2. Ta có: A = x2 - 6x + 5 = (x2 - 6x + 9) - 4 = (x - 3)2 - 4
Ta luôn có: (x - 3)2 \(\ge\)0 \(\forall\)x
=> (x - 3)2 - 4 \(\ge\)-4 \(\forall\)x
Dấu "=" xảy ra <=> x - 3 = 0 <=> x = 3
Vậy MinA = -4 tại x = 3
Ta có: B = 4x2 - 8x + 7 = 4(x2 - 2x + 1) + 3 = 4(x - 1)2 + 3
Ta luôn có: 4(x - 1)2 \(\ge\)0 \(\forall\)x
=> 4(x - 1)2 + 3 \(\ge\)3 \(\forall\)x
Dấu "=" xảy ra <=> x - 1 = 0 <=> x = 1
vậy MinB = 3 tại x = 1
Ta có: C = 2x2 + 4x - 6 = 2(x2 + 2x + 1) - 8 = 2(x + 1)2 - 8
Ta luôn có: 2(x + 1)2 \(\ge\)0 \(\forall\)x
=> 2(x + 1)2 - 8 \(\ge\)-8 \(\forall\)x
Dấu "=" xảy ra <=> x + 1 = 0 <=> x = -1
Vậy MinC = -8 tại x = -1
1/
\(A=x^2-6x+5\)
\(A=x^2-2\cdot3x+3^2-3^2+5\)
\(A=\left(x-3\right)^2-3^2+5\)
\(A=\left(x-3\right)^2-9+5\)
\(A=\left(x-3\right)^2-4\)
mà \(\left(x-3\right)^2\ge0\Rightarrow\left(x-3\right)^2-4\ge-4\)
\(\Rightarrow GTNNA\left(x^2-6x+5\right)=-4\)
với \(\left(x-3\right)^2=0;x=3\)
\(B=4x^2-8x+7\)
\(B=4\left(x^2-2x+\frac{7}{4}\right)\)
\(B=4\left(x^2-2\cdot1x+1-1+\frac{7}{4}\right)\)
\(B=4\left(x-1\right)^2+3\)
\(\left(x-1\right)^2\ge0\Rightarrow4\left(x^2-1\right)^2+3\ge3\)
\(\Rightarrow GTNNB=3\)
với \(\left(x-1\right)^2=0;x=1\)
\(C=2x^2+4x-6\)
\(C=2\left(x^2+2x-3\right)\)
\(C=2\left(x^2+2\cdot1x+1-1-3\right)\)
\(C=\left(x+1\right)^2-8\)
có\(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2-8\ge-8\)
\(\Rightarrow GTNNC=-8\)
với \(\left(x+1\right)^2=0;x=-1\)
3
\(A=-x^2+2x-3.\)
\(A=-\left(x^2-2x+3\right)\)
\(A=-\left(x^2-2\cdot1x+1-1+3\right)\)
\(A=-\left(x-1\right)^2-2\)
có \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2-2\ge-2\)
\(\Rightarrow GTLNA=-2\) với \(\left(x-1\right)^2=0;x=1\)
\(B=-9x^2+6x-4\)
\(B=-9\left(x^2-\frac{2}{3}x+\frac{4}{9}\right)\)
\(B=-9\left(x^2-2\cdot\frac{1}{3}x+\frac{1}{9}-\frac{1}{9}+\frac{4}{9}\right)\)
\(B=-9\left(x^2-2\cdot\frac{1}{3}x+\frac{1}{9}-\frac{3}{9}\right)\)
\(B=-9\left(x-\frac{1}{3}\right)^2+3\)
có \(\left(x-\frac{1}{3}\right)^2\ge0\Rightarrow-9\left(x-\frac{1}{3}\right)^2+3\ge3\)
\(\Rightarrow GTLNB=3\)với \(\left(x-\frac{1}{3}^2=0;\right)x=\frac{1}{3}\)
