\(A=x^4+4x^3+10x^2+12x=x^4+4x^2+9+4x^3+12x+6x^2-9\)
<=>\(A=x^4+4x^2+9+4x^3+12x+6x^2-9\)
<=>\(A=\left(x^2\right)^2+\left(2x\right)^2+3^2+2.x^2.2x+2.2x.3+2.x^2.3-9\)
<=>\(A=\left(x^2+2x+3\right)^2-9\)
<=>\(A=\left[\left(x+1\right)^2+2\right]^2-9\)
Vì \(\left(x+1\right)^2\ge0\Leftrightarrow\left(x+1\right)^2+2\ge2\Leftrightarrow\left[\left(x+1\right)^2+2\right]^2\ge4\)\(\Leftrightarrow A=\left[\left(x+1\right)^2+2\right]^2-9\ge-5\)
=>Amin=-5 <=> x=-1
Vậy Amin=5 tại x=-1
\(\frac{x}{3}=\frac{y}{4}\)
\(\Rightarrow4x=3y\)
\(\Rightarrow\frac{x}{y}=\frac{3}{4}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}y=4\\y=-4\end{cases}}\)
\(\frac{x}{3}=\frac{y}{4}\Rightarrow4x=3y\Rightarrow4x-3y=0\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{4x}{12}=\frac{3y}{12}=\frac{4x-3y}{12-12}=\frac{0}{0}\)
Sai chỗ nào nhỉ,mình cho 3 .
\(A=x\left(x^3+4x^2+10x+12\right)=x\left[x\left(x^2+4x+4\right)+\left(6x+12\right)\right]\)
\(=x\left[x\left(x+2\right)^2+6\left(x+2\right)\right]=x\left(x+2\right)\left(x^2+2x+6\right)\)
\(=\left(x^2+2x\right)\left(x^2+2x+6\right)\)
Ta có : \(x^2+2x=\left(x^2+2x+1\right)-1=\left(x+1\right)^2-1\ge-1\)
\(\Rightarrow A=\left(x^2+2x\right)\left(x^2+2x+6\right)\ge-1.\left(-1+6\right)=-5\)
Dấu "=" xảy ra khi x = -1
Vậy Min A = -5 <=> x = -1
Mình không chắc lắm. Bạn nào có cách khác thì đăng lên nhé :)
A = x4 + 4x3 + 10x2 + 12x = x4 + 4x2 + 9 + 4x3 + 12x + 6x2 - 9 = (x2 + 2x + 3)2 - 9 = [(x + 1)2 + 2]2 - 9
\(\left(x+1\right)^2\ge0\Rightarrow\left(x+1\right)^2+2\ge2\Rightarrow\left[\left(x+1\right)^2+2\right]^2\ge4\Rightarrow A_{min}=4-9=-5\)
