Ta có :

\(B\left(x\right)=3x^4+x^5-2\left(x^3+4\right)-10x^2+9x\)

\(=x^5+3x^4-2x^3-10x^2+9x-8\)

\(C\left(x\right)=A\left(x\right)-B\left(x\right)\)

\(=x^5+3x^4-2x^3-9x^2+11x-6-\left(x^5+3x^4-2x^3-10x^2+9x-8\right)\)

\(=x^5+3x^4-2x^3-9x^2+11x-6-x^5-3x^4+2x^3+10x^2-9x+8\)

\(=x^2-2x+2\)

Người đánh cắp hũ bạc là Hải.Vì ngọn đồi và đường thẳng là A

\(a=34,\left(12\right)\)

\(=34\dfrac{12}{99}=34\dfrac{4}{33}=\dfrac{1126}{33}\)

\(\Delta'=\left[-\left(m+1\right)\right]^2-1\cdot\left(-3\right)\)

\(=m^2+2m+1+3\\ =m^2+2m+4\)

Để pt có nghiệm kép \(\Leftrightarrow\Delta'=0\\ \Leftrightarrow m^2+2m+4=0\) (1)

\(\Delta'_m=1^2-1\cdot4=1-4=-3\)

Vì \(\Delta'< 0\) nên pt (1) vô nghiệm

Vậy ko có giá trị nào của m để pt đã cho có nghiệm kép

\(2x-8\sqrt{2x-1}=21\)

\(\Leftrightarrow8\sqrt{2x-1}=2x-21\\ \Leftrightarrow64\left(2x-1\right)=\left(2x-21\right)^2\)

\(\Leftrightarrow128x-64=4x^2-84x+441\\ \Leftrightarrow4x^2-84x+441-128x+64=0\)

\(\Leftrightarrow4x^2-212x+502=0\)

\(\Leftrightarrow2x^2-106x+251\)

\(\Delta'=\left(-53\right)^2-2\cdot251=2809-502=2307\)

Vì \(\Delta'>0\) nên pt có 2 nghiệm phân biệt

\(\Rightarrow x_1=\dfrac{53+\sqrt{2307}}{2}\)

\(x_2=\dfrac{53-\sqrt{2307}}{2}\)

Vậy..............................

\(A=\frac{40}{9}+\frac{10\sqrt{5}}{18\sqrt{\sqrt{6}}}\)

\(\left(\dfrac{-2}{5}+\dfrac{3}{7}\right)-\left(\dfrac{4}{9}+\dfrac{12}{20}-\dfrac{13}{35}\right)+\dfrac{7}{35}\)

\(=-\dfrac{2}{5}+\dfrac{3}{7}-\dfrac{4}{9}-\dfrac{3}{5}+\dfrac{13}{35}+\dfrac{7}{35}\\ =\left(-\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{13}{35}+\dfrac{7}{35}+\dfrac{3}{7}\right)-\dfrac{4}{9}\\ =-1+\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{4}{9}\\ =-1+1-\dfrac{4}{9}\\ =-\dfrac{4}{9}\)

Hoặc \(x^3+3x-4=0\)

\(\left(x^2+4x\right)-x-4=0\\ x\left(x+4\right)-\left(x+4\right)=0\\ \left(x+4\right)\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

Nhận xét : \(a+b+c=1+3-4=0\)

\(\Rightarrow x_1=1\)

\(x_2=\dfrac{-4}{1}=-4\)

\(\Delta=\left(2m-1\right)^2-4\cdot2\cdot\left(m-1\right)\)

\(=4m^2-4m+1-8m+8\\ =4m^2-12m+9\\ =\left(2m-3\right)^2\ge0\forall x\)

Vì pt luôn có nghiệm với mọi x , theo vi-ét ta có :

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

Ta có : \(4x_1^2+2x_1x_2+4x_2^2=1\)

\(\Leftrightarrow\left(4x_1^2+8x_1x_2+4x_2^2=1\right)-6x_1x_2=1\\ \Leftrightarrow4\left(x_1+x_2\right)^2-6x_1x_2=1\)

\(\Leftrightarrow4\left(-2m+1\right)^2-6\cdot\left(m-1\right)=1\\ \Leftrightarrow4\left(4m^2-4m+1\right)-6m+6=1\)

\(\Leftrightarrow16m^2-16m+4-6m+6-1=0\\ \Leftrightarrow16m^2-24m+5=0\)

\(\Delta'_m=\left(-12\right)^2-16\cdot5=144-80=64\)

\(\Rightarrow\sqrt{\Delta'_m}=8\)

Vì \(\Delta'>0\) nên pt có 2 nghiệm phân biệt

\(\Rightarrow x_1=\dfrac{12+8}{16}=\dfrac{5}{4}\)

\(x_2=\dfrac{12-8}{16}=\dfrac{1}{4}\)

Vậy..............................