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=\(\frac{1}{x+1}-\frac{1}{x+2}\)+\(\frac{1}{x+2}-\frac{1}{x+3}\)+\(\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}\)-\(\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}\)
=\(\frac{1}{x+1}-\frac{1}{x+6}\)
=\(\frac{x+6-x-1}{\left(x+1\right)\left(x+6\right)}\)
=\(\frac{5}{\left(x+1\right)\left(x+6\right)}\)
1/ (2x+3)(x-4)+(x+5)(x-2)=(3x-5)(x-4)
<=> 2x2 - 8x + 3x - 12 + x2 - 2x + 5x - 10 - 3x2 + 12x + 5x - 20 = 0
<=> 15x - 20 = 0
<=> 15x = 20
<=> x = 4/3
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a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)
=>(x+5)(x-3)+8=x^2-1
=>x^2+2x-15+8=x^2-1
=>2x-7=-1
=>x=3(loại)
b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)
=>(x-4)(x+1)+x^2+3+5(x-1)=0
=>x^2-3x-4+x^2+3+5x-5=0
=>2x^2+2x-6=0
=>x^2+x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
e: =>x^2-2x+1+2x+2=5x+5
=>x^2+3=5x+5
=>x^2-5x-2=0
=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)
g: (x-3)(x+4)*x=0
=>x=0 hoặc x-3=0 hoặc x+4=0
=>x=0;x=3;x=-4
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
