\(x+\frac{1}{2}+x+\frac{1}{4}+x+\frac{1}{8}+x+\frac{1}{16}=\frac{23}{16}=>4x=\frac{23}{16}-\frac{15}{16}=\frac{1}{2}=>x=\frac{1}{8}\)

(xx4)+1/2+1/4+1/8+1/16=23/16

xx4=23/16-(1/2+1/4+1/8+1/16)

xx4=1/2

x=1/2:4

x=1/8

Đúng 100% nhé Nhớ k cho mình đấy

Ơ !!??!?!!?!! chả nhìn thấy x ở đâu cả

\(\frac{10}{3}:\frac{5}{2}=\frac{10}{3}.\frac{2}{5}=\frac{20}{15}=\frac{3}{4}\)

=3/4.ung ho mk nghe

\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

\(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

\(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

=> x-23=0

x=0+23

x=23. Vậy x=23

Chúc bạn học tốt!^_^

\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\) 

=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

=>( 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

mình làm bài này từ hồi lớp 6

\(pt\Leftrightarrow\frac{29}{21}-\frac{x}{21}+\frac{27}{23}-\frac{x}{23}+\frac{25}{25}-\frac{x}{25}+\frac{23}{27}-\frac{x}{27}+\frac{21}{29}-\frac{x}{29}=-5\Leftrightarrow-x\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}\Leftrightarrow-x=\frac{-5-\frac{29}{21}-\frac{27}{23}-\frac{25}{25}-\frac{23}{27}-\frac{21}{29}}{\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}}=-50\Leftrightarrow x=50\\ \Rightarrow S=\left\{50\right\}\)

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

\(\frac{x+11}{115}+\frac{x+22}{104}=\frac{x+33}{93}+\frac{x+44}{82}\)

\(\Leftrightarrow\frac{1+\left(x+11\right)}{115}+\frac{1+\left(x+22\right)}{104}=\frac{1+\left(x+33\right)}{93}+\frac{1+\left(x+44\right)}{82}\)

\(\Leftrightarrow\frac{x+126}{115}+\frac{x+126}{104}=\frac{x+126}{93}+\frac{x+126}{82}\)

\(\Leftrightarrow\left(x+126\right).\left(\frac{1}{115}+\frac{1}{104}+\frac{1}{93}+\frac{1}{82}\right)=0\)

\(\Leftrightarrow x+126=6\Leftrightarrow x=-126\)vì\(\frac{1}{115}+\frac{1}{104}+\frac{1}{93}+\frac{1}{82}\ne0\)

vậy x=-126

- 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: \(S=\left\{23\right\}\)

chuyen ve nhom x-23 la nhan tu chung

#Tajia: Bạn sai rồi 

\(a,⇔\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)

\(⇔(x-23)(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27})=0\)

\(⇔x-23=0\) (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 \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}<0\))

\(⇔x=-2013\)