0\\ \Leftrightarrow6x+8=0\\ \Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)
thank
\(=\dfrac{2x^2-x-x-1+2-x^2}{x-1}=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)
\(\dfrac{x-1}{x-2}+\dfrac{x+3}{x-4}=\dfrac{2}{-x^2+6x-8}\left(đk:x\ne2,x\ne4\right)\Leftrightarrow\dfrac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{-2}{x^2-6x+8}\Leftrightarrow\dfrac{2x^2-4x-2}{x^2-6x+8}=\dfrac{-2}{x^2-6x+8}\Leftrightarrow2x^2-4x-2=-2\Leftrightarrow2x^2-4x=0\Leftrightarrow2x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)\(\Leftrightarrow x=0\)( do x≠2)
2)Biện luận PT
`m(mx-1)=x+1`
`<=>m^2x-m=x+1`
`<=>x(m^2-1)=m+1`
PT vô nghiệm `<=>{(m^2-1=0),(m+1\ne0):}<=>m=1`
PT vô số nghiệm `<=>{(m^2-1=0),(m+1=0):}<=>m=-1`
PT có nghiệm duy nhất `m^2-1\ne0<=>m^2\ne1<=>m\ne+-1=>x=(m+1)/(m^2-1)=1/(m-1)`
1) \(\dfrac{x^2-4}{x^2+2x+1}:\dfrac{4-2x}{2x+2}=\dfrac{\left(x-2\right)\left(x+2\right)2\left(x+1\right)}{\left(x+1\right)^22\left(2-x\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x+1\right)}{-2\left(x-2\right)\left(x+1\right)\left(x+1\right)}=\dfrac{-\left(x+2\right)}{x+1}=\dfrac{-x-2}{x+1}\)
2) \(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)=\dfrac{x+1}{x+2}:\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x+3\right)\left(x+3\right)}{\left(x+2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}=\dfrac{x^2+6x+9}{x^2+4x+4}\)
Ta có:
\(x^2+2x=x\left(x+2\right)\)
\(x^2+6x+8=x^2+2x+4x+8=x\left(x+2\right)+4\left(x+2\right)=\left(x+2\right)\left(x+4\right)\)
\(x^2+10x+24=x^2+4x+6x+24=x\left(x+4\right)+6\left(x+4\right)=\left(x+4\right)\left(x+6\right)\)
\(x^2+14x+48=x^2+6x+8x+48=x\left(x+6\right)+8\left(x+6\right)=\left(x+6\right)\left(x+8\right)\)
Phương trình trở thành:
\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+8\right)}=3\)
\(\Leftrightarrow2\left(\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+...+\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=3\)
\(\Leftrightarrow2\left(\dfrac{1}{x}-\dfrac{1}{x+8}\right)=3\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{3}{2}\Leftrightarrow3x\left(x+8\right)=16\Leftrightarrow x^2+8x=\dfrac{16}{3}\Leftrightarrow x=0,6188021535\)