\(\left(x-1\right)^2+\left|2y-x\right|=0\)

có \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left|2y-x\right|\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x-1=0\\2y-x=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\2y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{2}\end{cases}}}\)

vậy_

cảm ơn bạn nha bạn học lớp mấy thế năm nay mình lên lớp 7

• Khoảng 1: x<2013x is less than 2013𝑥<2013
• • |x−2013|=−(x−2013)=2013−xthe absolute value of x minus 2013 end-absolute-value equals negative open paren x minus 2013 close paren equals 2013 minus x|𝑥−2013|=−(𝑥−2013)=2013−𝑥
  • |x−2014|=−(x−2014)=2014−xthe absolute value of x minus 2014 end-absolute-value equals negative open paren x minus 2014 close paren equals 2014 minus x|𝑥−2014|=−(𝑥−2014)=2014−𝑥
  • |x−2015|=−(x−2015)=2015−xthe absolute value of x minus 2015 end-absolute-value equals negative open paren x minus 2015 close paren equals 2015 minus x|𝑥−2015|=−(𝑥−2015)=2015−𝑥
  • B=(2013−x)+(2014−x)+(2015−x)=6042−3xcap B equals open paren 2013 minus x close paren plus open paren 2014 minus x close paren plus open paren 2015 minus x close paren equals 6042 minus 3 x𝐵=(2013−𝑥)+(2014−𝑥)+(2015−𝑥)=6042−3𝑥 (Giảm dần)
• Khoảng 2: 2013≤x<20142013 is less than or equal to x is less than 20142013≤𝑥<2014
• • |x−2013|=x−2013the absolute value of x minus 2013 end-absolute-value equals x minus 2013|𝑥−2013|=𝑥−2013
  • |x−2014|=−(x−2014)=2014−xthe absolute value of x minus 2014 end-absolute-value equals negative open paren x minus 2014 close paren equals 2014 minus x|𝑥−2014|=−(𝑥−2014)=2014−𝑥
  • |x−2015|=−(x−2015)=2015−xthe absolute value of x minus 2015 end-absolute-value equals negative open paren x minus 2015 close paren equals 2015 minus x|𝑥−2015|=−(𝑥−2015)=2015−𝑥
  • B=(x−2013)+(2014−x)+(2015−x)=2016−xcap B equals open paren x minus 2013 close paren plus open paren 2014 minus x close paren plus open paren 2015 minus x close paren equals 2016 minus x𝐵=(𝑥−2013)+(2014−𝑥)+(2015−𝑥)=2016−𝑥 (Giảm dần)
• Khoảng 3: 2014≤x<20152014 is less than or equal to x is less than 20152014≤𝑥<2015
• • |x−2013|=x−2013the absolute value of x minus 2013 end-absolute-value equals x minus 2013|𝑥−2013|=𝑥−2013
  • |x−2014|=x−2014the absolute value of x minus 2014 end-absolute-value equals x minus 2014|𝑥−2014|=𝑥−2014
  • |x−2015|=−(x−2015)=2015−xthe absolute value of x minus 2015 end-absolute-value equals negative open paren x minus 2015 close paren equals 2015 minus x|𝑥−2015|=−(𝑥−2015)=2015−𝑥
  • B=(x−2013)+(x−2014)+(2015−x)=x−2012cap B equals open paren x minus 2013 close paren plus open paren x minus 2014 close paren plus open paren 2015 minus x close paren equals x minus 2012𝐵=(𝑥−2013)+(𝑥−2014)+(2015−𝑥)=𝑥−2012 (Tăng dần)
• Khoảng 4: x≥2015x is greater than or equal to 2015𝑥≥2015
• • |x−2013|=x−2013the absolute value of x minus 2013 end-absolute-value equals x minus 2013|𝑥−2013|=𝑥−2013
  • |x−2014|=x−2014the absolute value of x minus 2014 end-absolute-value equals x minus 2014|𝑥−2014|=𝑥−2014
  • |x−2015|=x−2015the absolute value of x minus 2015 end-absolute-value equals x minus 2015|𝑥−2015|=𝑥−2015
  • B=(x−2013)+(x−2014)+(x−2015)=3x−6042cap B equals open paren x minus 2013 close paren plus open paren x minus 2014 close paren plus open paren x minus 2015 close paren equals 3 x minus 6042𝐵=(𝑥−2013)+(𝑥−2014)+(𝑥−2015)=3𝑥−6042 (Tăng dần) 

1. Tìm giá trị nhỏ nhất:
2. • Giá trị B giảm đến x=2014x equals 2014𝑥=2014 (B = 2016 - 2014 = 2) rồi bắt đầu tăng.
   • Giá trị nhỏ nhất của B là 2, đạt được khi xx𝑥 nằm trong khoảng [2014,2015]open bracket 2014 comma 2015 close bracket[2014,2015]. 

mới học lớp 6 à

Có: \(\left(x-2y+1\right)^2\ge0\forall x;y\)

\(\left|y+1\right|\ge0\forall y\)

\(\Rightarrow\left(x-2y+1\right)^2+\left|y+1\right|\ge0\forall x;y\)

\(\Rightarrow\left(x-2y+1\right)^2+\left|y+1\right|+17\ge17\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-2y+1\right)^2=0\\\left|y+1\right|=0\end{cases}}\)

\(\left|y+1\right|=0\Leftrightarrow y+1=0\Leftrightarrow y=-1\)

\(\left(x-2y+1\right)^2=0\Leftrightarrow x-2y+1=0\Leftrightarrow x-2.\left(-1\right)+1=0\Leftrightarrow x+2+1=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)

Vậy GTNN của A = 17 \(\Leftrightarrow\left(x;y\right)=\left(-3;-1\right)\)

Ta có : \(-\left|2,5-x\right|+1,3=0\)

=> \(-\left|2,5-x\right|=-1,3\)

=> \(\left|2,5-x\right|=1,3\)

\(\Leftrightarrow\orbr{\begin{cases}2,5-x=1,3\\2,5-x=-1,3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2,5-1,3\\x=2,5+1,3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1,2\\x=3,8\end{cases}}\)

Th1 :\(\left|2,5-x\right|=2,5-x\) khi \(2,5-x\ge0\Leftrightarrow x\le2,5\) ta có 

      \(-\left(2,5-x\right)+1,3=0\Leftrightarrow-2,5+x+1,3=0\)

\(\Leftrightarrow x+1,3=2,5\Leftrightarrow x=1,2\left(tm\right)\)

Th2 : \(\left|2,5-x\right|=-\left(2,5-x\right)=-2,5+x\) khi \(2,5-x< 0\Leftrightarrow x>2,5\) ta có

      \(-\left(-2,5+x\right)+1,3=0\Leftrightarrow2,5-x+1,3\)

\(\Leftrightarrow-x+1,3=-2,5\Leftrightarrow-x=-3,8\Leftrightarrow x=3,8\left(tm\right)\)

vậy pt có tập nghiệm S={1,2 ; 3,8}

\(A=\left|x+12\right|+\left(y+2\right)^2+11\ge11\)

ta có \(\hept{\begin{cases}\left|x+12\right|\ge0\\\left(y+2\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left|x+12\right|+\left(y+2\right)^2+11\ge11\)

\(\Rightarrow A_{min}=11\Leftrightarrow\hept{\begin{cases}x+12=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=-12\\y=-2\end{cases}}}\)