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a) \(P=\left|x-2016\right|+\left|x-2017\right|+\left|x-2018\right|\)
*TH1: \(x< 2016\):
\(P=2016-x+2017-x+2018-x=6051-3x>6051-3\cdot2016=3\)
*TH2: \(2016\le x< 2017\):
\(P=x-2016+2017-x+2018-x=2019-x>2019-2017=2\)
*TH3: \(2017\le x< 2018\):
\(P=x-2016+x-2017+2018-x=x-2015\ge2017-2015=2\)(Dấu "=" xảy ra khi x = 2017)
*TH4: \(x\ge2018\):
\(P=x-2016+x-2017+x-2018=3x-6051\ge3\cdot2018-6051=3\)(Dấu "=" xảy ra khi x = 2018)
Vậy GTNN của P là 2 khi x = 2017.
b) \(x-2xy+y-3=0\)
\(\Leftrightarrow x\left(1-2y\right)+y-\frac{1}{2}-\frac{5}{2}=0\)
\(\Leftrightarrow2x\left(\frac{1}{2}-y\right)-\left(\frac{1}{2}-y\right)=\frac{5}{2}\)
\(\Leftrightarrow\left(2x-1\right)\left(\frac{1}{2}-y\right)=\frac{5}{2}\)
\(\Leftrightarrow\left(2x-1\right)\left(1-2y\right)=5\)
| 2x-1 | 5 | -5 | 1 | -1 |
| 1-2y | 1 | -1 | 5 | -5 |
| x | 3 | -2 | 1 | 0 |
| y | 0 | 1 | -2 | 3 |
a)\(A=12-\left|x-3\right|-\left|y+7\right|\)
\(-\left|x-3\right|\le0;-\left|y+7\right|\le0\)
\(\Rightarrow A\le12-0-0=12\)
Vậy Max A = 12 <=> x = 3 ; y = -7
b)\(B=-\left(x-2018\right)^6-1\)
\(-\left(x-2018\right)^6\le0\)
\(B\le0-1=-1\)
Vậy Max B = -1 <=> x = 2018
a) \(A=12-\left|x-3\right|-\left|y+7\right|\)
Nhận thấy: \(\left|x-3\right|\ge0;\)\(\left|y+7\right|\ge0\)
suy ra: \(A=12-\left|x-3\right|-\left|y+7\right|\le12\)
Vậy MIN A = 12
Dấu "=" xảy ra <=> \(x=3;y=-7\)
b) \(B=-\left(x-2018\right)^6-1\)
Nhận thấy: \(\left(x-2018\right)^6\ge0\)
suy ra: \(B=-\left(x-2018\right)^2-1\le-1\)
Vậy MIN B = -1
Dấu "=" xảy ra <=> \(x=2018\)
c) \(C=\frac{20}{7}-\left|x+8\right|-\left(3y+7\right)^{2016}\)
Nhận thấy: \(\left|x+8\right|\ge0\) \(\left(3y+7\right)^{2016}\ge0\)
suy ra: \(C=\frac{20}{7}-\left|x+8\right|-\left(3y+7\right)^{2016}\le\frac{20}{7}\)
Vậy MIN C = 20/7
Dấu "=" xảy ra <=> \(x=-8;y=-\frac{7}{3}\)
- 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)
- Tìm giá trị nhỏ nhất:
- 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].
Khi có tổng các giá trị tuyệt đối dạng $|x-a| +
\(\Leftrightarrow\orbr{\begin{cases}x\cdot\left(x-3\right)=x\\x\cdot\left(x-3\right)=-x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-3=\frac{x}{x}\\x-3=-\frac{x}{x}\end{cases}\Leftrightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1+3\\x=-1+3\end{cases}}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)
Vậy x=2 hoặc x=4