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min a nếu x = 0
=>0 + 0 - 0 + 2038
=> A = 2038
Hội con 🐄 chúc bạn học tốt!!!
\(A=x^4+6x^2+3^2+x^2-4x+2^2+2025.\)
\(A=\left(x^2+3\right)^2+\left(x-2\right)^2+2025\)
Vì \(\hept{\begin{cases}\left(x^2+3\right)^2\ge0\forall x\\\left(x-2\right)^2\ge0\forall x\end{cases}}\Rightarrow\left(x^2+3\right)^2+\left(x-2\right)^2+2025\ge2025\forall x\)
Dấu '' = " xảy ra khi
\(\left(x^2+3\right)^2=0\) hoặc \(\left(x-2\right)^2=0\)
\(\Rightarrow x=\pm\sqrt{3}\) \(\Rightarrow x=2\)
Vậy \(Min_A=2025\Leftrightarrow x=\pm\sqrt{3};x=2\)
Study well
\(B=3x^2+3x-1\)
\(=3\left(x^2+x-\dfrac{1}{3}\right)\)
\(=3\left(x^2+x+\dfrac{1}{4}-\dfrac{7}{12}\right)\)
\(=3\left(x+\dfrac{1}{2}\right)^2-\dfrac{7}{4}>=-\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi x+1/2=0
=>\(x=-\dfrac{1}{2}\)
\(C=-2x^2+7x+3\)
\(=-2\left(x^2-\dfrac{7}{2}x-\dfrac{3}{2}\right)\)
\(=-2\left(x^2-2\cdot x\cdot\dfrac{7}{4}+\dfrac{49}{16}-\dfrac{73}{16}\right)\)
\(=-2\left(x-\dfrac{7}{4}\right)^2+\dfrac{73}{8}< =\dfrac{73}{8}\forall x\)
Dấu '=' xảy ra khi x-7/4=0
=>x=7/4
\(A=x^2-6x-4=x^2-6x+9-13=\left(x-3\right)^2-13\ge-13\)
Vậy \(A_{min}=-13\Leftrightarrow x=3\)
\(B=x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(B_{min}=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
\(A=x^4-4x^3+9x^2-20+22\\ A=x^4-4x^3+4x^2+5x^2-20x+20+2\\ A=x^2\left(x^2-4x+4\right)+5\left(x^2-4x+4\right)\\ A=\left(x^2+5\right)\left(x-2\right)^2+2\)
Nhận xét:
\(x^2+5>0\\ \left(x-2\right)^2\ge0\\ \Rightarrow\left(x^2+5\right)\left(x-2\right)^2\ge0\\ \Rightarrow A=\left(x^2+5\right)\left(x-2\right)^2+2\ge2\)
Dấu "=" xảy ra khi:
\(\left(x^2+5\right)\left(x-2\right)^2=0\\ \Rightarrow\left(x-2\right)^2=0\left(vì.x^2+5>0\right)\\ \Rightarrow x-2=0\\ x=2\)
Vậy MinA = 2 khi x = 2