Quên mọe dạng rồi nên làm vớ vẩn 😊😊😊

Sai 100% :)))

\(\left|x-2010\right|+\left|x-2012\right|+\left|x-2014\right|=4\)

\(\Leftrightarrow\hept{\begin{cases}\left|x-2010\right|=4\\\left|x-2012\right|=4\\\left|x-2014\right|=4\end{cases}}\)

Từ đó cứ giải bth nhá :)))

Ta có: 

| x - 1010 | + | x - 2012 | + | x - 2014 | 

= (| x - 1010 | + | 2014 - x | )+ | x - 2012 |

\(\ge\)| x - 1010 + 2014 - x | + | x - 2012 | 

= 4 + | x - 2012 |

 \(\ge4\)

Mà theo bài ra thì  | x - 1010 | + | x - 2012 | + | x - 2014 | = 4 

Do đó: ( x - 1010 ) ( 2014 - x )\(\ge\)0 và x - 2012 = 0 

<=> x = 2012 thỏa mãn 

Vậy x = 2012.

<=> \(\frac{\left(x+2014\right)}{2011}+1+\frac{\left(x+2013\right)}{2012}+1=\frac{\left(x+2012\right)}{2013}+1+\frac{\left(x+2011\right)}{2014}+1\)

\(\Rightarrow\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}=\frac{\left(x+4025\right)}{2013}+\frac{\left(x+4025\right)}{2014}\)

=> \(\frac{\left(x+4025\right)}{2011}+\frac{\left(x+4025\right)}{2012}-\frac{\left(x+4025\right)}{2013}-\frac{\left(x+4025\right)}{2014}=0\)

=> \(\left(x+4025\right)\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack=0\)

vì \(\left(\frac{1}{2011}+\frac{1}{2012}\right)>\left(\frac{1}{2013}+\frac{1}{2014}\right)\)

=> \(\left\lbrack\left(\frac{1}{2011}+\frac{1}{2012}\right)-\left(\frac{1}{2013}+\frac{1}{2014}\right)\right\rbrack>0\) hay ≠0

=> \(x+4025=0\)

\(x=-4025\)

phân số sẽ bằng 1 

vì tử = mẫu 

nhé !

Đ/s : =1

2012 + 2013 x 2014 / 2014 x 2015 -2016 = 1 

mình trả lời đầu tiên nha

bn giải tri tiết giùm mình được không

\(2014^2+\left(2014+2\right)\left(2014-2\right)=2014^2+2014^2-4=2\times2014^2-4\)

2014 x2014+2016 x 2012

=2014+2016 x 2012

=(2014+2016) x 2012

=2012 x 4036

????????

ah hihihi do ngoc

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)

\(\Leftrightarrow\frac{yz\left(x+y+z\right)+xz\left(x+y+z\right)+xy\left(x+y+z\right)-xyz}{xyz\left(x+y+z\right)}=0\)

\(\Leftrightarrow\)\(xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz-xyz=0\)

\(\Leftrightarrow\)\(\left(xyz+y^2z\right)+\left(xyz+x^2z\right)+\left(xz^2+yz^2\right)+\left(xy^2+x^2y\right)=0\)

\(\Leftrightarrow yz\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+xy\left(x+y\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(yz+xz+xy+z^2\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y\\x+z=0\end{cases}}=0\)  hoặc y+z=0

Do đó ta có B=0