a: 

b: Phương trình hoành độ giao điểm là:

3x-2=-x-1

=>3x+x=-1+2

=>4x=1

=>\(x=\dfrac{1}{4}\)

Thay x=1/4 vào y=-x-1, ta được:

\(y=-\dfrac{1}{4}-1=-\dfrac{5}{4}\)

Vậy: \(E\left(\dfrac{1}{4};-\dfrac{5}{4}\right)\)

c: Tọa độ A là:

\(\left\{{}\begin{matrix}y=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Tọa độ B là:

\(\left\{{}\begin{matrix}x=0\\y=3\cdot0-2=-2\end{matrix}\right.\)

Vậy: \(A\left(\dfrac{2}{3};0\right);B\left(0;-2\right)\)

\(OA=\sqrt{\left(\dfrac{2}{3}-0\right)^2+\left(0-0\right)^2}=\dfrac{2}{3}\)

\(OB=\sqrt{\left(0-0\right)^2+\left(-2-0\right)^2}=2\)

Vì Ox\(\perp\)Oy

nên OA\(\perp\)OB

=>ΔOAB vuông tại O

=>\(S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB=\dfrac{2}{3}\)

Phương trình hoành độ giao điểm là:

\(\dfrac{1}{2}x^2=2x-m+1\)

=>\(\dfrac{1}{2}x^2-2x+m-1=0\)

\(\Delta=\left(-2\right)^2-4\cdot\dfrac{1}{2}\left(m-1\right)\)

\(=4-2\left(m-1\right)=4-2m+2=-2m+6\)

Để phương trình có hai nghiệm phân biệt thì \(\Delta>0\)

=>-2m+6>0

=>-2m>-6

=>m<3

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{2}{\dfrac{1}{2}}=4\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{m-1}{\dfrac{1}{2}}=2\left(m-1\right)\end{matrix}\right.\)

\(x_1x_2\left(y_1+y_2\right)+48=0\)

=>\(\dfrac{1}{2}\left(x_1^2+x_2^2\right)\cdot x_1x_2+48=0\)

=>\(\dfrac{1}{2}\cdot2\cdot\left(m-1\right)\cdot\left[\left(x_1+x_2\right)^2-2x_1x_2\right]+48=0\)

=>\(\left(m-1\right)\cdot\left[4^2-2\cdot2\left(m-1\right)\right]+48=0\)

=>\(\left(m-1\right)\left(16-4m+4\right)+48=0\)

=>\(\left(m-1\right)\left(-4m+20\right)+48=0\)

=>\(\left(m-1\right)\left(-m+5\right)+12=0\)

=>\(-m^2+5m+m-5+12=0\)

=>\(-m^2+6m+7=0\)

=>\(m^2-6m-7=0\)

=>(m-7)(m+1)=0

=>\(\left[{}\begin{matrix}m=7\left(loại\right)\\m=-1\left(nhận\right)\end{matrix}\right.\)

ĐKXĐ: x<>0 và y<>0

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\y-x=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+9\\\dfrac{1}{x}+\dfrac{1}{x+9}=\dfrac{1}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+9\\\dfrac{x+9+x}{x\left(x+9\right)}=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x+9\\x\left(x+9\right)=6\left(2x+9\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+9\\x^2+9x-12x-54=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=x+9\\x^2-3x-54=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x+9\\\left(x-9\right)\left(x+6\right)=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-9=0\\y=x+9\end{matrix}\right.\\\left\{{}\begin{matrix}x+6=0\\y=x+9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=9\\y=9+9=18\end{matrix}\right.\\\left\{{}\begin{matrix}x=-6\\y=-6+9=3\end{matrix}\right.\end{matrix}\right.\)

Ta có: M nằm trên cung AB

=>\(sđ\stackrel\frown{AM}+sđ\stackrel\frown{BM}=sđ\stackrel\frown{AB}\)

=>\(sđ\stackrel\frown{BM}=100^0-60^0=40^0\)

ĐKXĐ: \(x\notin\left\{0;-20\right\}\)

\(\left\{{}\begin{matrix}y=20+x\\\dfrac{100}{x}-\dfrac{100}{20+x}=\dfrac{5}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\\dfrac{400}{x\left(x+20\right)}=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x+20\\x\left(x+20\right)=2400\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\x^2+20x-2400=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\\left(x+60\right)\left(x-40\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=x+20\\\left[{}\begin{matrix}x+60=0\\x-40=0\end{matrix}\right.\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=x+20\\x\in\left\{-60;40\right\}\end{matrix}\right.\)(nhận)

=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-60\\y=x+20=-60+20=-40\end{matrix}\right.\\\left\{{}\begin{matrix}x=40\\y=x+20=60\end{matrix}\right.\end{matrix}\right.\)

Để hệ có nghiệm duy nhất thì \(\dfrac{m-1}{2}\ne\dfrac{-m}{-1}=m\)

=>\(2m\ne m-1\)

=>\(m\ne-1\)(1)

\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\2x-y=m+5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(m-1\right)x-my=3m-1\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(m-1\right)x-m\left(2x-m-5\right)=3m-1\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(m-1\right)-2mx+m^2+5m-3m+1=0\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(-m-1\right)+m^2+2m+1=0\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(m+1\right)=\left(m+1\right)^2\\y=2x-m-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=m+1\\y=2\left(m+1\right)-m-5=2m+2-m-5=m-3\end{matrix}\right.\)

\(x^2-y^2< 4\)

=>\(\left(m+1\right)^2-\left(m-3\right)^2< 4\)

=>\(m^2+2m+1-m^2+6m-9< 4\)

=>8m-8<4

=>8m<12

=>\(m< \dfrac{3}{2}\)

Kết hợp (1), ta được: \(\left\{{}\begin{matrix}m< \dfrac{3}{2}\\m\ne-1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3x=y^2+y+1\left(1\right)\\3y=x^2+x+1\left(2\right)\end{matrix}\right.\)

Lấy (1) trừ (2), suy ra: \(x^2-2x+1=y^2-2y+1\)

\(\Leftrightarrow\left(x-1\right)^2=\left(y-1\right)^2\Leftrightarrow\left|x-1\right|=\left|y-1\right|\left(3\right)\)

Trường hợp 1: \(x,y\ge1\), khi đó: \(\left(3\right)\Leftrightarrow x=y\).

Thay lại vào hệ thì \(x=y=1\) (nhận).

Trường hợp 2: \(\left\{{}\begin{matrix}x\ge1\\y< 1\end{matrix}\right.\), khi đó \(\left(3\right)\Leftrightarrow x+y=2\Leftrightarrow y=2-x\).

Thay lại vào (2), suy ra: \(\left[{}\begin{matrix}x=1\left(N\right)\\x=-5\left(L\right)\end{matrix}\right.\), suy ra \(y=1\) (loại).

Trường hợp 3: \(\left\{{}\begin{matrix}x< 1\\y\ge1\end{matrix}\right.\), khi đó \(\left(3\right)\Leftrightarrow x+y=2\), tương tự như trường hợp 2, loại.

Trường hợp 4: \(x,y< 1\), khi đó \(\left(3\right)\Leftrightarrow x=y\), tương tự như trường hợp 1 thì \(x=y=1\left(L\right)\).

Tổng quát, hệ phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(1;1\right)\)

\(A=\dfrac{a^2}{a\sqrt{a^2+9bc}}+\dfrac{b^2}{b\sqrt{b^2+9ca}}+\dfrac{c^2}{c\sqrt{c^2+9ab}}\)

\(A\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+9bc}+b\sqrt{b^2+9ca}+c\sqrt{c^2+9ab}}\)

Áp dụng Bunhiacopxki:

\(\sqrt{a}.\sqrt{a^3+9abc}+\sqrt{b}.\sqrt{b^3+9abc}+\sqrt{c}.\sqrt{c^3+9abc}\le\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+27abc\right)}\)

\(\Rightarrow A\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+27abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+27abc}}\) (1)

Ta có:

\(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\right)+6abc\)

\(\dfrac{1}{10}\left(a^3+b^3+c^3\right)\ge\dfrac{3}{10}abc\)

\(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2\ge6\sqrt[6]{a^6b^6c^6}=6abc\)

\(\Rightarrow\left(a+b+c\right)^3\ge\dfrac{9}{10}\left(a^3+b^3+c^3\right)+\dfrac{3}{10}abc+18abc+6abc\)

\(\Rightarrow\left(a+b+c\right)^3\ge\dfrac{9}{10}\left(a^3+b^3+c^3+27abc\right)\) (2)

(1);(2) \(\Rightarrow A\ge\sqrt{\dfrac{\dfrac{9}{10}\left(a^3+b^3+c^3+27abc\right)}{a^3+b^3+c^3+27abc}}=\dfrac{3\sqrt{10}}{10}\)

Dấu "=" xảy ra khi \(a=b=c\)

Đặt \(f\left(x\right)=ax^3+bx^2+c\)

Do \(f\left(x\right)\) chia hết \(x+2\Rightarrow f\left(-2\right)=0\)

\(\Rightarrow-8a+4b+c=0\) (1)

Do \(f\left(x\right)\) chia \(x^2-1\) dư 5

\(\Rightarrow f\left(x\right)=g\left(x\right).\left(x^2-1\right)+5\) với \(g\left(x\right)\) là 1 đa thức bậc nhất nào đó

\(\Rightarrow ax^3+bx^2+c=g\left(x\right)\left(x^2-1\right)+5\) (*)

Thay \(x=1\) vào (*) \(\Rightarrow a+b+c=5\) (2)

Thay \(x=-1\) vào (*) \(\Rightarrow-a+b+c=5\) (3)

(1);(2);(3) \(\Rightarrow\left\{{}\begin{matrix}-8a+4b+c=0\\a+b+c=5\\-a+b+c=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{3}\\c=\dfrac{20}{3}\end{matrix}\right.\)

ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(x^2-6x+2=4\sqrt{2x-1}-2x\sqrt{2x-1}\)

\(\Leftrightarrow x^2+2x\sqrt{2x-1}+2x-1=4\left(2x-1\right)+4\sqrt{2x-1}+1\)

\(\Leftrightarrow\left(x+\sqrt{2x-1}\right)^2=\left(2\sqrt{2x-1}+1\right)^2\)

Do \(x\ge\dfrac{1}{2}\Rightarrow\left\{{}\begin{matrix}x+\sqrt{2x-1}>0\\2\sqrt{2x-1}+1>0\end{matrix}\right.\)

Nên pt tương đương:

\(x+\sqrt{2x-1}=2\sqrt{2x-1}+1\)

\(\Leftrightarrow x-1=\sqrt{2x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\\left(x-1\right)^2=2x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+2=0\end{matrix}\right.\)

\(\Rightarrow x=2+\sqrt{2}\)