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A = | x - 2015 | +| x - 2016 |
A = | x - 2015 | + | 2016 - x |
A = | x - 2015 | + | 2016 - x | \(\ge\)| x - 2015 + 2016 - x |
A = | x - 2015 | + | 2016 - x | \(\ge\)1
Dấu = xảy ra\(\Leftrightarrow\)x - 2015 = 0 ; 2016 - x = 0
\(\Rightarrow\)x = 2015 hoặc x = 2016
Min A = 1 \(\Leftrightarrow\)x = 2015 hoặc x = 2016
\(a,A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-2018\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-2018\)
Đặt \(x^2+5x=a\)
\(\Rightarrow A=\left(a-6\right)\left(a+6\right)-2018=a^2-2054\)
\(\Rightarrow A_{min}=2054\Leftrightarrow a=0\)
\(\Rightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow x\in\left\{0;-5\right\}\)
\(b,B=\left(x-1\right)\left(x-4\right)\left(x-5\right)\left(x-8\right)+2018.\)
\(=\left(x^2-9x+8\right)\left(x^2-9x+20\right)+2018\)
Đặt \(x^2-9x+14=a\)
\(\Rightarrow B=\left(a-6\right)\left(a+6\right)+2018\)
\(=a^2-36+2018=a^2+1982\)
\(\Rightarrow B_{min}=1982\Leftrightarrow a^2=0\Rightarrow a=0\)
\(\Rightarrow x^2-9x+14=0\)
\(\Rightarrow x^2-2x-7x+14=0\)
\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-7\right)=0\)
\(\Rightarrow x\in\left\{2;7\right\}\)
xin loi , may tinh minh hong unikey
Dat \(\frac{x}{2017}=\frac{y}{2018}=\frac{z}{2019}=k\)
Suy ra \(x=2017k;y=2018k;z=2019k\)
Khi đó 4.(x-y).(y-z) = \(4.\left(2017k-2018k\right).\left(2018k-2019k\right)=4.\left(-k\right).\left(-k\right)=4k^2\)
\(\left(z-x\right)^2=\left(2019k-2017k\right)^2=\left(2k\right)^2=4k^2\)
Nen \(4.\left(x-y\right).\left(y-z\right)=\left(z-x\right)^2\)
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 |