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\(a,A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-2018\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-2018\)
Đặt \(x^2+5x=a\)
\(\Rightarrow A=\left(a-6\right)\left(a+6\right)-2018=a^2-2054\)
\(\Rightarrow A_{min}=2054\Leftrightarrow a=0\)
\(\Rightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow x\in\left\{0;-5\right\}\)
\(b,B=\left(x-1\right)\left(x-4\right)\left(x-5\right)\left(x-8\right)+2018.\)
\(=\left(x^2-9x+8\right)\left(x^2-9x+20\right)+2018\)
Đặt \(x^2-9x+14=a\)
\(\Rightarrow B=\left(a-6\right)\left(a+6\right)+2018\)
\(=a^2-36+2018=a^2+1982\)
\(\Rightarrow B_{min}=1982\Leftrightarrow a^2=0\Rightarrow a=0\)
\(\Rightarrow x^2-9x+14=0\)
\(\Rightarrow x^2-2x-7x+14=0\)
\(\Leftrightarrow x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-7\right)=0\)
\(\Rightarrow x\in\left\{2;7\right\}\)
a) Ta có:
\(\sqrt{x}\ge0\Rightarrow\frac{1}{2}+\sqrt{x}\ge\frac{1}{2}+0=\frac{1}{2}\Rightarrow P_{min}=\frac{1}{2}\) khi và chỉ khi \(\sqrt{x}=0\Rightarrow x=0\)
b) Ta có:
\(2.\sqrt{x-1}\ge0\Rightarrow7-2.\sqrt{x-1}\le7-2.0=7\Rightarrow Q_{max}=7\)khi và chỉ khi \(2.\sqrt{x-1}=0\Rightarrow\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
Bài 1a)
\(P\left(x\right)=x^{2018}+4x^2+10\)
VÌ \(x^{2018}\ge0\forall x;4x^2\ge0\forall x\)
\(\Rightarrow x^{2018}+4x^2+10\ge10\forall x\)
Hay \(P\left(x\right)\ge10\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Bài 1b)
\(M\left(x\right)=x^2+x+1\)
\(M\left(x\right)=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(M\left(x\right)=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{-1}{2}\)
Bài 2a)
\(Q\left(x\right)=-x^4-1\)
Vì \(-x^4\le0\forall x\)
\(\Rightarrow-x^4-1\le-1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
b) \(N\left(x\right)=-x^2+2x-2\)
\(N\left(x\right)=-\left(x^2-2x+2\right)\)
\(N\left(x\right)=-\left(x^2-2x+1+1\right)\)
\(N\left(x\right)=-\left[\left(x-1\right)^2+1\right]\)
\(N\left(x\right)=-1-\left(x-1\right)^2\le-1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=1\)
a) Thấy \(\hept{\begin{cases}x^{2018}\ge0\\4x^2\ge0\end{cases}\Rightarrow x^{2018}+4x^2+10\ge10}\)
Dấu = xảy ra khi \(\hept{\begin{cases}x^{2018}=0\\4x^2=0\end{cases}\Rightarrow x=0}\)
Vậy Min p(x)=10 khi x=0
b) \(M\left(x\right)=x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Dấu = xảy ra khi x+1/2=0 <=> x=1/2
Vậy min M(x)=3/4 khi x=1/2
2. a)\(Q\left(x\right)=-x^4-1=-\left(x^4+1\right)\)
Vì \(x^4\ge0\Rightarrow x^4+1\ge1\Rightarrow Q\left(x\right)=-\left(x^4+1\right)\le-1\)
Dấu = xảy ra khi x4=0 <=> x=4
Vậy Max Q(x)=-1 khi x=4
b) \(N\left(x\right)=-\left(x^2-2x+2\right)=-\left(x^2-2x+1\right)-1=-\left(x-1\right)^2-1\le-1\)
Dấu = xảy ra khi x-1=0 <=> x=1
Vậy Max N(x)=-1 khi x=1
kim namjoon kí hiệu \(\forall\)có nghĩa là " với mọi "