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\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)
b:
\(D=-25x^2+10x-1-10\)
\(=-\left(25x^2-10x+1\right)-10\)
\(=-\left(5x-1\right)^2-10< =-10\)
Dấu = xảy ra khi x=1/5
\(E=-9x^2-6x-1+20\)
\(=-\left(9x^2+6x+1\right)+20\)
\(=-\left(3x+1\right)^2+20< =20\)
Dấu = xảy ra khi x=-1/3
\(F=-x^2+2x-1+1\)
\(=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1< =1\)
Dấu = xảy ra khi x=1
1. \(x^2+x-6=0\)
\(x^2-2x+3x-6=0\)
\(x\left(x-2\right)+3\left(x-2\right)=0\)
\(\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
2.f(x)=\(x^2-2.2x+4+6\)
\(=\left(x-2\right)^2+6\)
Vì \(\left(x-2\right)^2\ge0\forall x\)
->\(\left(x+2\right)^2+6\ge6\)
Dấu = xẩy ra khi x+2=0 <=>x=2
Ta có: \(A=\left(x-3\right)^2+\left(11-x\right)^2\)
\(=x^2-6x+9+x^2-22x+121\)
\(=2x^2-28x+130\)
\(=2\left(x^2-14x+49+16\right)\)
\(=2\left(x-7\right)^2+32\ge32\forall x\)
Dấu '=' xảy ra khi x=7
\(B=\left(x-3\right)^2+\left(x-11\right)^2\ge0\)
\(MinB=0\Leftrightarrow\hept{\begin{cases}x-3=0\\x-11=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=3\\x=11\end{cases}}\)
C = (x + 1).(x - 2).(x - 3).(x - 6)
= [(x + 1)(x - 6)][(x - 2)(x - 3)]
= (x2 - 5x - 6)(x2 - 5x + 6)
Đặt x2 - 5x = t, ta có:
C = (t - 6)(t + 6) = t2 - 36
Vì t2 lớn hơn hoặc bằng 0 => t2 - 36 lớn hơn hoặc bằng -36
Dấu "=" xảy ra khi t2 = 0 => t = 0 => x2 - 5x = 0 => x(x - 5) = 0 => x = 0 hoặc x = 5
Vậy Min C = -36 tại x = 0 hoặc 5
`A=x^2-4x+1`
`=x^2-4x+4-3`
`=(x-2)^2-3>=-3`
Dấu "=" xảy ra khi x=2
`B=4x^2+4x+11`
`=4x^2+4x+1+10`
`=(2x+1)^2+10>=10`
Dấu "=" xảy ra khi `x=-1/2`
`C=(x-1)(x+3)(x+2)(x+6)`
`=[(x-1)(x+6)][(x+3)(x+2)]`
`=(x^2+5x-6)(x^2+5x+6)`
`=(x^2+5x)^2-36>=-36`
Dấu "=" xảy ra khi `x=0\or\x=-5`
`D=5-8x-x^2`
`=21-16-8x-x^2`
`=21-(x^2+8x+16)`
`=21-(x+4)^2<=21`
Dấu "=" xảy ra khi `x=-4`
`E=4x-x^2+1`
`=5-4+4-x^2`
`=5-(x^2-4x+4)`
`=5-(x-2)^2<=5`
Dấu "=" xảy ra khi `x=5`
a: Ta có: \(A=-x^2+4x+3\)
\(=-\left(x^2-4x+4-7\right)\)
\(=-\left(x-2\right)^2+7\le7\forall x\)
Dấu '=' xảy ra khi x=2
b: Ta có: \(B=-x^2+x\)
\(=-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
\(B1,a,A=x^2-6x+11\)
\(=\left(x^2-6x+9\right)+2\)
\(=\left(x-3\right)^2+2\ge2\)
Dấu "=" <=> x=3
Vậy ..........
\(b,B=x^2-20x+101\)
\(=\left(x^2-20x+100\right)+1\)
\(=\left(x-10\right)^2+1\ge1\)
Dấu "=" <=> x = 10
Vậy .
\(2,a,A=4x-x^2+3\)
\(=7-\left(x^2-4x+4\right)\)'
\(=7-\left(x-2\right)^2\le7\)
Dấu ''='' <=> x = 2
Vậy .
\(b,B=-x^2+6x-11\)
\(=-2-\left(x^2-6x+9\right)\)
\(=-2-\left(x-3\right)^2\le-2\)
Dấu ""=" <=> x = 3
Vậy..
\(B=\left(x-3\right)^2+\left(x-11\right)^2\)
\(=x^2-6x+9+x^2-22x+121\)
\(=2\left(x^2-14x+49\right)+32\)
\(=2\left(x-7\right)^2+32\)
Ta có: \(\left(x-7\right)^2\ge0\Leftrightarrow2\left(x-7\right)^2+32\ge32\)
Vậy \(MinB=32\Leftrightarrow x=7\)
\(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
\(=\left(x+1\right)\left(x-6\right)\left(x-2\right)\left(x-3\right)\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
Đặt \(c=x^2-5x\)lúc này \(C\)thành: \(C=\left(c-6\right)\left(c+6\right)=c^2-36\)
Mà: \(c^2\ge0\forall c\Leftrightarrow c^2-36\ge-36\Leftrightarrow C\ge-36\)
Dấu '' = '' xảy ra: \(c=0\)
\(\Leftrightarrow x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
Vậy \(MinC=-36\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)