Tìm GTNN A=|x-9|+8-x
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\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\Leftrightarrow\left(4-x\right)\left(4-x\right)=-5\times-5\)
\(\Rightarrow\left(4-x\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
\(\dfrac{4-x}{-5}=\dfrac{-5}{4-x}\)
\(\Rightarrow\left(4-x\right).\left(4-x\right)=\left(-5\right).\left(-5\right)\)
\(\Rightarrow\left(4-x\right)^2=25\)
\(\Rightarrow\left(4-x\right)^2=5^2\)
\(\Rightarrow4-x=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4-5\\x=4-\left(-5\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)
\(A=x^4+x^2-6x+9=\left(x^4-2x^2+1\right)+\left(3x^2-6x+3\right)+5\)
\(=\left(x^2-1\right)^2+3\left(x-1\right)^2+5\ge5\)
\(B=\left(x-4\right)\left(x-1\right)\left(x-5\right)\left(x-8\right)+2017\)
\(=\left(x^2-9x+8\right)\left(x^2-9x+20\right)+2017\)
Đặt \(x^2-9x+8=a\)
\(\Rightarrow B=a\left(a+12\right)+2017=a^2+12a+36+1981\)
\(=\left(a+36\right)^2+1981\ge1981\)
a, Vì |x-3| \(\ge\)0
=>A=|x-3|+50\(\ge\)50
Dấu "=" xảy ra khi x=3
Vậy GTNN của A = 50 khi x=3
b, Vì |x+8| \(\ge0\)
=>B=2014-|x+8|\(\le2014\)
Dấu "=" xảy ra khi x=-8
Vậy GTLN của B = 2014 khi x=-8
c, Vì \(\hept{\begin{cases}\left|x-100\right|\ge0\\\left|y+2014\right|\ge0\end{cases}}\)
\(\Rightarrow\left|x-100\right|+\left|y+2014\right|\ge0\)
\(\Rightarrow C=\left|x-100\right|+\left|y+2014\right|-2015\ge-2015\)
Dấu "=" xảy ra khi x=100,y=-2014
Vậy GTNN của C=-2015 khi x=100,y=-2014
\(x^2-6x+y^2-10y-15\)
\(=x^2-6x+y^2-10y+9+25-49\)
\(=\left(x^2-6x+9\right)+\left(y^2-10y+25\right)-49\)
\(=\left(x-3\right)^2+\left(y-5\right)^2-49\ge-49\)
Vậy GTNN của bt là -49\(\Leftrightarrow\hept{\begin{cases}x-3=0\\y-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=5\end{cases}}\)
\(x^2-6x+y^2-10y-15\)\
\(=\left(x^2-9x+9\right)+\left(y^2-10y+25\right)-49\)
\(=\left(x-3\right)^2+\left(y-5\right)^2-49\)\(\ge49\)
vậy GTNN là 49
\(B=\left|x-7\right|+\left|x+8\right|\)
\(\Rightarrow B=\left|7-x\right|+\left|x+8\right|\)
\(\Rightarrow B\ge\left|7-x+x+8\right|\)
\(\Rightarrow B\ge\left|15\right|\)
\(\Rightarrow B\ge15\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-7\right)\left(x+8\right)\ge0\)
Vậy \(B_{min}=15\Leftrightarrow\left(x-7\right)\left(x+8\right)\ge0\)
\(K\left(x\right)=L\left(x\right)\)
\(\Rightarrow x^2-3x+2=x^2+px+q+1\)
\(\Rightarrow-3x+2=px+q+1\)
-Áp dụng PP hệ số bất định:
\(\Rightarrow p=-3;q+1=2\Rightarrow q=1\)
Ta có : \(\left|x-\frac{2}{5}\right|\ge0\forall x\in R\)
=> A = \(\left|x-\frac{2}{5}\right|+\frac{8}{9}\ge\frac{8}{9}\forall x\in R\)
Vậy GTNN của A là : \(\frac{8}{9}\) khi x = \(\frac{2}{5}\)
\(x+2\dfrac{1}{4}=4\dfrac{5}{6}\)
\(x=4\dfrac{5}{6}-2\dfrac{1}{4}\)
\(x=4+\dfrac{5}{6}-2-\dfrac{1}{4}\)
\(x=2+\dfrac{7}{12}\)
\(x=\dfrac{31}{12}\)
_____________
\(9\dfrac{3}{5}-x=3\dfrac{3}{8}\)
\(x=9\dfrac{3}{5}-3\dfrac{3}{8}\)
\(x=9+\dfrac{3}{5}-3-\dfrac{3}{8}\)
\(x=6+\dfrac{9}{40}\)
\(x=\dfrac{249}{40}\)
\(x+2\dfrac{1}{4}=4\dfrac{5}{6}\)
\(=>x+\dfrac{9}{4}=\dfrac{29}{6}\)
\(=>x=\dfrac{29}{6}-\dfrac{9}{4}=\dfrac{58}{12}-\dfrac{27}{12}\)
\(=>x=\dfrac{31}{12}\)
_______
\(9\dfrac{3}{5}-x=3\dfrac{3}{8}\)
\(=>\dfrac{48}{5}-x=\dfrac{27}{8}\)
\(=>x=\dfrac{48}{5}-\dfrac{27}{8}=\dfrac{384}{40}-\dfrac{135}{40}\)
\(=>x=\dfrac{249}{40}\)
Ta có:
|x-9|\(\ge\)0\(\forall\)x
|x-9|+8\(\ge\)0\(\forall\)x
|x-9|+8-x\(\ge\)-x\(\forall\)x
A\(\ge\)-x\(\forall\)x
Dấu "=" xảy ra khi và chỉ khi |x-9|=0
\(\Leftrightarrow\)x-9=0
\(\Leftrightarrow\)x=9
Vậy MIN A=-x với x=9