Question

[Nguyễn Trần An Như]

Tìm giá trị nhỏ nhất của

a) A=(x²-x)(x²+3x+2)

b)B=x⁴+(x-2)⁴+6x²(x-2)²

Answer 1

[Nguyễn Minh Đăng]

Bài làm:

a) \(A=\left(x^2-x\right)\left(x^2+3x+2\right)=x\left(x-1\right)\left(x+1\right)\left(x+2\right)\)

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

Đặt \(x^2+x-1=t\)\(\Rightarrow A=\left(t-1\right)\left(t+1\right)=t^2-1\ge-1\left(\forall t\right)\)

Dấu "=" xảy ra khi: \(t^2=0\Leftrightarrow\left(x^2+x-1\right)^2=0\Leftrightarrow x^2+x-1=0\)

\(\Leftrightarrow\left(x^2+x+\frac{1}{4}\right)-\frac{5}{4}=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(x+\frac{1}{2}+\frac{\sqrt{5}}{2}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}-\frac{\sqrt{5}}{2}=0\\x+\frac{1}{2}+\frac{\sqrt{5}}{2}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{\sqrt{5}-1}{2}\\x=\frac{\sqrt{5}+1}{2}\end{cases}}\)

b) Ta có: \(B=x^4+\left(x-2\right)^4+6x^2\left(x-2\right)^2=\left[x^4+2x^2\left(x-2\right)^2+\left(x-2\right)^4\right]+4x^2\left(x-2\right)^2\)

\(=\left[x^2+\left(x-2\right)^2\right]^2+\left[2x\left(x-2\right)\right]^2\ge2\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Phần b hơi sai sai nên bn xem phần a thôi nhé

Answer 2

[Nguyễn Linh Chi]

Sửa lại câu b 

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

\(=x^4+x^4-8x^3+24x^2-32x+16+6x^4-24x^3+24x^2\)

\(=8x^4-32x^3+48x^2-32x+16\)

\(=8\left(x^4-4x^3+6x^2-4x+1\right)+8\)

\(=8\left(x-1\right)^4+8\ge8\)

=> min B = 8 tại x = 1