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\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\).Do \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\ne0\)
\(\Rightarrow x=-1\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=0-1\)
\(\Rightarrow x=-1\)
Vậy x=-1
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow x+1.\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\div\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Ta có :
Vậy x = -1
P/s : mk ko coppy của do not đâu nhá mk làm rồi tải thời gian tải ảnh lên và trả lwoif bằng hình ảnh mk mwosi chậm thôi
https://olm.vn/hoi-dap/question/995768.html Sky Ntt trả lwoif sau mik ok mk ko coppy ai !
a) (x+1)/10+(x+1)/11+(x+1)/12=(x+1)/13+(x+1)/14
(x+1)/10+(x+1)/11+(x+1)/12-(x+1)/13-(x+1)/14=0
(x+1)(1/10+1/11+1/12-1/13-1/14)=0 (1)
thay 1/10>1/13
1/11>1/14
1/12>0
suy ra 1/10+1/11+1/12>1/13+1/14
suy ra 1/10+1/11+1/12-1/13-1/14>0
suy ra 1/10+1/11+1/12-1/13-1/14 khac 0
nền (1) tương dương x+1=0
tương dương x=-1
Vay x=-1
b) (x+4)/2000+(x+3)/2001=(x+2)/2002+(x+1)/2003
(x+4)/2000+(x+3)/2001-(x+2)/2002-(x+1)/2003=0
[(x+4)/2000+1]+[(x+3)/2001+1]-[(x+2)/2002+1]-[(x+1)/2003+1]=0
(x+2004)/2000+(x+2004)/2001-(x+2004)/2002-(x+2004)/2003=0
(x+2004)(1/2000+1/2001-1/2002/1/2003)=0 (2)
thay 1/2000>1/2002
1/2001>1/2003
suy ra 1/2000+1/2001>1/2002+1/2003
suy ra 1/2000+1/2001-1/2002-1/2003>0
suy ra 1/2000+1/2001-1/2002-1/2003 khac 0
nen (2) tuong duong x+2004=0
tuong duong x=-2004
Vay x=-2004
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)nên \(x+1=0\)
\(\Rightarrow x=-1\)
Ta có : \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
=> \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà : \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
Nên : x + 1 = 0
=> x = -1
sai dấu kìa bạn
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}+\frac{x+1}{13}+\frac{x+1}{14}=0\)
\(=>\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)=0\)
\(=>\left(x+1\right)=0\)
\(=>x=-1\)