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\(A\left(x\right)=8-5x+3x^2-15-3x+16=3x^2-8x+9\)
\(B\left(x\right)=5x-2x^2+4x-1-x^2-3x=-3x^2+6x-1\)
\(C\left(x\right)=B\left(x\right)-A\left(x\right)=\left(-3x^2+6x-1\right)-\left(3x^2-8x+9\right)\)
\(C\left(x\right)=-6x^2+14x-10\)

\(2x^3-50=0\)
\(\Rightarrow2\left(x^3-25\right)=0\)
\(\Rightarrow x^3-25=0\Rightarrow x^3=25\)
\(\Rightarrow x=\sqrt[3]{25}\)
\(x^2-5x=-6\)
\(\Rightarrow x\left(x-5\right)=-6\)
Xét ước
\(\left(2x-1\right)^2-\left(3x+5\right)=0\)
\(\Rightarrow4x^2-4x+1-3x-5=0\)
\(\Rightarrow4x^2-4-7x=0\)
\(\Rightarrow4x^2-7x=4\)
\(\Rightarrow x\left(4x-7\right)=4\)
Xét ước
\(4x^2-20x+25=0\)
\(\Rightarrow\left(2x-5\right)^2=0\)
\(\Rightarrow2x=5\Rightarrow x=\dfrac{5}{2}\)
\(\left(3x-1\right)^2-\left(x-2\right)^2=0\)
\(\Rightarrow\left(3x-1\right)^2=\left(x-2\right)^2\)
\(\Rightarrow\left|3x-1\right|=\left|x-2\right|\)
Xét dấu:v

1. => x=(3/4)5:(3/4)3
=> x=(3/4)2
=> x=9/16
2. |(5x+12)2| >=0
=> GTNN của a là -21,5
<=> 5x+12=0
<=> 5x=-12
<=> x=-12/5

1, f(x)=x3-x2+x-1
\(f\left(x\right)=x^2\left(x-1\right)+\left(x-1\right)=\left(x^2+1\right)\left(x-1\right)\)
f(x) = 0 \(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2+1=0\\x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2=-1\\x=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x\in\phi\\x=1\end{array}\right.\)
\(\Leftrightarrow x=1\)
4, P(x)=x2+5x
\(\Rightarrow P\left(x\right)=x\left(x+5\right)\)
\(P\left(x\right)=0\Leftrightarrow\)\(\left[\begin{array}{nghiempt}x=0\\x+5=0\Rightarrow x=-5\end{array}\right.\)
6, Q(x)=3x2-4x
\(\Rightarrow Q\left(x\right)=x\left(3x-4\right)\)
7, H(x)= 5x5+10x
x + (5x + 2)^2 = 25 (x^2 + 1)
x + 25x^2 + 20x + 4 = 25 x^2 +25
21x + 25x^2 - 25x^2 = 25 - 4
21x = 21
x = 1
Vậy x = 1