\(pt\Leftrightarrow64x^3+144x^2+108x+27+125-525x+735x^2-343x^3+27-216x+576x^2-512x^3=0\)

\(\Leftrightarrow-791x^3+1455x^2-633x+179=0\)

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 \(\left(x+1\right)^2=0;x=-1\)

2.

c) \(C=2x^2+4x-6=2\left(x^2+2x+1\right)-8\)

\(=2\left(x+1\right)^2-8\ge-8\forall x\)

Dấu"=" xảy ra<=> \(2\left(x+1\right)^2=0\Leftrightarrow x=-1\)

3.

c) \(C=-3x^2-6x+9=-3\left(x^2+2x+1\right)+12\)

\(=-3\left(x+1\right)^2+12\le12\forall x\)

Dấu "=" xảy ra<=> \(-3\left(x+1\right)^2=0\Leftrightarrow x=-1\)

\(2,GTNN\)

\(A=x^2-6x+5=x^2+6x+9-4\)

\(=\left(x+3\right)^2-4\ge-4\)

\(A_{min}=-4\Leftrightarrow\left(x+3\right)^2=0\Rightarrow x=-3\)

\(B=4x^2-8x+7=4\left(x^2-2x+\frac{7}{4}\right)\)

\(=4\left(x^2-2x+1+\frac{3}{4}\right)=4\left(x-1\right)^2+3\ge3\)

\(\Rightarrow B_{min}=3\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)

\(C=2x^2+4x-6=2\left(x^2+2x-3\right)\)

\(=2\left(x^2+2x+1-4\right)=2\left(x+1\right)^2-8\ge-8\)

\(\Rightarrow C_{min}=-8\Leftrightarrow\left(x+1\right)^2=0\Rightarrow x=-1\)

\(3,GTLN\)

\(A=-x^2+2x-3=-\left(x^2-2x+3\right)\)

\(=-\left(x^2-2x+1-4\right)=-\left(x-1\right)^2+4\le4\)

\(A_{max}=4\Leftrightarrow-\left(x-1\right)^2=0\Rightarrow x=1\)

\(B=-9x^2+6x-4=-\left[9x^2-6x+4\right]\)

\(=-\left[\left(3x\right)^2-6x+1+3\right]=-\left(3x-1\right)^2-3\)

\(B_{max}=-3\Leftrightarrow-\left(3x-1\right)^2=0\Rightarrow x=\frac{1}{3}\)

\(C=-3x^2-6x+9=-3\left(x^2+2x-3\right)\)

\(=-3\left(x^2+2x+1-4\right)=-3\left(x+1\right)^2+12\)

\(C_{max}=12\Leftrightarrow-3\left(x+1\right)^2=0\Rightarrow x=-1\)

Sửa lại đề \(x^3-6x^2-x+30=0\)

\(PT\Leftrightarrow x^3+2x^2-8x^2-16x+15x+30=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-8x+15\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x-5\right)=0\)

Đến đây tự giải nha :)

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\(3=a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)\(\Leftrightarrow\)\(a+b+c\le3\)

\(P=\frac{\sqrt{4\left(a+3\right)}+\sqrt{4\left(b+3\right)}+\sqrt{4\left(c+3\right)}}{2}\le\frac{a+b+c+21}{4}\le6\)

Dấu "=" xảy ra khi \(a=b=c=1\)