mn giúp e vs ạ
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bài 2
Ta có:
\(A=\left|x-102\right|+\left|2-x\right|\Rightarrow A\ge\left|x-102+2-x\right|=-100\Rightarrow GTNNcủaAlà-100\)đạt được khi \(\left|x-102\right|.\left|2-x\right|=0\)
Trường hợp 1: \(x-102>0\Rightarrow x>102\)
\(2-x>0\Rightarrow x< 2\)
\(\Rightarrow102< x< 2\left(loại\right)\)
Trường hợp 2:\(x-102< 0\Rightarrow x< 102\)
\(2-x< 0\Rightarrow x>2\)
\(\Rightarrow2< x< 102\left(nhận\right)\)
Vậy GTNN của A là -100 đạt được khi 2<x<102.
a) \(M=2022-\left|x-9\right|\le2022\)
\(maxM=2022\Leftrightarrow x=9\)
b) \(N=\left|x-2021\right|+2022\ge2022\)
\(minN=2022\Leftrightarrow x=2021\)
Bài 1:
Ta có: \(6.|3x-12|\ge0\forall x\)
\(\Rightarrow23+6.|3x-12|\ge23+0\forall x\)
Hay \(A\ge23\forall x\)
Dấu"=" xảy ra \(\Leftrightarrow3x-12=0\)
\(\Leftrightarrow x=4\)
Vậy Min A=23 \(\Leftrightarrow x=4\)
Bài 2:
Ta có: \(5.|14-7x|\ge0\forall x\)
\(\Rightarrow-5.|14-7x|\le0\forall x\)
\(\Rightarrow2019-5.|14-7x|\le2019-0\forall x\)
Hay \(B\le2019\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow14-7x=0\)
\(\Leftrightarrow x=2\)
Vậy Max B=2019 \(\Leftrightarrow x=2\)
\(A=5-8x+x^2=-8x+x^2+6-11\)
\(=\left(x-4\right)^2-11\)
Vì \(\left(x-4\right)^2\ge0\forall x\)\(\Rightarrow\left(x-4\right)^2-11\ge-11\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)
Vậy Amin = - 11 <=> x = 4
\(B=\left(2-x\right)\left(x+4\right)=-x^2-2x+8\)
\(=-\left(x^2+2x+1\right)+9=-\left(x+1\right)^2+9\)
Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+1\right)^2+9\le9\)
Dấu "=" xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy Bmax = 9 <=> x = - 1
2: B=|x+5|-|x-2|<=|x+5-x+2|=7
Dấu = xảy ra khi -5<=x<=2
Bài 15:
1: \(A=4x-x^2+1\)
\(=-\left(x^2-4x-1\right)\)
\(=-\left(x^2-4x+4-5\right)\)
\(=-\left(x-2\right)^2+5< =5\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
2: \(B=3-4x-x^2\)
\(=-\left(x^2+4x-3\right)\)
\(=-\left(x^2+4x+4-7\right)\)
\(=-\left(x+2\right)^2+7< =7\forall x\)
Dấu '=' xảy ra khi x+2=0
=>x=-2
3: \(C=8-x^2-5x\)
\(=-\left(x^2+5x-8\right)\)
\(=-\left(x^2+2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{57}{4}\right)\)
\(=-\left(x+\dfrac{5}{2}\right)^2+\dfrac{57}{4}< =\dfrac{57}{4}\forall x\)
Dấu '=' xảy ra khi \(x+\dfrac{5}{2}=0\)
=>\(x=-\dfrac{5}{2}\)
4: \(D=-x^2+6x-4\)
\(=-\left(x^2-6x+4\right)\)
\(=-\left(x^2-6x+9-5\right)\)
\(=-\left(x-3\right)^2+5< =5\forall x\)
Dấu '=' xảy ra khi x-3=0
=>x=3
5: \(E=-10-x^2-6x\)
\(=-\left(x^2+6x+10\right)=-\left(x^2+6x+9+1\right)\)
\(=-\left(x+3\right)^2-1< =-1\forall x\)
Dấu '=' xảy ra khi x+3=0
=>x=-3
6: \(F=-x^2+13x+1\)
\(=-\left(x^2-13x-1\right)\)
\(=-\left(x^2-2\cdot x\cdot\dfrac{13}{2}+\dfrac{169}{4}-\dfrac{173}{4}\right)\)
\(=-\left(x-\dfrac{13}{2}\right)^2+\dfrac{173}{4}\le\dfrac{173}{4}\forall x\)
Dấu '=' xảy ra khi x-13/2=0
=>x=13/2
7: \(G=-4x^2+8x-7\)
\(=-\left(4x^2-8x+7\right)\)
\(=-\left(4x^2-8x+4+3\right)\)
\(=-\left(2x-2\right)^2-3< =-3\forall x\)
Dấu '=' xảy ra khi 2x-2=0
=>2x=2
=>x=1
8: \(H=-4x^2-12x\)
\(=-\left(4x^2+12x\right)\)
\(=-\left(4x^2+12x+9-9\right)\)
\(=-\left(2x+3\right)^2+9< =9\forall x\)
Dấu '=' xảy ra khi 2x+3=0
=>x=-3/2
9: \(I=3x-9x^2-1\)
\(=-9\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
\(=-9\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{1}{12}\right)\)
\(=-9\left(x-\dfrac{1}{6}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi x-1/6=0
=>x=1/6
10: \(K=7-9x^2-8x\)
\(=-9\left(x^2+\dfrac{8}{9}x-\dfrac{7}{9}\right)\)
\(=-9\left(x^2+2\cdot x\cdot\dfrac{4}{9}+\dfrac{16}{81}-\dfrac{79}{81}\right)=-9\left(x+\dfrac{4}{9}\right)^2+\dfrac{79}{9}< =\dfrac{79}{9}\forall x\)
Dấu '=' xảy ra khi x+4/9=0
=>x=-4/9