\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
Những câu hỏi liên quan
a) frac{x-23}{24}+frac{x-23}{25}frac{x-23}{26}+frac{x-23}{27}Leftrightarrowfrac{x-23}{24}+frac{x-23}{25}-frac{x-23}{26}-frac{x-23}{27}0Leftrightarrowleft(x-23right)left(frac{1}{24}+frac{1}{25}-frac{1}{26}-frac{1}{27}right)0Leftrightarrow x-230( vì frac{1}{24}+frac{1}{25}-frac{1}{26}-frac{1}{27}ne0)Leftrightarrow x23Vậy nghiệm của pt x23
Đọc tiếp
a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
\(\Leftrightarrow x-23=0\)( vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\ne0\))
\(\Leftrightarrow x=23\)
Vậy nghiệm của pt x=23
Tìm x biết \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> x-23=0
x=0+23
x=23. Vậy x=23
Chúc bạn học tốt!^_^
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\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=>( x-13)(\(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\) = 0
ta thấy 1/24>1/25>1/26>1/27 => 1/24+1/25 - 1/ 26 - 1/17 > 0
=> x -13 = -
=> x=13
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Xem thêm câu trả lời
Giải các phương trình sau:
a) \(x+\frac{2x\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
b) \(\frac{x-23}{24}+\frac{x+23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
Giải các phương trình sau:
a) \(x\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
b) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
- Ở câu a thì bạn chỉ cần quy đồng mẫu ở các vế cho bằng nhau, rồi bỏ mẫu. Bạn cứ thế mà thực hiện phép tính thôi.
- Còn câu b thì giải như vầy:
<=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
<=>\(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}\right)=0\)
Vì \(\left(\frac{1}{24}+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}\right)\ne0\)
<=> \(x-23=0\)
<=>\(x=23\)
Vậy phương trình có tập nghiệm: \(S=\left\{23\right\}\)
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Giải các phương trình sau bằng cách đưa về dạng ax + b = 0:
\(a,\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(b,\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
Giải PT
a,\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
b, \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
c, \(\frac{x+1}{2012}+\frac{x+2}{2011}=\frac{x+3}{2010}+\frac{x+4}{2009}\)
\(a,⇔\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
\(⇔(x-23)(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27})=0\)
\(⇔x-23=0\) (vì \(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\))
\(⇔x=23\)
\(b,⇔\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}+\frac{x+100}{95}=0\)
\(⇔(x+100)(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95})=0\)
\(⇔x+100=0\) (vì \(\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}>0\))
\(⇔x=-100\)
\(c,⇔(\frac{x+1}{2012}+1)+(\frac{x+2}{2011}+1)=(\frac{x+3}{2010}+1)+(\frac{x+4}{2009}+1)\)
\(⇔\frac{x+2013}{2012}+\frac{x+2013}{2011}-\frac{x+2013}{2010}-\frac{x+2013}{2009}=0\)
\(⇔(x+2013)(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009})=0\)
\(⇔x+2013=0\) (vì \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}<0\))
\(⇔x=-2013\)
Giải các phương trình sau bằng cách đưa về dạng ax + b = 0:
\(a,\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(b,\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
a. \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
\(\Leftrightarrow x=23\) (Vì \(\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)\ne0\) )
b. \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
\(\Leftrightarrow x=-100\) (Vì \(\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)\ne0\) )
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a) frac{x-23}{24}+frac{x-23}{25}frac{x-23}{26}+frac{x-23}{27} b) frac{x+1}{2004}+frac{x+2}{2003}frac{x+3}{2002}+frac{x+4}{2001}
d) frac{x+1}{9}+frac{x+2}{8}frac{x+3}{7}+frac{x+4}{6}
c) frac{x-45}{55}+frac{x-47}{53}frac{x-55}{45}+frac{x-53}{47}
Đọc tiếp
a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\) b) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
d) \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
c) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
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
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giải các phương trình
a)\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
b) \(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
c) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
d)\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)
Các câu na ná chắc nên mk làm mẫu 2 bài thui nha !
a, pt <=> x-23/24 + x-23/25 - x-23/26 - x-23/27 = 0
<=> (x-23).(1/24+1/25-1/26-1/27) = 0
<=> x-23=0 ( vì 1/24+1/25-1/26-1/27 > 0 )
<=> x=23
b, pt <=> (201-x/99 + 1)+(203-x/97 + 1)+(205-x/95 + 1) = 0
<=> 300-x/99 + 300-x/97 + 300-x/95 = 0
<=> (300-x).(1/99+1/97+1/95) = 0
<=> 300-x = 0 ( vì 1/99+1/97+1/95 > 0 )
<=> x=300
Tk mk nha
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