a) A=(x-4)²+ |y-1|-6

Ta thấy:

(x-4)² ≥ 0 ∀ x

|y-1| ≥ 0 ∀ y

⇒ (x-4)²+ |y-1|  ≥ 0 ∀ x, y

⇒ (x-4)²+ |y-1|-6  ≥ -6 ∀ x, y

⇒ A ≥ -6 ∀ x, y

Dấu '=' xảy ra khi: \(\left[{}\begin{matrix}x-4=0\\y-1=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)

Vậy Min A = -6 tại x=4, y = 1

b) B= (x²-1)⁴+2.|2y-4|-3

Ta thấy:

(x²-1)⁴ ≥ 0 ∀ x

|2y-4| ≥ 0 ∀ y

⇒ 2|2y-4| ≥ 0 ∀ y

⇒ (x²-1)⁴+2.|2y-4| ≥ 0 ∀ x, y

⇒ (x²-1)⁴+2.|2y-4|-3 ≥ -3 ∀ x, y

Vậy Min B = -3 tại x=\(\pm\)1, y = 2

a: Ta có: \(x^2-x+1\)

\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)

b: Ta có: \(x^2+y^2-4x+y+5\)

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

\(=\left(x-2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x,y\)

Dấu '=' xảy ra khi x=2 và \(y=-\dfrac{1}{2}\)

\(A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{x^2+y^2+2xy}+\frac{1}{\frac{\left(x+y\right)^2}{2}}=4+2=6\)

dấu "=" xảy ra khi x=y=1/2

\(A=\left|x\right|+\left|8-x\right|\)

Áp dụng bđt:

\(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

\(A\ge\left|x+8-x\right|\)

\(A\ge8\)

Dấu "=" xảy ra khi:

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\8-x\ge0\Rightarrow x\le8\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\8-x< 0\Rightarrow x>8\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow0\le x\le8\)

Bài a đề gì kì zậy mik ko hỉu

đề toán trương2 mình đó

bí quá ! giúp đi
 

Bài 2: 

a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)

\(\Leftrightarrow x-2=4-x\)

\(\Leftrightarrow2x=6\)

hay x=3

b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)

\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)

\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)