Question

[Nguyễn Bảo Gia Huy]

Tìm GTNN của biểu thức:

a) A=\(25x^2+3y^2-10x+11\)

b) B=\(\left(x-3\right)^2+\left(x-11\right)^2\)

c) C=\(\left(x+1\right).\left(x-2\right).\left(x-3\right).\left(x-6\right)\)

Answer 1

[Trần Thanh Phương]

a) \(A=25x^2+3y^2-10x+11\)

\(A=\left(5x-1\right)^2+3y^2+11\ge11\forall x;y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{5}\\y=0\end{matrix}\right.\)

b) \(B=\left(x-3\right)^2+\left(x-11\right)^2\)

\(B=2\left(x^2-14x+65\right)\)

\(B=2\left[\left(x-7\right)^2+16\right]\)

\(B=2\left(x-7\right)^2+32\ge32\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=7\)

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

\(C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)

Đặt \(x^2-5x-6=a\)

\(C=a\left(a+12\right)\)

\(C=a^2+12a+36-36\)

\(C=\left(a+6\right)^2-36\ge-36\)

Dấu "=" xảy ra \(\Leftrightarrow a=-6\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

Answer 2

[Như Trần]

\(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\\ C=\left(x+1\right)\left(x-6\right)\left(x-2\right)\left(x-3\right)\\ C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\\ C=\left(x^2-5x\right)^2-6^2\\ C=\left(x^2-5x\right)^2-36\)

Ta có:

\(\left(x^2-5x\right)^2\ge0\\ \Rightarrow C=\left(x^2-5x\right)^2-36\ge-36\)

Dấu "=" xảy ra khi và chỉ khi:

(x² - 5x)² = 0 => x² - 5x = 0 => x(x - 5) = 0

=> x = 5 hoặc x = 0

Vậy MinC = -36 <=> x = 5; x = 0