\(\frac{x+1}{X^2+x+1}\)-\(\frac{x-1}{x^2-x+1}\)=\(\frac{3}{x\left(x^4+x^2+1\right)}\)
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HQ
Hà Quang Minh
Giáo viên
9 tháng 9 2023
a)
\(\begin{array}{l}\frac{2}{{3{\rm{x}}}} + \frac{x}{{x - 1}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{2}{{3{\rm{x}}}} - \frac{x}{{1 - x}} + \frac{{6{{\rm{x}}^2} - 4}}{{2{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4\left( {1 - x} \right) - 6{{\rm{x}}^2} + 3\left( {6{{\rm{x}}^2} - 4} \right)}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{4 - 4{\rm{x}} - 6{{\rm{x}}^2} + 18{{\rm{x}}^2} - 12}}{{6{\rm{x}}\left( {1 - x} \right)}}\\ = \frac{{12{{\rm{x}}^2} - 4{\rm{x}} - 8}}{{6{\rm{x}}\left( {1 - x} \right)}}\end{array}\)
b)
\(\begin{array}{l}\frac{{{x^3} + 1}}{{1 - {x^3}}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1}}{{{x^3} - 1}} + \frac{x}{{x - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}}\\ = \frac{{ - {x^3} - 1 + x\left( {{x^2} + x + 1} \right) - \left( {{x^2} - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{{ - {x^3} - 1 + {x^3} + {x^2} + x - {x^2} + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \frac{x}{{{x^3} - 1}}\end{array}\)
c)
\(\begin{array}{l}\left( {\frac{2}{{x + 2}} - \frac{2}{{1 - x}}} \right).\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2\left( {1 - x} \right) - 2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{2 - 2{\rm{x}} - 2{\rm{x}} - 4}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{ - 4{\rm{x - 2}}}}{{\left( {x + 2} \right)\left( {1 - x} \right)}}.\frac{{{x^2} - 4}}{{4{{\rm{x}}^2} - 1}}\\ = \frac{{\left( { - 4{\rm{x}} - 2} \right)\left( {x - 2} \right)\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 8{\rm{x}} - 2{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1} \right)}}\\ = \frac{{ - 4{{\rm{x}}^2} + 6{\rm{x}} + 4}}{{\left( {1 - x} \right)\left( {4{{\rm{x}}^2} - 1} \right)}}\end{array}\)
d)
\(\begin{array}{l}1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}\left( {\frac{1}{{1 - x}} - \frac{1}{{1 - {x^2}}}} \right)\\ = 1 + \frac{{{x^3} - x}}{{{x^2} + 1}}.\frac{{1 + x - 1}}{{1 - {x^2}}}\\ = 1 + \frac{{x\left( {{x^2} - 1} \right)}}{{{x^2} + 1}}.\frac{x}{{1 - {x^2}}}\\ = 1 + \frac{{ - {x^2}\left( {{x^2} - 1} \right)}}{{\left( {{x^2} + 1} \right)\left( {{x^2} - 1} \right)}}\\ = 1 + \frac{{ - {x^2}}}{{{x^2} + 1}}\\ = \frac{{{x^2} + 1 - {x^2}}}{{{x^2} + 1}}\\ = \frac{1}{{{x^2} + 1}}\end{array}\)
27 tháng 5 2020
d) \(\frac{1}{\left(x+2\right)}+\frac{1}{\left(x+3\right)}=\frac{1}{\left(x+2\right)\left(x+3\right)}\)
ĐKXĐ : \(x\ne-2;x\ne-3\)
\(\Leftrightarrow x+3+x+2=1\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\) (không nhận)
Vậy : \(S=\varnothing\)
27 tháng 5 2020
Giai phương trình sau :
a) \(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{3}{1-x}=\frac{5}{x+5}\)
ĐKXĐ : \(x\ne1;x\ne-5\)
Với điều kiện trên ta có :
\(\Leftrightarrow\)\(\frac{10}{\left(x+5\right)\left(x-1\right)}+\frac{-3}{x-1}=\frac{5}{x+5}\)
\(\Leftrightarrow10-3\left(x+5\right)=5\left(x-1\right)\)
\(\Leftrightarrow10-3x-15=5x-5\)
\(\Leftrightarrow-8x=0\)
\(\Leftrightarrow x=0\) (nhận)
Vậy : \(S=\left\{0\right\}\)
OM
28 tháng 4 2020
Giải phương trình
a) \(\frac{2x}{x-1}-\frac{x}{x-2}=\frac{x^2}{\left(x-1\right)\left(x-2\right)}\left(x\ne1,x\ne2\right)\)
\(\Leftrightarrow\frac{2x\left(x-2\right)-x\left(x-1\right)-x^2}{\left(x-1\right)\left(x-2\right)}=0\)
\(\Rightarrow2x^2-x^2-x^2-4x+x=0\)
\(\Leftrightarrow-3x=0\Leftrightarrow x=0\left(tm\right)\)
KL: Vậy...
b)\(\frac{1}{x+2}-\frac{6}{x-1}+\frac{8}{\left(x+2\right)\left(x-1\right)}=0\left(x\ne-2,x\ne1\right)\)
\(\Leftrightarrow\frac{\left(x-1\right)-6\left(x+2\right)+8}{\left(x+2\right)\left(x-1\right)}=0\)
\(\Rightarrow x-1-6x-12+8=0\)
\(\Leftrightarrow-5x=-7\Leftrightarrow x=\frac{7}{5}\left(tm\right)\)
c) \(\frac{x+2}{x+3}-\frac{x+1}{x-1}=\frac{4}{\left(x+3\right)\left(x-1\right)}\left(x\ne-3,x\ne1\right)\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-1\right)-\left(x+1\right)\left(x+3\right)-4}{\left(x+3\right)\left(x-1\right)}=0\)
\(\Rightarrow x^2+x-2-x^2-4x-3-4=0\)
\(\Leftrightarrow-3x=9\Leftrightarrow x=-3\left(ktm\right)\)
2 tháng 12 2016
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
\(=\frac{1}{x}\)
2 tháng 12 2016
ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{x+5}\)
=\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}\)
= \(\frac{1}{x}\)
28 tháng 5 2020
a,
\(\frac{10}{(x+5)(x-1)}+\frac{3}{x-1}=\frac{5}{x+5}\)
\(\frac{10}{(x+5)(x-1)}+\frac{3(x+5)}{(x+5)(x-1)}=\frac{5(x-1)}{(x+5)(x-1)}\)
\(\Leftrightarrow\)10+3x+15=5x-5
\(\Leftrightarrow\)25+3x=5x-5
\(\Leftrightarrow\)25+3x-5x+5=0
\(\Leftrightarrow\)30-2x=0
\(\Leftrightarrow\)-2x=-30
\(\Leftrightarrow\)x=15
vậy pt có ngiệm là x=15
Bài này dễ, chỉ cần quy đồng thôi
Để ý cái mẫu
ĐKXĐ: \(\forall x\in Z\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\)\(\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}-\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}=\frac{3}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow\)\(x\left(x+1\right)\left(x^2-x+1\right)-x\left(x-1\right)\left(x^2+x+1\right)=3\)
\(\Leftrightarrow\)\(x\left(x^3+1\right)-x\left(x^3-1\right)=3\)
\(\Leftrightarrow\)\(x^4+x-x^4+x=3\)
\(\Leftrightarrow\)\(2x=3\)
\(\Leftrightarrow\)\(x=1,5\)(t/m ĐKXĐ)
Vậy....