\(A=\left|x-5\right|+\left|x+3\right|\ge\left|5-x+x+3\right|=8\)

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-5\ge0\\x+3\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge5\\x\ge-3\end{cases}\Rightarrow}x\ge5}\)

Vậy,..........

\(P=2x^2+x+1\)

\(P=\left(\sqrt{2}.x\right)^2+2.\sqrt{2}.\frac{1}{2}x+\frac{1}{2}-\frac{1}{2}+1\)

\(P=\left(\sqrt{2}.x+\sqrt{\frac{1}{2}}\right)^2-\left(\frac{1}{2}-1\right)\)

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

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

\(P=\left(x\sqrt{2}+\sqrt{\frac{1}{2}}-\sqrt{\frac{1}{2}}\right)\left(x\sqrt{2}+\sqrt{\frac{1}{2}}-\sqrt{\frac{1}{2}}\right)\)

\(P=\left(x\sqrt{2}\right)\left(x\sqrt{2}\right)\)

\(P=\left(x\sqrt{2}\right)^2\)

\(P=2x^2\)

1. \(\left(A+B\right)^2=A^2+2AB+B^2\)

2. \(\left(A-B\right)^2=A^2-2AB+B^2\)

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

4. \(A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)\)

5. \(A^2+B^2=\left(A+B\right)\left(A^2-AB+B^2\right)\)

6. \(\left(A+B\right)^3=A^3+3A^2B+3AB^2+B^3\)

7. \(\left(A-B\right)^2=A^3-3A^2B+3AB^2-B^3\)

Sai cái thứ 5 và 7 mà

a) \(A=2x^2-15\ge-15\forall x\)

\(minA=-15\Leftrightarrow x=0\)

b) \(B=2\left(x+1\right)^2-17\ge-17\forall x\)

\(minB=-17\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

nó dễ ợt mà -_- 

x²y²+4xy+4=(xy+2)² xong :))