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h.
\(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)
\(\Leftrightarrow\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
Vì: \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\)
Suy ra: 2004 - x = 0
Vậy x = 2004
\(a,\dfrac{x-23}{24}+\dfrac{x-23}{25}=\dfrac{x-23}{26}+\dfrac{x-23}{27}\)
\(\Leftrightarrow\dfrac{x-23}{24}+\dfrac{x-23}{25}-\dfrac{x-23}{26}-\dfrac{x-23}{27}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\right)=0\)
\(\Leftrightarrow x-23=0\) ( vì \(\dfrac{1}{24}+\dfrac{1}{25}-\dfrac{1}{26}-\dfrac{1}{27}\ne0\) )
\(\Leftrightarrow x=23\)
Vậy pt có tập nghiệm S = { 23 }
\(b,\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy pt có tập nghiệm S = { 100 }
\(c,\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\)
\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+1+2004}{2004}+\dfrac{x+2+2003}{2003}-\dfrac{x+3+2002}{2002}-\dfrac{x+4+2001}{2001}=0\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}-\dfrac{x+2005}{2002}-\dfrac{x+2005}{2001}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
Vậy pt có tập nghiệm S = { 2005 }
\(d,\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Leftrightarrow\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}+\dfrac{205-x+95}{95}=0\)
\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
\(\Leftrightarrow300-x=0\)
\(\Leftrightarrow x=300\)
Vậy pt có tập nghiệm S = { 300 }
\(e,\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
\(\Leftrightarrow\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-45-55}{55}+\dfrac{x-47-53}{53}-\dfrac{x-55-45}{45}-\dfrac{x-53-47}{47}=0\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
Vậy pt có tập nghiệm S = { 100 }
\(f,\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)
\(\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\)
\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)
\(\Leftrightarrow x+10=0\)
\(\Leftrightarrow x=-10\)
Vậy pt có tập nghiệm S = { 10 }
\(h,\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
\(\Leftrightarrow\dfrac{2-x}{2002}=\dfrac{1-x}{2003}+\dfrac{-x}{2004}+1\)
\(\Leftrightarrow\dfrac{2-x}{2002}+1=\dfrac{1-x}{2003}+1+\dfrac{-x}{2004}+1\)
\(\Leftrightarrow\dfrac{2-x+2002}{2002}-\dfrac{1-x+2003}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
\(\Leftrightarrow2004-x=0\)
\(\Leftrightarrow x=2004\)
Vậy pt có tập nghiệm S = { 2004 }
\(g,\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
\(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}-\dfrac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy pt có tập nghiệm S = { -100 }
a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)
\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)
b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)
\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)
\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)
\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)
Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)
c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)
a)\(\dfrac{201-x}{99}+\dfrac{203-x}{97}=\dfrac{205-x}{95}+3=0\)
<=>\(\left(\dfrac{201-x}{99}+1\right)+\left(\dfrac{203-x}{97}+1\right)+\left(\dfrac{205-x}{95}+1\right)=0\)
<=>\(\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}=\dfrac{205-x+95}{95}=0\)
<=> \(\dfrac{300-x}{99}+\dfrac{300-x}{97}=\dfrac{300-x}{95}=0\)
<=> \(\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
<=> 300 - x = 0
<=> x = 300
b) \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
<=> \(\dfrac{2-x}{2002}+1=\left(\dfrac{1-x}{2003}+1\right)+\left(\dfrac{x}{2004}+1\right)\){Cộng cả hai vế của phương trình với 2}
<=> \(\dfrac{2-x+2002}{2002}=\dfrac{1-x+2003}{2003}+\dfrac{-x+2004}{2004}\)
<=> \(\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)
<=> \(\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
<=> \(\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
<=> 2004 - x = 0
<=> x = 2004.
b) x-45/55 + x-47/53 = x-55/45 + x-53/47
<=>x-45/55 -1 + x-47/53 -1= x-55/45 -1 + x-53/47 - 1
<=>x-100/55 + x-100/53 = x-100/45 + x-100/47
<=>(x-100)(1/55+1/53-1/45-1/47)=0
<=>x-100=0
<=>x=100
Vậy x = 100
a. \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
<=> \(5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-6\cdot5\)
<=> \(25x+10-80x+10=24x+12-30\)
<=> \(25x-80x-24x=12-30-10-10\)
<=> \(-79x=-38\)
<=> \(x=\dfrac{-38}{-79}\)
\(x=\dfrac{38}{79}\)
b. \(x-\dfrac{2x-5}{5}+\dfrac{x+8}{6}=7+\dfrac{x-1}{3}\)
<=> \(30\cdot x-6\left(2x-5\right)+5\left(x+8\right)=30\cdot7+10\left(x-1\right)\)
<=> \(30x-12x+30+5x+40=210+10x-10\)
<=> \(30x-12x+5x-10x=210-10-30-40\)
<=> \(13x=130\)
<=> \(x=\dfrac{130}{13}\)
\(x=10\)
c. \(\dfrac{x+1}{15}+\dfrac{x+2}{7}+\dfrac{x+4}{4}+6=0\)
<=> \(28\left(x+1\right)+60\left(x+2\right)+105\left(x+4\right)+420\cdot6=0\)
<=> \(28x+28+60x+120+105x+420+2520=0\)
<=> \(28x+60x+105x=-28-120-420-2520\)
<=> \(193x=-3088\)
<=> \(x=\dfrac{-3088}{193}\)
\(x=-16\)
d. \(\dfrac{x-342}{15}+\dfrac{x-323}{17}+\dfrac{x-300}{19}+\dfrac{x-273}{21}=10\)
<=> \(6783\left(x-342\right)+5985\left(x-323\right)+5355\left(x-300\right)+4845\left(x-273\right)=101745\cdot10\)
<=> \(6783x-2319786+5985x-1933155+5355x-1606500+4845x-1322685=1017450\)
<=> \(6783x+5985x+5355x+4845x=1017450+2319786+1933155+1606500+1322685\)
<=> \(22968x=8199576\)
<=> \(x=\dfrac{8199576}{22968}\)
\(x=357\)
\(\dfrac{x+1}{94}+\dfrac{x+2}{93}+\dfrac{x+3}{92}=\dfrac{x+4}{91}+\dfrac{x+5}{90}+\dfrac{x+6}{89}\)
\(\Rightarrow\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\)
\(\Rightarrow\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}=\dfrac{x+95}{91}+\dfrac{x+95}{90}+\dfrac{x+95}{89}\)
\(\Rightarrow\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\)
\(\Rightarrow\left(x+95\right)\left(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\right)=0\)
Vì \(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\ne0\) nên \(x+95=0\Leftrightarrow x=-95\)
Mk làm luôn nhé , không chép lại đề đâu !!! Ahihi
\(\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\)⇔\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\)
⇔ \(\left(x+95\right)\)\(\left(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\right)\) = 0
⇔\(x+95=0\)
⇔ \(x=-95\)
Vậy , ......
Giải phương trình sau:
\(\dfrac{x}{50}\) +\(\dfrac{x_{ }-1}{49}\)+\(\dfrac{x-2}{48}\)+\(\dfrac{x-3}{47}\)+\(\dfrac{x-150}{25}\)= 0
⇔ \(\dfrac{\left(x-50\right)+50}{50}\)+\(\dfrac{\left(x-50\right)+49}{49}\)+\(\dfrac{\left(x-50\right)+48}{48}\)+\(\dfrac{\left(x-50\right)-100}{25}\)= 0
⇔\(\dfrac{x-50}{50}\)+ 1 + \(\dfrac{x-50}{49}\)+1+\(\dfrac{x-50}{48}\)+1+\(\dfrac{x-50}{47}\)+1+\(\dfrac{x-50}{25}\)-4 = 0
⇔\(\dfrac{x-50}{50}\)+\(\dfrac{x-50}{49}\)+\(\dfrac{x-50}{48}\)+\(\dfrac{x-50}{47}\)+\(\dfrac{x-50}{25}\)= 0
⇔ (x - 50 ) ( \(\dfrac{1}{50}\)+ \(\dfrac{1}{49}\)+\(\dfrac{1}{48}\)+\(\dfrac{1}{47}\)+\(\dfrac{1}{25}\)) = 0
⇔ x-50 =\(\dfrac{0}{\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}}\)
⇔ x- 50 = 0
⇔ x = 50
vậy S = \(\left\{50\right\}\)
a) \(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{5x-1}{3x+2}=\dfrac{5x-7}{3x-1}\)
\(=\dfrac{5x-1-5x+7}{3x+2-3x+1}\)
\(=\dfrac{-1+7}{2+1}\)
\(=\dfrac{6}{3}\)
\(=2\)
Với \(\dfrac{5x-1}{3x+2}=2\)
\(\Rightarrow5x-1=2\left(3x+2\right)\)
\(\Rightarrow5x-1-2\left(3x+2\right)=0\)
\(\Rightarrow5x-1-6x-4=0\)
\(\Rightarrow-x-5=0\)
\(\Rightarrow x=-5\)
Pt\(\Leftrightarrow\dfrac{x+98}{2}+1+\dfrac{x+96}{4}+1+\dfrac{x+65}{35}+1=\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+49}{51}+1\)
\(\Leftrightarrow\dfrac{x+100}{2}+\dfrac{x+100}{4}+\dfrac{x+100}{35}-\dfrac{x+100}{97}-\dfrac{x+100}{95}-\dfrac{x+100}{51}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{35}-\dfrac{1}{97}-\dfrac{1}{35}-\dfrac{1}{51}\right)=0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
Vậy...