\(1,\\ b,=\left(x+1\right)^2-y=\left(84+1\right)^2-15=85^2-15=7210\\ 2,\\ a,A:B=\left(x^3+3x^2+3x-2\right):\left(x+1\right)\\ =\left[\left(x+1\right)^3-1\right]:\left(x+1\right)=\left(x+1\right)^2\left(\text{dư }-1\right)\\ b,A⋮B\Leftrightarrow-1⋮B\Leftrightarrow x+1\inƯ\left(-1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-2;0\right\}\)

\(x+y=3\\ \Rightarrow x=3-y\\ \Rightarrow A=\left(3-y\right)^2+y^2=2y^2-6y+9\\ \Rightarrow A=2\left(y^2-2\cdot\dfrac{3}{2}y+\dfrac{9}{4}\right)+\dfrac{9}{2}=2\left(y-\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\)

Vậy \(A_{min}=\dfrac{9}{2}\Leftrightarrow y=\dfrac{3}{2}\Leftrightarrow x=3-\dfrac{3}{2}=\dfrac{3}{2}\)

Đáp án A

\(x^2\left(y-1\right)-4\left(y-1\right)\\ =\left(y-1\right)\left(x^2-4\right)=\left(y-1\right)\left(x-2\right)\left(x+2\right)\)

\(\dfrac{2x+4}{-5}=\dfrac{6-x}{3}\\ \Rightarrow5\left(x-6\right)=3\left(2x+4\right)\\ \Rightarrow5x-30=6x+12\\ \Rightarrow x=-42\)

Hệ số tỉ lệ là \(a=x\cdot y=4\cdot10=40\)

Gọi hệ số tỉ lệ là a

\(\Rightarrow a=xy=8\cdot5=40\\ \Rightarrow y=\dfrac{40}{x}\\ \text{Thay }x=10\Rightarrow y=\dfrac{40}{10}=4\)

\(=\sqrt{9a^2\left(b-2\right)^2}=9\left|a\left(b-2\right)\right|\)

Vì \(b-2=-\sqrt{3}-2< 0;a=2>0\Leftrightarrow a\left(b-2\right)< 0\)

\(\Leftrightarrow9\left|a\left(b-2\right)\right|=9a\left(2-b\right)=18\left(2-\sqrt{3}\right)=36-18\sqrt{3}\)

Áp dụng tc dtsbn:

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}\Rightarrow\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a^2-b^2}{c^2-d^2}\\ \Rightarrow\dfrac{a^2+b^2}{a^2-b^2}=\dfrac{c^2+d^2}{c^2-d^2}\)

\(=\left(6x^2-9x+10x-15\right):\left(2x-3\right)\\ =\left[3x\left(2x-3\right)+5\left(2x-3\right)\right]:\left(2x-3\right)\\ =3x+5\)

PT có 2 nghiệm \(\Leftrightarrow\Delta=\left(4m+1\right)^2-8\left(m-4\right)\ge0\)

\(\Leftrightarrow16m^2+33\ge0\left(\text{luôn đúng}\right)\)

Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=4m+1\\x_1x_2=-2\left(m-4\right)\end{matrix}\right.\)

\(B=\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=\left(4m+1\right)^2+8\left(m-4\right)\\ B=16m^2+16m-31=4\left(4m^2+4m+1\right)-35=4\left(2m+1\right)^2-35\ge-35\)

Vậy \(B_{min}=-35\Leftrightarrow m=-\dfrac{1}{2}\)