Question

giúp mình với 1 tìm giá trị nhỏ nhất

A= (x-2)^2+(2x-3)^2

B=(x-3)^2+(x-1)^2

C=(x-1).(x+3).(x+2).(x+6)

D=(x+1).(x-2).(x-3).(x-6)

please huhu

Answer 1

a: \(=x^2-4x+4+4x^2-12x+9\)

\(=5x^2-16x+13\)

\(=5\left(x^2-\dfrac{16}{5}x+\dfrac{13}{5}\right)\)

\(=5\left(x^2-2\cdot x\cdot\dfrac{8}{5}+\dfrac{64}{25}+\dfrac{1}{25}\right)\)

\(=5\left(x-\dfrac{8}{5}\right)^2+\dfrac{1}{5}>=\dfrac{1}{5}\)

Dấu '=' xảy ra x=8/5

b: \(=x^2-6x+9+x^2-2x+1\)

\(=2x^2-8x+10\)

\(=2\left(x^2-4x+5\right)=2\left(x-2\right)^2+2>=2\)

Dấu '=' xảy ra khi x=2

Answer 2

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

\(=x^2-6x+9+4x^2-12x+9\)

\(=5x^2-18x+18\)

\(=5\left(x^2-\dfrac{18}{5}+\dfrac{81}{25}\right)+18-\dfrac{81}{5}\)

\(=5\left(x-\dfrac{9}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{9}{5}\)

- Vậy \(MinA=\dfrac{19}{5}\), đạt tại \(x=\dfrac{9}{5}\).

\(B\) làm tương tự như \(A\).

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

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

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

- Đặt \(t=x^2+5x-6\), ta có:

\(C=t\left(t+12\right)=t^2+12t=\left(t^2-12t+36\right)-36=\left(t+6\right)^2-36\ge-36\)

- Dấu "=" xảy ra khi \(t+6=0\Leftrightarrow x^2+5x-6+6=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

- Vậy \(MinC=-36\), đạt tại \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(D\) làm tương tự như \(C\)

Answer 3

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

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

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

- Đặt \(t=x^2+5x-6\), ta có:

\(C=t\left(t+12\right)=t^2+12t=\left(t^2+12t+36\right)-36=\left(t+6\right)^2-36\ge-36\).

- Dấu "=" xảy ra khi \(t+6=0\Leftrightarrow x^2+5x-6+6=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

- Vậy \(MinC=-36\), đạt tại \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\).

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

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

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

- Đặt \(t=x^2-5x-6\), ta có:

\(D=t\left(t+12\right)=\left(t^2+12t+36\right)-36=\left(t+6\right)^2-36\ge-36\)

- Dấu "=" xảy ra khi \(t+6=0\Leftrightarrow x^2-5x-6+6=0\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

- Vậy \(MinD=-36\), đạt tại \(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)