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20 tháng 2 2023

Cần gấp ^^

\(\Leftrightarrow\left(\dfrac{x-11}{111}+1\right)+\left(\dfrac{x-12}{112}+1\right)=\left(\dfrac{x-23}{123}+1\right)+\left(\dfrac{x-24}{124}+1\right)\)

=>x+100=0

=>x=-100

4 tháng 1 2019

Bạn vất vả r

8 tháng 2 2018

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

8 tháng 2 2018

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

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

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

\(\Leftrightarrow x-23=0\) ( 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 }

1 tháng 3 2022

9092 = 0

1 tháng 3 2022

cái này + mỗi phân số vs 1 á 

 

8 tháng 2 2018

a.

\(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)

\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+3}{4}=3-\dfrac{x+2}{3}\)

\(\Leftrightarrow\dfrac{\left(x+1\right).6}{12}+\dfrac{\left(x+3\right).3}{12}=\dfrac{36}{12}-\dfrac{\left(x+2\right).4}{12}\)

\(\Leftrightarrow6x+6+3x+9=36-4x-8\)

\(\Leftrightarrow9x+15=28-4x\)

\(\Leftrightarrow9x+4x=28-15\)

\(\Leftrightarrow13x=13\)

\(\Leftrightarrow x=1\)

8 tháng 2 2018

a) \(\dfrac{1}{2}\left(x+1\right)+\dfrac{1}{4}\left(x+3\right)=3-\dfrac{1}{3}\left(x+2\right)\)

\(\Leftrightarrow\dfrac{6\left(x+1\right)+3\left(x+3\right)}{12}=\dfrac{36-4\left(x+2\right)}{12}\)

\(\Leftrightarrow6\left(x+1\right)+3\left(x+3\right)=36-4\left(x+2\right)\)

\(\Leftrightarrow6x+6+3x+9=36-4x-8\)

\(\Leftrightarrow9x+15=-4x+28\)

\(\Leftrightarrow9x+4x=28-15\)

\(\Leftrightarrow13x=13\)

\(\Leftrightarrow x=1\)

Vậy ................................

14 tháng 12 2017

a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

\(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)

b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)

\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)

\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)

\(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)

c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

\(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)

3 tháng 3 2017

a. \(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}=\dfrac{x+100}{94}+\dfrac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

c. \(\Leftrightarrow3x^2+3x-x-1=0\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow\left[\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

3 tháng 2 2019

Câu a)

Giải phương trình,(x + 1)/2004 + (x + 2)/2003 = (x + 3)/2002 + (x + 4)/2001,Toán học Lớp 8,bà i tập Toán học Lớp 8,giải bà i tập Toán học Lớp 8,Toán học,Lớp 8

3 tháng 2 2019

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

31 tháng 3 2017

a)\(\dfrac{1}{2}\)(x+1)+\(\dfrac{1}{4}\)(x+3)=3-\(\dfrac{1}{3}\)(x+2)

\(\Leftrightarrow\)\(\dfrac{1}{2}\)x+\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)x+\(\dfrac{3}{4}\)=3-\(\dfrac{1}{3}\)x-\(\dfrac{2}{3}\)

\(\Leftrightarrow\)\(\dfrac{1}{2}\)x+\(\dfrac{1}{4}\)x+\(\dfrac{1}{3}\)x=-\(\dfrac{1}{2}\)-\(\dfrac{3}{4}\)+3-\(\dfrac{2}{3}\)

\(\Leftrightarrow\)\(\dfrac{13}{12}\)x=\(\dfrac{13}{12}\)

\(\Leftrightarrow\)x=1

Vậy nghiệm của pt là x=1

b)\(\dfrac{x+2}{98}\)+\(\dfrac{x+4}{96}\)=\(\dfrac{x+6}{94}\)+\(\dfrac{x+8}{92}\)

\(\Leftrightarrow\)\(\dfrac{x+2}{98}\)+\(\dfrac{x+4}{96}\)-\(\dfrac{x+6}{94}\)-\(\dfrac{x+8}{92}\)=0

\(\Leftrightarrow\)(\(\dfrac{x+2}{98}\)+1)+(\(\dfrac{x+4}{96}\)+1)-(\(\dfrac{x+6}{94}\)+1)-(\(\dfrac{x+8}{92}\)+1)=0

\(\Leftrightarrow\)\(\dfrac{x+2+98}{98}\)+\(\dfrac{x+4+96}{96}\)-\(\dfrac{x+6+94}{94}\)-\(\dfrac{x+8+92}{92}\)=0

\(\Leftrightarrow\)\(\dfrac{x+100}{98}\)+\(\dfrac{x+100}{96}\)-\(\dfrac{x+100}{94}\)-\(\dfrac{x+100}{92}\)=0

\(\Leftrightarrow\)(x+100)(\(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\))=0

\(\Leftrightarrow\)x+100=0(vì\(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\)\(\ne\)0)

\(\Leftrightarrow\)x=-100

Vậy nghiệm của pt là x=-100

11 tháng 1 2018

a ) \(\dfrac{x+1}{100}+\dfrac{x+2}{99}+\dfrac{x+4}{97}=-4\)

\(\Leftrightarrow\dfrac{x+101}{100}+\dfrac{x+101}{99}+\dfrac{x+101}{97}=-1\)

\(\Leftrightarrow\left(x+101\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{97}\right)=-1\)

Vô lí => Phương trình trên vô nghiệm .

b ) \(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)

\(\Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-73}{16}=0\)

\(\Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)

\(\Leftrightarrow x=89\)

Vậy x = 89.

11 tháng 1 2018

wrong