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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 :<
\(\dfrac{x-1}{99}-1+\dfrac{x-3}{97}-1+\dfrac{x-5}{95}-1=\dfrac{x-2}{98}-1+\dfrac{x-4}{96}-1\)
\(\Leftrightarrow\dfrac{x-100}{99}+\dfrac{x-100}{97}+\dfrac{x-100}{95}=\dfrac{x-100}{98}+\dfrac{x-100}{96}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}\ne0\right)=0\Leftrightarrow x=100\)
\(\dfrac{x+2}{89}+\dfrac{x+5}{86}>\dfrac{x+8}{83}+\dfrac{x+11}{80}\)
\(\Leftrightarrow\dfrac{x+2}{89}+1+\dfrac{x+5}{86}+1>\dfrac{x+8}{83}+1+\dfrac{x+11}{80}+1\)
\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}>\dfrac{x+91}{83}+\dfrac{x+91}{80}\)
\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}-\dfrac{x+91}{83}-\dfrac{x+91}{80}>0\)
\(\Leftrightarrow\left(x+91\right)\left(\dfrac{1}{89}+\dfrac{1}{86}-\dfrac{1}{83}-\dfrac{1}{80}\right)>0\)
Ta có: \(\dfrac{1}{89}+\dfrac{1}{86}+\dfrac{1}{83}+\dfrac{1}{80}< 0\)
\(\Leftrightarrow x+91< 0\)
\(\Leftrightarrow x< -91\)
Vậy...........
\(\Leftrightarrow\left(\dfrac{x-11}{111}+1\right)+\left(\dfrac{x-12}{112}+1\right)=\left(\dfrac{x-23}{123}+1\right)+\left(\dfrac{x-24}{124}+1\right)\)
=>x+100=0
=>x=-100
\(\dfrac{x+2}{89}+\dfrac{x+5}{86}>\dfrac{x+8}{83}+\dfrac{x+11}{80}\)
\(\Leftrightarrow\dfrac{x+91}{89}+\dfrac{x+91}{86}>\dfrac{x+91}{83}+\dfrac{x+91}{80}\)
\(\Leftrightarrow\left(x+91\right)\left(\dfrac{1}{89}+\dfrac{1}{86}\right)>\left(x+91\right)\left(\dfrac{1}{83}+\dfrac{1}{80}\right)\)
Mà \(\dfrac{1}{89}+\dfrac{1}{86}< \dfrac{1}{83}+\dfrac{1}{80}\)
Nên \(x+91< 0\Leftrightarrow x< -91\)
Ta có x+1/99 + x+2/98 + x+3/97 = x+4/96 + x+5/95 + x+10/90
=> x+1/99 + x+2/98 + x+3/97 - x+4/96 - x+5/95 - x+10/90=0
=> (x+1/99 + 1) + (x+2/98 + 1) + (x+3/97 +1) - (x+4/96 + 1) - (x+5/95 + 1) - (x+10/90 + 1) = 0
=> x+100/99 + x+100/98 + x+100/97 - x+100/96 - x+100/95 - x+100/90 =0
=> (x+100)(1/99+1/98+1/97-1/96-1/95-1/90) = 0
Mà 1/99+1/98+1/97-1/96-1/95-1/90 khác 0
=> x+100=0 => x=-100
Vậy phương trình có nghiệm là x=-100
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+10}{99}\)
\(\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1-\left(\frac{x+4}{96}+1+\frac{x+5}{95}+1+\frac{x+10}{99}+1\right)=0\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}-\left(\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{90}\right)=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{90}\right)=0\)
Mà\(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{90}\ne0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
a, Ta có : \(\frac{x+2}{89}+\frac{x+5}{86}=\frac{x+8}{83}+\frac{x+11}{80}\)
=> \(\frac{x+2}{89}+1+\frac{x+5}{86}+1=\frac{x+8}{83}+1+\frac{x+11}{80}+1\)
=> \(\frac{x+91}{89}+\frac{x+91}{86}=\frac{x+91}{83}+\frac{x+91}{80}\)
=> \(\frac{x+91}{89}+\frac{x+91}{86}-\frac{x+91}{83}-\frac{x+91}{80}=0\)
=> \(\left(x+91\right)\left(\frac{1}{89}+\frac{1}{86}-\frac{1}{83}-\frac{1}{80}\right)=0\)
=> \(x+91=0\)
=> \(x=-91\)
Vậy phương trình trên có nghiệm là \(S=\left\{-91\right\}\)
b, Ta có : \(\frac{x-11}{99}+\frac{x-12}{98}=\frac{x-3}{97}+\frac{x-4}{96}\)
=> \(\frac{x-11}{99}-1+\frac{x-12}{98}-1=\frac{x-3}{97}-1+\frac{x-4}{96}-1\)
=> \(\frac{x-110}{99}+\frac{x-110}{98}=\frac{x-110}{97}+\frac{x-110}{96}\)
=> \(\frac{x-110}{99}+\frac{x-110}{98}-\frac{x-110}{97}-\frac{x-110}{96}=0\)
=> \(\left(x-110\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
=> \(x-110=0\)
=> \(x=110\)
Vậy phương trình trên có nghiệm là \(S=\left\{110\right\}\)
rất cảm ơn bn