a) \(A=x^4+4x+7=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)

\(minA=3\Leftrightarrow x=-2\)

b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxC=7\Leftrightarrow x=2\)

d) \(D=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxD=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)

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\(m_{NaOH}=a\left(g\right),V_{dd_{NaOH}}=b\left(l\right)\)

\(n_{NaOH}=\dfrac{a}{40}\left(mol\right)\)

\(n_{NaOH}=0.5b\left(mol\right)\)

\(n_{NaOH\left(2M\right)}=2.5\cdot2=5\left(mol\right)\)

\(\Rightarrow\dfrac{a}{40}+0.5b=5\left(1\right)\)

\(m_{dd_{NaOH}}=2500\cdot1.06=2650\left(g\right)\)

\(\Rightarrow a+1000b\cdot1.06=2650\left(2\right)\)

\(\left(1\right),\left(2\right):\)

Số lẻ quá em ơi :<

\(\)

mNaOH=a(g),VddNaOH=b(l)mNaOH=a(g),VddNaOH=b(l)

⇒a40+0.5b=5(1)⇒a40+0.5b=5(1)

mddNaOH=2500⋅1.06=2650(g)mddNaOH=2500⋅1.06=2650(g)

Bài 2:

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)

Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)

=>\(4⋮\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)

=>\(\sqrt{x}\in\left\{1;2;4;5;7\right\}\)

=>\(x\in\left\{1;4;16;25;49\right\}\)