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NH
0
BH
23 tháng 2 2019
pt đầu \(\Leftrightarrow x+1+\frac{1}{x+1}+x+7+\frac{7}{x+7}=x+3+\frac{3}{x+3}+x+5+\frac{5}{x+5}\)
\(\Rightarrow\frac{1}{x+1}+\frac{7}{x+7}=\frac{3}{x+3}+\frac{5}{x+5}\\ \Rightarrow\frac{8x+14}{x^2+8x+7}=\frac{8x+30}{x^2+8x+15}\)
\(\Leftrightarrow\left(4x+7\right)\left(x^2+8x+15\right)=\left(4x+15\right)\left(x^2+8x+7\right)\)
Đặt a=4x+7
b=x2 +8x+7
như vậy ta được pt mới có dạng \(a\left(b+8\right)=b\left(a+8\right)\Leftrightarrow ab+8a=ab+8b\Rightarrow a=b\)
hay\(4x+7=x^2+8x+7\Rightarrow x^2+4x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
T
30 tháng 4 2019
a,\(2x\left(x-3\right)=x-3.\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\frac{1}{2}\)
Vậy .....
b, \(\frac{x+2}{x-2}-\frac{5}{x}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow\frac{\left(x+2\right)\cdot x}{\left(x-2\right)\cdot x}-\frac{5\left(x-2\right)}{x\left(x-2\right)}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow\frac{x^2+2x-\left(5x-10\right)}{\left(x-2\right)x}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow\frac{x^2+2x-5x+10}{x^2-2x}=\frac{8}{x^2-2x}\)
\(\Leftrightarrow x^2+2x-5x+10=8\)
\(\Leftrightarrow x^2-3x+10-8=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy ....
CQ
1
SF
25 tháng 4 2020
\(\frac{1}{x+2}=\frac{5}{2-x}+\frac{12+x}{x^2-4}\) (1)
đkxđ: \(x\ne\pm2\)
(1)\(\Leftrightarrow\frac{\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{-5\left(x+2\right)+12+x}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow x-2=-5\left(x+2\right)+12+x\)
\(\Leftrightarrow5x=4\)
\(\Leftrightarrow x=\frac{5}{4}\)(thỏa mãn đkxđ)
NT
2
27 tháng 3 2020
a) \(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)
<=> \(\frac{x}{4}+\frac{5}{4}-\frac{2x}{3}+1=\frac{6x}{8}-\frac{1}{8}+\frac{2x}{12}-\frac{1}{12}\)
<=> \(-\frac{4}{3}x=-\frac{59}{24}\)
<=> \(x=\frac{59}{32}\)
Vậy S = { 59/32}
b) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)
<=> \(\frac{x^2+14x+40}{12}-\frac{-x^2-2x+8}{4}=\frac{x^2+8x-20}{3}\)
<=> \(\left(\frac{x^2}{12}+\frac{x^2}{4}-\frac{x^2}{3}\right)+\left(\frac{14}{12}x+\frac{2}{4}x-\frac{8}{3}x\right)=-\frac{20}{8}+\frac{8}{4}-\frac{40}{12}\)
<=> \(-x=-8\)
<=> x = 8
Vậy S = { 8 }
\(\frac{240}{x-12}-\frac{240}{x}=\frac{5}{3}\)
\(\Leftrightarrow\frac{240}{x-12}-\frac{240}{x}=\frac{5}{3}\left(x\ne12,x\ne0\right)\)
\(\Leftrightarrow\frac{240}{x-12}-\frac{240}{x}-\frac{5}{3}=0\)
\(\Leftrightarrow\frac{720x-720x+8640-5x^2+60x}{3x\left(x-12\right)}=0\)
\(\Leftrightarrow8640-5x^2+60x=0\)
\(\Leftrightarrow5\left(1728-x^2+12x\right)=0\)
\(\Leftrightarrow5\left(-x^2+12x+1728\right)=0\)
\(\Leftrightarrow5\left(-x^2+48x-36x+1728\right)=0\)
\(\Leftrightarrow5\left[-x\left(x-48\right)-36\left(x-48\right)\right]=0\)
\(\Leftrightarrow5\left[-\left(x-48\right)\right].\left(x+36\right)=0\)
\(\Leftrightarrow-5\left(x-48\right).\left(x+36\right)=0\)
\(\Leftrightarrow\left(x-48\right)\left(x+36\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-48=0\\x+36=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=48\\x=-36\end{matrix}\right.,x\ne12,x\ne0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=48\\x=-36\end{matrix}\right.\)