x+1
3
=
30
x+2
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\(1,x^2+4x-2=\left(x+2\right)^2-6\ge6\)
Dấu \("="\Leftrightarrow x=-2\)
\(2.x^2+7x+1=\left(x+\dfrac{7}{2}\right)^2-\dfrac{45}{4}\ge-\dfrac{45}{4}\)
Dấu \("="\Leftrightarrow x=-\dfrac{7}{2}\)
\(3,25x^2+30x+11=\left(5x+3\right)^2+2\ge2\)
Dấu \("="\Leftrightarrow x=-\dfrac{3}{5}\)
x = 31 => 30 = x-1
\(\Rightarrow C=x^6-\left(x-1\right)x^5-\left(x-1\right)x^4-\left(x-1\right)x^3-\left(x-1\right)x^2-\left(x-1\right)x+70\)
\(=x^6-\left(x^6-x^5\right)-\left(x^5-x^4\right)-\left(x^4-x^3\right)-\left(x^3-x^2\right)-\left(x^2-x\right)+70\)
\(=x+70=31+70=101\)
\(\left\{{}\begin{matrix}26⋮x\\x\ge13\end{matrix}\right.\Rightarrow x\in\left\{13;26\right\}\)
\(\left\{{}\begin{matrix}16⋮x\\x< 8\end{matrix}\right.\Rightarrow x\in\left\{1;2;4\right\}\)
\(\left\{{}\begin{matrix}18⋮x\\0< x< 40\end{matrix}\right.\Rightarrow x\in\left\{1;2;3;6;9;18\right\}\)
\(\left\{{}\begin{matrix}x⋮15\\30< x< 40\end{matrix}\right.\Rightarrow x\in\varnothing\)
\(\left\{{}\begin{matrix}x⋮12\\22\le5x\le50\end{matrix}\right.\Rightarrow x\in\varnothing\)
\(\left\{{}\begin{matrix}x⋮4\\16\le x\le36\end{matrix}\right.\Rightarrow x\in\left\{16;20;24;28;32;36\right\}\)
a, \(A=x^3-30x^2-31x+1\)
\(=x^3-31x^2+x^2-31x+1\)
\(=x^2\left(x-31\right)+x\left(x-31\right)+1\)
\(=\left(x^2+x\right)\left(x-31\right)+1\)
Thay x = 31 \(\Rightarrow A=1\)
Vậy A = 1 khi x = 31
b, tách ra làm tương tự phần a