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\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+14}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+14}\)
\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{\left(x+14\right)-\left(x+2\right)}{\left(x+2\right)\left(x+14\right)}\)
\(\Leftrightarrow x=\left(x+14\right)-\left(x+2\right)\)
\(\Leftrightarrow x=x+14-x-2\)
\(\Leftrightarrow x=\left(x-x\right)+\left(14-2\right)\)
\(\Leftrightarrow x=0+12\)
\(\Leftrightarrow x=12\)
a) \(2\left(x-5\right)-3\left(x+7\right)=14\)
\(\Leftrightarrow2x-10-3x-21=14\)
\(\Leftrightarrow-x-31=14\)
\(\Leftrightarrow-x=45\Leftrightarrow x=-45\)
b) \(5\left(x-6\right)-2\left(x+3\right)=12\)
\(\Leftrightarrow5x-30-2x-6=12\)
\(\Leftrightarrow3x-36=12\)
\(\Leftrightarrow3x=48\Leftrightarrow x=16\)
c) \(3\left(x-4\right)-\left(8-x\right)=12\)
\(\Leftrightarrow3x-12-8+x=12\)
\(\Leftrightarrow4x-20=12\)
\(\Leftrightarrow4x=32\Leftrightarrow x=8\)
d) \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
\(\Leftrightarrow-21x+35+14x-28=28\)
\(\Leftrightarrow-7x+35=0\Leftrightarrow x=5\)
=> \(\frac{1}{x+2}-\frac{1}{x+14}=\frac{1}{16}\)
=> \(\frac{x+14}{\left(x+2\right).\left(x+14\right)}-\frac{x+2}{\left(x+2\right).\left(x+14\right)}=\frac{1}{16}\)
=> \(\frac{\left(x+14\right)-\left(x+2\right)}{\left(x+2\right).\left(x+14\right)}=\frac{1}{16}\)
=> \(\frac{x+14-x-2}{x\left(2+14\right)}=\frac{1}{16}\)
=> \(\frac{12}{16x}=\frac{1}{16}\)
=> x = 12
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}=\dfrac{-1}{4}\)
\(\Leftrightarrow\left(x-1\right)\left(-1\right)=4\)
\(\Leftrightarrow x-1=-4\)
\(\Leftrightarrow x=-3\left(tm\right)\)
Vậy ..............
\(\frac{1}{x+2}-\frac{1}{x+5}+...+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+7}=\frac{x}{\left(x+2\right)\left(x+7\right)}\)
\(\Rightarrow x=1\)
Th1: x<-4
Pt sẽ là 2-x-x-4=8
=>-2x-2=8
=>-2x=10
=>x=-5(nhận)
TH2: -4<=x<2
Pt sẽ là x+4+2-x=8
=>6=8(loại)
TH3: x>=2
Pt sẽ là x-2+x+4=8
=>2x+2=8
=>x=3(nhận)