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10) \(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+17}{83}+\frac{x+116}{4}=0\)
\(\Leftrightarrow\)\(\frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow\)\(\frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/86 + 1/85 + 1/84 + 1/83 + 1/4 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy....
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sửa đề đến đây thôi bạn nhé, do nếu thêm vào thì mình cũng ko biết có quy luật gì nữa :<
\(\dfrac{x-1}{99}-1+\dfrac{x-3}{97}-1+\dfrac{x-5}{95}-1=\dfrac{x-2}{98}-1+\dfrac{x-4}{96}-1\)
\(\Leftrightarrow\dfrac{x-100}{99}+\dfrac{x-100}{97}+\dfrac{x-100}{95}=\dfrac{x-100}{98}+\dfrac{x-100}{96}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}\ne0\right)=0\Leftrightarrow x=100\)
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\(\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)
\(\Leftrightarrow\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
=> x + 100 = 0
=> x = -100
Vậy x = -100
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1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)
ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow-7x+3=-4x-15\)
\(\Leftrightarrow-7x+4x=-15-3\)
\(\Leftrightarrow-3x=-18\)
\(\Leftrightarrow x=6\)( tmđk )
Vậy x = 6 là nghiệm của phương trình
2) 2x + 3 < 6 - ( 3 - 4x )
<=> 2x + 3 < 6 - 3 + 4x
<=> 2x - 4x < 6 - 3 - 3
<=> -2x < 0
<=> x > 0
Vậy nghiệm của bất phương trình là x > 0
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(x+1)/99+(x+2)/98+(x+3)/97=(x+4)/96+(x+5)/95+(x+6)/94
[(x+1)/99 +1]+[(x+2)/98 +1]+[(x+3)/97 +1]-3=[(x+4)/96 +1]+[(x+5)/95 +1]+[(x+6)/94 +1]-3
[(x+1+99)/99+(x+2+98)/98+(x+3+97)/97]-3=[(x+4+96)/96+(x+5+95)/95+(x+6+94)/94]-3
(x+100)/99+(x+100)/98+(x+100)/97=(x+100)/96+(x+100)/95+(x+100)/94
(x+100)(1/99+1/98+1/97)=(x+100)(1/96+1/95+1/94)
(x+100)(1/99+1/98+1/97)-(x+100)(1/96+1/95+1/94)=0
(x+100)(1/99+1/98+1/97-1/96-1/95-1/94)=0
Ma : 1/99+1/98+1/97-1/96-1/95-1/94 \(\ne\)0
=>x+100=0
=>x=-100
k mk nha khong hieu noi mk nha.
1/3x-1/2=(3/5-4x)15/7
1/3x-1/2=9/7-60/7x
1/3x+60/7x=1/2+9/7
187/21x=25/14
x=75/374
k mk nha ban.
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7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+36}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (vì 1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy.....
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7) \(\frac{x+25}{75}+\frac{x+30}{70}=\frac{x+35}{65}+\frac{x+40}{60}\)
\(\Leftrightarrow\)\(\frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\) (1/75 + 1/70 - 1/65 - 1/60 \(\ne\)0)
\(\Leftrightarrow\)\(x=-100\)
Vậy...
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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...
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a: =>5(2-x)<3(3-2x)
=>10-5x<9-6x
=>x<-1
b: =>2/9x+5/3>=1/5x-1/5+1/3x
=>2/9x+5/3>=8/15x-1/5
=>-14/45x>=-28/15
=>x<=6
\(\dfrac{x+1}{98}+\dfrac{x+2}{97}+\dfrac{x+90}{9}+\dfrac{x+84}{15}>-4\\ \Leftrightarrow\left(\dfrac{x+1}{98}+1\right)+\left(\dfrac{x+2}{97}+1\right)+\left(\dfrac{x+90}{9}+1\right)+\left(\dfrac{x+84}{15}+1\right)>0\\ \Leftrightarrow\dfrac{x+99}{98}+\dfrac{x+99}{97}+\dfrac{x+99}{9}+\dfrac{x+99}{15}>0\\ \Leftrightarrow\left(x+99\right)\left(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}\right)>0\)
Vì \(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}>0\Rightarrow x+99>0\Rightarrow x>-99\)
cho tớ hỏi ông