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\(\dfrac{x-1}{99}-1+\dfrac{x-3}{97}-1+\dfrac{x-5}{95}-1=\dfrac{x-2}{98}-1+\dfrac{x-4}{96}-1\)
\(\Leftrightarrow\dfrac{x-100}{99}+\dfrac{x-100}{97}+\dfrac{x-100}{95}=\dfrac{x-100}{98}+\dfrac{x-100}{96}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}\ne0\right)=0\Leftrightarrow x=100\)
pạn -1 vào mỗi phân số là xong. Rùi ra x\(\frac{x-2015}{1986}\)+\(\frac{x-2015}{1988}\)+ \(\frac{x-2015}{1990}\)+...+\(\frac{x-2015}{x1996}\)-\(\frac{x-2015}{29}\)-\(\frac{x-2015}{27}\)-...\(\frac{x-2015}{19}\)=0
<=>(x-2015)(\(\frac{1}{1986}\)+\(\frac{1}{1988}\)+... -\(\frac{1}{19}\))=0...(mà \(\frac{1}{1986}\)+...- \(\frac{1}{19}\) khác 0)
=>x-2015=0
<=> x=2015
Ta có : \(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}\)\(=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
\(\Rightarrow\left(\frac{x-1991}{9}-1\right)+\left(\frac{x-1993}{7}-1\right)+\left(\frac{x-1995}{5}-1\right)+\left(\frac{x-1997}{3}-1\right)+\left(\frac{x-1999}{1}-1\right)\)
\(=\left(\frac{x-9}{1991}-1\right)+\left(\frac{x-7}{1993}-1\right)+\left(\frac{x-5}{1995}-1\right)+\left(\frac{x-3}{1997}-1\right)+\left(\frac{x-1}{1999}\right)\)
\(\Rightarrow\frac{x-2000}{9}+\frac{x-2000}{7}+\frac{x-2000}{5}+\frac{x-2000}{3}\)
\(=\frac{x-2000}{1991}+\frac{x-2000}{1993}+\frac{x-2000}{1995}+\frac{x-2000}{1997}+\frac{x-2000}{1999}\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)=\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(x-2000\right)\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)=0\)
\(\Rightarrow\left(x-2000\right)\left[\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\right]=0\)
Vì \(\left(\frac{1}{9}+\frac{1}{7}+\frac{1}{5}+\frac{1}{3}\right)-\left(\frac{1}{1991}+\frac{1}{1993}+\frac{1}{1995}+\frac{1}{1997}+\frac{1}{1999}\right)\ne0\)
=> x - 2000 = 0
=> x = 2000
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
mấy câu này dễ mà :V câu a+c lấy mỗi phân số trừ cho 1 ra tử chung rút ra thì tính b+d thì cộng một tử chung rồi lại tính tiếp thôi
\(\begin{array}{l}a)\frac{1}{x} + \frac{2}{{x + 1}} + \frac{3}{{x + 2}} - \frac{1}{x} - \frac{2}{{x - 1}} - \frac{3}{{x + 2}}\\ = \left( {\frac{1}{x} - \frac{1}{x}} \right) + \left( {\frac{2}{{x + 1}} - \frac{2}{{x - 1}}} \right) + \left( {\frac{3}{{x + 2}} - \frac{3}{{x + 2}}} \right)\\ = 0 + \frac{2}{{x + 1}} - \frac{2}{{x - 1}} + 0\\ = \frac{{2\left( {x - 1} \right) - 2\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{2{\rm{x}} - 2 - 2{\rm{x}} - 2}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = \frac{{ - 4}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\end{array}\)
\(\begin{array}{l}b)\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{3}{{{x^2} - 9}} + \frac{{1 - 2{\rm{x}}}}{x} + \frac{{x - 1}}{{2{\rm{x}} + 1}} - \frac{3}{{x + 3}}\\ = \left( {\frac{{2{\rm{x}} - 1}}{x} + \frac{{1 - 2{\rm{x}}}}{x}} \right) + \left( {\frac{{1 - x}}{{2{\rm{x}} + 1}} + \frac{{x - 1}}{{2{\rm{x}} + 1}}} \right) + \left( {\frac{3}{{{x^2} - 9}} - \frac{3}{{x + 3}}} \right)\\ = 0 + 0 + \frac{3}{{\left( {x + 3} \right)\left( {x - 3} \right)}} - \frac{3}{{x + 3}}\\ = \frac{{3 - 3\left( {x - 3} \right)}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} = \frac{{12 - 3{\rm{x}}}}{{\left( {x + 3} \right)\left( {x - 3} \right)}}\end{array}\)
\(\frac{x-971}{972}+\frac{x-973}{970}+\frac{x-975}{968}+\frac{x-977}{966}=\frac{x-972}{971}+\frac{x-970}{973}+\frac{x-968}{975}+\frac{x-966}{977}\)
\(\Rightarrow\frac{x-971}{972}-1+\frac{x-973}{970}-1+\frac{x-975}{968}-1+\frac{x-977}{966}-1=\frac{x-972}{971}-1+\frac{x-970}{973}-1+\frac{x-968}{975}-1+\frac{x-966}{977}-1\)\(\Rightarrow\frac{x-1943}{972}+\frac{x-1943}{970}+\frac{x-1943}{968}+\frac{x-1943}{966}=\frac{x-1943}{971}+\frac{x-1943}{973}+\frac{x-1943}{975}+\frac{x-966}{977}\)
\(\Leftrightarrow\frac{x-1943}{972}+\frac{x-1943}{970}+\frac{x-1943}{968}+\frac{x-1943}{966}-\frac{x-1943}{971}-\frac{x-1943}{973}-\frac{x-1943}{975}-\frac{x-966}{977}=0\)
\(\Leftrightarrow\left(x-1943\right).\left(\frac{1}{972}+\frac{1}{970}+\frac{1}{968}+\frac{1}{966}-\frac{1}{971}-\frac{1}{973}+\frac{1}{975}+\frac{1}{977}\right)=0\)
\(\Leftrightarrow\left(x-1943\right)=0\Leftrightarrow x=1943\)
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