Question

Tìm x biết: \(x^3-8-\left(x-2\right).\left(x-12\right)=0\)
Tìm giá trị nhỏ nhất : \(A=x^2-2x+9\) ; \(B=x^2+6x-3\) ; \(C=\left(x-1\right).\left(x-3\right)+9\)
Tìm giá trị lớn nhất : \(D=-x^2-4x+7\) ; \(E=5-4x^2+4\)

Answer 1

\(x^3-8-\left(x-2\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\)

Lại có : \(x^2+x+16=\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}>0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy...

b/ \(A=x^2-2x+9=\left(x-1\right)^2+8\ge8\)

Dấu "=" xảy ra khi : \(x=1\)

Vậy...

c/ \(B=x^2+6x-3=\left(x^2+6x+9\right)-12=\left(x+3\right)^2-12\ge-12\)

Dấu "=" xảy ra khi \(x=-3\)

Vậy...

d/ \(C=\left(x-1\right)\left(x-3\right)+9=\left(x^2-4x+3\right)+9-\left(x^2-4x+4\right)+8=\left(x-2\right)^2+8\ge8\)

Dấu "=" xảy ra khi : \(x=2\)

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e/ \(D=-x^2-4x+7=-\left(x^2+4x+4\right)+3=-\left(x+2\right)^2+3\le3\)

Dấu "=" xảy ra khi \(x=-2\)

Vậy...

g/ \(E=5-4x^2+4x=-\left(4x^2-4x+4\right)+9=-\left(x-2\right)^2+9\le9\)

Dấu "=" xảy ra khi \(x=2\)

Vậy...