ĐKXĐ: \(x\ne0\)

\(\frac{x+1}{2x}=1\Rightarrow x+1=2x\Rightarrow x=1\) (thỏa mãn ĐKXĐ)

Vậy x = 1

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

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

\(a^3-8a^2+16a=a\left(a^2-8a+16\right)=a\left(a-4\right)^2\)

\(a,x^3+2x^2-4x-8=x^2\left(x+2\right)-4\left(x+2\right)\)

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

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

\(b,a^3-8a^2+16a=a\left(a^2-8a+16\right)\)

\(=a\left(a-4\right)^2\)

1.B= -(x^2 - 4x - 3)
= -(x^2 - 2x2 + 4 - 7)
= -(x - 2)^2 + 7 ≤ 7 
 Dấu "=" xảy ra khi x - 2 = 0 <=> x = 2
=>Amax = 7 khi x=2
2. chịu tự đi mà làm ngốc thật

2.ĐK: \(x\ne-1\)

 \(Q=\frac{2x^2+2}{\left(x+1\right)^2}=\frac{\left(x-1\right)^2+\left(x+1\right)^2}{\left(x+1\right)^2}=\frac{\left(x-1\right)^2}{\left(x+1\right)^2}+1\ge1\forall x\)

Dấu "=" xảy ra khi: \(x-1=0\Rightarrow x=1\)

Vậy GTNN của Q là 1 khi x = 1

1. \(B=4x-x^2+3=-x^2+4x-4+7=-\left(x-2\right)^2+7\le7\forall x\)

Dấu "=" xảy ra khi \(x-2=0\Rightarrow x=2\)

Vậy GTLN của B là 7 khi x = 2

\(B=x^3-6x^2+12x-8-x\left(x^2-4x+3\right)+3x^2-9x+4\)

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

\(A=x^2+y^2+xy-6x-6y+2\)

\(\Rightarrow4A=4x^2+4y^2+4xy-24x-24y+8\)

           \(=\left(4x^2+4xy+y^2\right)+3y^2-24x-24y+8\)

            \(=\left[\left(2x+y\right)^2-12\left(2x+y\right)+36\right]+3y^2-12y-28\)

           \(=\left(2x+y-6\right)^2+3\left(y^2-4y+4\right)-40\)

            \(=\left(2x+y-6\right)^2+3\left(y-2\right)^2-40\ge-40\)

\(\Rightarrow4A\ge-40\)

\(\Rightarrow A\ge-10\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x+y-6=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=6-y\\y=2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=2\end{cases}}}\)

Vậy \(A_{min}=-10\Leftrightarrow x=y=2\)

P/S: cách giải trên gọi là cách chung riêng !