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\(2020^2-2019^2+2018^2-2017^2+...+2^2-1^2\)
\(=\left(2020-2019\right)\left(2020+2019\right)+\left(2018-2017\right)\left(2018+2017\right)++\left(2-1\right)\left(2+1\right)\)
\(=2019+2018+2017+...+2+1\)
\(=\frac{\left(2019+1\right)2019}{2}\)
\(=2039190\)
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A= 2006 X 2008 - 20072
A = 2006 . 2008 - 2007 . 2007
A = 2006 . ( 2007 + 1 ) - 2007 . ( 2006 + 1 )
A = 2006 . 2007 + 2006 - 2007 . 2006 + 2007
A = -1
B= 2016 X 2018 - 20172
B= 2016 . 2018 - 2017 . 2017
B = 2016 . ( 2017 + 1 ) - 2017 . ( 2016 + 1 )
B = 2016 . 2017 + 2016 - 2017 . 2016 + 2017
B = -1
Ta có : |x - 2| ; |x - 5| ; |x - 18| ≥0∀x∈R≥0∀x∈R
=> |x - 2| + |x - 5| + |x - 18| ≥0∀x∈R≥0∀x∈R
=> D có giá trị nhỏ nhất khi x = 2;5;18
Mà x ko thể đồng thời nhận 3 giá trị
Nên GTNN của D là : 16 khi x = 5 ok nha bạn
x^2/x-1 = x^2-4x+4/x-1 + 4 = (x-2)^1/x-1 + 4 >= 4
Dấu "=" xảy ra <=> x-2 = 0 <=> x = 2 (tm)
Vậy GTNN của x^2/x-1 = 4 <=> x= 2
k mk nha
\(x^2-x+y^2-y\)
\(=x.x-x+y.y-y\)
\(=2x+2y\)
\(2\left(x+y\right)\)
Ta có :
2(xy - x^2 - y + 1008) = y^2 + 2018
<=> 2xy - 2x^2 - 2y + 2016 = y^2 + 2018
<=> 2xy - 2x^2 - 2y = y^2 + 2
<=> 2xy - 2x^2 - 2y - y^2 - 2 = 0
<=> -(2x^2 - 2xy + y^2/2) - y^2/2 - 2y - 2 = 0
<=> -2(x^2 - xy + y^2/4) - 2(y^2/4 + y + 1) = 0
<=> -2(x-y/2)^2 - 2(y/2 + 1)^2 = 0
<=> 2(x-y/2)^2 + 2(y/2 + 1)^2 = 0
Dấu " = " xảy ra <=> x - y/2 = 0 ; y/2 + 1 = 0
<=> x = y/2 ; y = -2
<=> x = -1 ; y = -2
Vậy x = -1 ; y = -2
Ta có :
2(xy - x^2 - y + 1008) = y^2 + 2018
<=> 2xy - 2x^2 - 2y + 2016 = y^2 + 2018
<=> 2xy - 2x^2 - 2y = y^2 + 2
<=> 2xy - 2x^2 - 2y - y^2 - 2 = 0
<=> -(2x^2 - 2xy + y^2/2) - y^2/2 - 2y - 2 = 0
<=> -2(x^2 - xy + y^2/4) - 2(y^2/4 + y + 1) = 0
<=> -2(x-y/2)^2 - 2(y/2 + 1)^2 = 0
<=> 2(x-y/2)^2 + 2(y/2 + 1)^2 = 0
Dấu " = " xảy ra <=> x - y/2 = 0 ; y/2 + 1 = 0
<=> x = y/2 ; y = -2
<=> x = -1 ; y = -2
Vậy x = -1 ; y = -2
\(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
Ta có: \(\left(x-2\right)^{2020}\ge0\forall x\)
\(\left(y+5\right)^{2018}\ge0\forall y\)
Do đó: \(\left(x-2\right)^{2020}+\left(y+5\right)^{2018}\ge0\forall x,y\)
Dấu '=' xảy ra khi (x,y)=(2;-5)