a: \(=\dfrac{x^5}{3x}+\dfrac{12x^2}{3x}+\dfrac{5x}{3x}=\dfrac{1}{3}x^4+4x+\dfrac{5}{3}\)

b: \(=\dfrac{-5x^4}{-5x}-\dfrac{15x^3}{5x}+\dfrac{18x}{5x}=x^3-3x^2+\dfrac{18}{5}\)

c: \(=\dfrac{-x^6}{0.5x}+\dfrac{5x^4}{0.5x}-\dfrac{2x^3}{0.5x}=-2x^5+10x^3-4x^2\)

\(5x\left(3x+7\right)-15x^2=70\)

\(\Leftrightarrow15x^2+35x-15x^2=70\)

\(\Leftrightarrow35x=70\)

\(\Leftrightarrow x=2\)

\(2\left(x+1\right)=5x+7\\ \Leftrightarrow2x+2=5x+7\\\Leftrightarrow 2x-5x=-2+7\\\Leftrightarrow -3x=5\\ \Leftrightarrow x=-\frac{5}{3}\)

Vậy phương trình trên có nghiệm là \(-\frac{5}{3}\)

\(3x-1=x+3\\ \Leftrightarrow3x-x=1+3\\ \Leftrightarrow2x=4\\\Leftrightarrow x=2\)

Vậy phương trình trên có nghiệm là \(2\)

\(15-7x=9-3x\\\Leftrightarrow -7x+3x=-15+9\\\Leftrightarrow -4x=-6\\ \Leftrightarrow x=\frac{3}{2}\)

Vậy phương trình trên có nghiệm là \(\frac{3}{2}\)

\(2x+1=15x-5\\ \Leftrightarrow2x-15x=-1-5\\ \Leftrightarrow-13x=-6\\ \Leftrightarrow x=\frac{6}{13}\)

Vậy phương trình trên có nghiệm là \(\frac{6}{13}\)

\(3x-2=2x+5\\ \Leftrightarrow3x-2x=2+5\\ \Leftrightarrow x=7\)

Vậy phương trình trên có nghiệm là \(7\)

Đặt \(x^2+3x=a\left(a>=-\dfrac{9}{4}\right)\)

BPT sẽ trở thành \(a>=2+\sqrt{5a+14}\)

=>\(a-2>=\sqrt{5a+14}\)

=>\(\sqrt{5a+14}< =a-2\)

=>\(\left\{{}\begin{matrix}a-2>=0\\5a+14< =\left(a-2\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\5a+14-a^2+4a-4< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\-a^2+9a+10< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\a^2-9a-10>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left(a-10\right)\left(a+1\right)>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a>=2\\\left[{}\begin{matrix}a>=10\\a< =-1\end{matrix}\right.\end{matrix}\right.\)

=>a>=10

=>\(x^2+3x>=10\)

=>\(x^2+3x-10>=0\)

=>(x+5)(x-2)>=0

=>\(\left[{}\begin{matrix}x>=2\\x< =-5\end{matrix}\right.\)

(5x-2)(3x+1)+(7-15x)(x+3)=-20

<=> 15x²+5x-6x-2+7x+21-15x²-45x+20=0

<=>39-39x=0

<=>39(1-x)=0

<=>1-x=0

=>x=1

(5x-2)(3x+1)+(7-15x)(x+3)=-20

=>\(15x^2-6x+5x-2+7x-15^2+21-45x=-20\)

=>\(-39x+19=-20\)

=>\(-39x=-39\)

=>\(x=1\)

vậy x=1

(5x-2)(3x+1)+(7-15x)(x+3)=-20

<=>\(15x^2+5x-6x-2+7x-15x^2+21-45x=-20\)

<=>\(-39x+19=0\)

<=>\(-39x=-19\)

<=>\(x=\dfrac{19}{39}\)

Vậy \(x=\dfrac{19}{39}\)

Answer:

\(3x^2.\left(2x^3-x+5\right)\)

\(=3x^2.2x^3+3x^2.(-x)+3x^2.5\)

\(=6x^5-3x^3+15x^2\)

\((4xy+3y-5x).x^2y\)

\(=4xy.x^2y+3y.x^2y-5x.x^2y\)

\(=4x^3+3x^2y^2-5x^3y\)

\(\Rightarrow6x-15x^2+15x^2-18=0\)

\(\Rightarrow6x-18=0\)

\(\Rightarrow6x=18\)

\(\Rightarrow x=18:6=3\)

\(3x\left(2-5x\right)+15x^2-18=0\)

\(6x-15x^2+15x^2-18=0\)

\(6x-18=0\)

\(x=18:6=3\)

Vậy.........

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