Giải:

a) \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)

\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)

\(\Leftrightarrow-12x+27x=30\)

\(\Leftrightarrow15x=30\)

\(\Leftrightarrow x=2\)

Vậy ...

b) \(2x\left(x-1\right)+x\left(5-2x\right)=15\)

\(\Leftrightarrow2x^2-2x+5x-2x^2=15\)

\(\Leftrightarrow-2x+5x=15\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=5\)

Vậy ...

\(7x+\left(-6\right)=0\\ \Leftrightarrow7x=6\\ \Leftrightarrow x=\dfrac{6}{7}\)

Vậy nghiệm của đa thức p(x) là \(x=\dfrac{6}{7}\)

Đa thức \(P\left(x\right)\) có nghiệm khi: 

\(P\left(x\right)=0\)

\(\Rightarrow7x+\left(-6\right)=0\)

\(\Rightarrow7x-6=0\)

\(\Rightarrow7x=6\)

\(\Rightarrow x=\dfrac{6}{7}\)

Vậy nghiệm của đa thức \(P\left(x\right)\) là \(\dfrac{6}{7}\)

ĐKXĐ: \(\dfrac{3}{2}\le x\le3\)

\(A=\sqrt{2x-3}+\sqrt{6-2x}+\left(2-\sqrt{2}\right)\sqrt{3-x}\)

\(A\ge\sqrt{2x-3+6-2x}+\left(2-\sqrt{2}\right)\sqrt{3-x}\ge\sqrt{3}\)

\(A_{min}=\sqrt{3}\) khi \(3-x=0\Rightarrow x=3\)

\(A=1.\sqrt{2x-3}+\sqrt{2}.\sqrt{6-2x}\le\sqrt{\left(1+2\right)\left(2x-3+6-2x\right)}=3\)

\(A_{max}=3\) khi \(2x-3=\dfrac{6-2x}{2}\Rightarrow x=2\)

=>KCl2

 =>  CuCl2

2Al+3Cl2=>2AlCl3

2Al+H2SO4=>3H2+Al2(SO4)3

C9H20+14O2=>10H2O+9CO2

x^4-4x^3+6=0

=>\(x\simeq1,3;x\simeq3,9\)

 Ta có: x² + 2x + 3

    = x² + x + x + 1 + 2

  = (x² + x) + (x + 1) + 2

  = x(x + 1) + (x + 1) + 2

  = (x + 1)(x + 1) + 2

  = (x + 1)² + 2 > 0 [ vì (x + 1)² \(\ge\)0; 2 >0)

=> đa thức f(x) ko có nghiệm

\(\left(x-3\right)^3-\left(x+3\right)^3\)

\(=x^3-3x^2.3+3x.3^2-3^3-\left(x^3+3x^23+3x3^2+3^3\right)\)

\(=x^3-9x^2+27x-9-x^3-9x^2-27x-9\)

\(=-18x^2-18=-18\left(x^2+1\right)\)