Câu hỏi của Bảo Uyên - Toán lớp 7 - Học toán với OnlineMath

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\(A=\left|x-2011\right|+\left|x-2012\right|+\left|x-2013\right|+\left|x-2014\right|+\left|x-2015\right|\)

\(=\left(\left|x-2011\right|+\left|x-2015\right|\right)+\left(\left|x-2012\right|+\left|x-2014\right|\right)+\left|x-2013\right|\)

Đặt \(B=\left|x-2011\right|+\left|x-2015\right|\)

\(=\left|x-2011\right|+\left|2015-x\right|\ge\left|x-2011+2015-x\right|=4\left(1\right)\)

Dấu"=" xảy ra \(\Leftrightarrow\left(x-2011\right)\left(2015-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-2011\ge0\\2015-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2011< 0\\2015-x< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge2011\\x\le2015\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2011\\x>2015\end{cases}\left(loai\right)}\)

\(\Leftrightarrow2011\le x\le2015\)

Đặt \(C=\left|x-2012\right|+\left|x-2014\right|\)

\(=\left|x-2012\right|+\left|2014-x\right|\ge\left|x-2012+2014-x\right|=2\left(2\right)\)

Dấu"="xảy ra \(\Leftrightarrow\left(x-2012\right)\left(2014-x\right)\ge0\)

\(\Leftrightarrow\hept{\begin{cases}x-2012\ge0\\2014-x\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2012< 0\\2014-x< 0\end{cases}}\) 

\(\Leftrightarrow\hept{\begin{cases}x\ge2012\\x\le2014\end{cases}}\)hoặc\(\hept{\begin{cases}x< 2012\\x>2014\end{cases}\left(loai\right)}\)

\(\Leftrightarrow2012\le x\le2014\)

Ta có: \(\left|x-2013\right|\ge0;\forall x\left(3\right)\)

Dấu"="Xảy ra \(\Leftrightarrow\left|x-2013\right|=0\)

                      \(\Leftrightarrow x=2013\)

Từ (1),(2) và (3) \(\Rightarrow B+C+\left|x-2013\right|\ge6\)

Hay \(A\ge6\)

Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}2011\le x\le2015\\2012\le x\le2014\\x=2013\end{cases}}\)\(\Leftrightarrow x=2013\)

Vậy \(A_{min}=6\Leftrightarrow x=2013\)

• 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]. 

Áp dụng \(|a|\ge0\)với \(\forall a\)Dấu "=" xảy ra khi \(a\ge0\)

Ta có: \(|x-2013|+|x-2015|=|x-2013|+|2015-x|\ge x-2013+2015-x=2với\forall x\)

Dâu "=" xảy ra khi \(x-2013\ge0\)và\(2015-x\ge0\)\(\Leftrightarrow\)\(2013\le x\le2015\)

Lại có: \(|x-2014|\ge0với\forall x\)

Dấu "=" xảy ra khi \(x-2014=0\Leftrightarrow x=2014\)

Do đó \(A\ge2+0=2với\forall x\)

Dấu "=" xảy ra khi \(2013\le x\le2015\)và \(x=2014\)\(\Leftrightarrow\)\(x=2014\)

Vậy \(minA=2\)khi\(x=2014\)

Ta có: \(\left|x-2013\right|+\left|x-2015\right|=\left|x-2013\right|+\left|2015-x\right|\ge\left|x-2013+2015-x\right|\)

                                                                         \(\left|x-2013\right|+\left|2015-x\right|\ge2\)\(\left(1\right)\)

                                                                   Và \(\left|2014-x\right|\ge0\)

                                                                  \(\Rightarrow\left|x-2013\right|+\left|2014-x\right|+\left|2015-x\right|\ge2\)

                                                                Mà \(\left|x-2013\right|+\left|2014-x\right|+\left|2015-x\right|=A\)

                                                                      \(\Rightarrow A\)có GTNN là 2

                                         Từ\(\left(1\right)\)

                                 \(\Rightarrow\)Dấu \("="\)xảy ra khi \(\left(x-2013\right)\left(2015-x\right)\ge0\)

                                                \(\Rightarrow2013\le x\le2015\)

                                                 \(\Rightarrow x=2014\)

                              Vậy, \(A\)có GTNN là 2 khi\(x=2014\)

giá trị nhỏ nhất = 1

rõ hơn đi bạn

có phải giá trị tuyệt đối ko

là 1 đấy

\(\frac{x+4}{2010}+\frac{x+3}{2011}=\frac{x+2}{2012}+\frac{x+1}{2013}\)

\(\Leftrightarrow\left(\frac{x+4}{2010}+1\right)+\left(\frac{x+3}{2011}+1\right)=\left(\frac{x+2}{2012}+1\right)+\left(\frac{x+1}{2013}+1\right)\)

\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}=\frac{x+2014}{2012}+\frac{x+2014}{2013}\)

\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}-\frac{x+2014}{2012}-\frac{x+2014}{2013}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

\(\Leftrightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)

V...

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