Tìm x :
a) \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)
Giúp mik với . Ai nhanh tay trả lời câu hỏi trên thì mik tick nhé . TKS trước !
Bài 1 : Tìm x thuộc Z biết : ( Trình bày rõ 2 ý => 2 likes )
a, \(\frac{x+3}{47}+\frac{x+2}{48}=-2\)
b, \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-1\)
tìm x biết:
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
tìm x biết:
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)
\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)
\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)
\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)
Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)
\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy \(x=100\)
\(\dfrac{x-1}{99}-\dfrac{x+1}{101}+\dfrac{x-2}{98}-\dfrac{x+2}{102}+\dfrac{x-3}{97}-\dfrac{x+3}{103}+\dfrac{x-4}{96}-\dfrac{x+4}{104}=0\)
<=> \(\dfrac{x-1}{99}-1-\dfrac{x+1}{101}-1+\dfrac{x-2}{98}-1-\dfrac{x-2}{102}-1+\dfrac{x-3}{97}-1-\dfrac{x+3}{103}-1+\dfrac{x-4}{96}-1-\dfrac{x+4}{104}=0\)
Tìm x \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)
\(\Leftrightarrow\frac{x+10}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)\(\Rightarrow x+100=0\Leftrightarrow x=-100\)
bài 1 tìm x :
a) \(\frac{x+1}{5}+\frac{x+1}{7}=\frac{x+1}{9}\)
b)\(\frac{x+4}{96}+\frac{x+3}{97}=\frac{x+2}{98}+\frac{x+1}{99}\)
a, \(\frac{x+1}{5}+\frac{x+1}{7}=\frac{x+1}{9}\)
\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{7}-\frac{x+1}{9}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
b, \(\frac{x+4}{96}+\frac{x+3}{97}=\frac{x+2}{98}+\frac{x+1}{99}\)
\(\Leftrightarrow\left(\frac{x+4}{96}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+1}{99}+1\right)\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}=\frac{x+100}{98}+\frac{x+100}{99}\)
\(\Leftrightarrow\frac{x+100}{96}+\frac{x+100}{97}-\frac{x+100}{98}-\frac{x+100}{99}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{96}+\frac{1}{97}+\frac{1}{98}+\frac{1}{99}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
a) x + 1/5 + x + 1/7 = x + 1/9
<=> 1/5x + 1/5 + 1/7x + 1/7 = 1/9x + 1/9
<=> (1/5x + 1/7x) + (1/5 + 1/7) = 1/9x + 1/9
<=> 12/35x + 12/35 = 1/9x + 1/9
<=> 12/35x + 12/35 - 1/9x = 1/9
<=> 73/315x + 12/35 = 1/9
<=> 73/315x = 1/9 - 12/35
<=> 73/315x = -73/315
<=> x = 73/315 : -73/315 = -1
=> x = -1
b) làm tương tự
Giải phương trình :
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+6}{94}\)
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+6}{94}\)
\(\Leftrightarrow\)\(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+\frac{x+6}{94}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{94}\)
\(\Leftrightarrow\)(x+100)(\(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\))=0
\(\Leftrightarrow\)x+100=0(vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\ne0\))
\(\Leftrightarrow\)x=-100
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+6}{94}\)
\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+6}{94}+1\right)\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}-\frac{x+100}{94}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\ne0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
Vậy \(x=-100\)
Tìm x:
\(a.\)\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(b.\)\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
Giải giúp ạ :3
\(a)\) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+4}{96}+1\right)=-4+4\)
\(\Leftrightarrow\)\(\frac{x+1+99}{99}+\frac{x+2+98}{98}+\frac{x+3+97}{97}+\frac{x+4+96}{96}=0\)
\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)
Nên \(x+100=0\)
\(\Rightarrow\)\(x=-100\)
Vậy \(x=-100\)
Chúc bạn học tốt ~
\(b)\) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(1-\frac{1}{x+1}=\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=1-\frac{2008}{2009}\)
\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{2009}\)
\(\Leftrightarrow\)\(x+1=2009\)
\(\Leftrightarrow\)\(x=2009-1\)
\(\Leftrightarrow\)\(x=2008\)
Vậy \(x=2008\)
Chúc bạn học tốt ~
a) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)+\)\(\left(\frac{x+4}{96}+1\right)=-4+4\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)
\(\Rightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Tìm x biết :
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
Ta có :
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=-4+4\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)nên \(x+100=0\)
\(\Rightarrow x=0-100=-100\)
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yk nha
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)hãy tìm x
(\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}\))+4=0
=)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}\)=0
=)\(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Mà \(\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)>0\)
=)\(x+100=0\)
=)\(x=-100\)
Vậy \(x=-100\)
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}+4=0\)
\(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)
\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
x+100=0 vì (1/99+1/98+1/97+1/96) khác 0
x=0-100
x=-100
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}=-4\)
\(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0\)
\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0\)
\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\)
nên x+100=0
x= -100