\(...\Rightarrow x^3-9x^2+27x-27-\left(x^3-27\right)+9\left(x^2+2x+1\right)=15\)

\(\Rightarrow x^3-9x^2+27x-27-x^3+27+9x^2+18x+9=15\)

\(\Rightarrow45x+9=15\Rightarrow45x=6\Rightarrow x=\dfrac{6}{45}=\dfrac{2}{15}\)

a) \(49-\left(3x-1\right)^2=0\)

\(\Leftrightarrow7^2-\left(3x-1\right)^2=0\)

\(\Leftrightarrow\left(7-3x+1\right)\left(7+3x-1\right)=0\)

\(\Leftrightarrow\left(8-3x\right)\left(6+3x\right)=0\)

\(\hept{\begin{cases}8-3x=0\\6+3x=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)

Vậy \(x=\frac{8}{3};x=-2\)

b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=3\left(1-x^2\right)\)

\(\Leftrightarrow\left(x-1\right)^3-\left(x^3+2^3\right)-3\left(1-x^2\right)=0\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-8-3+3x^2=0\)

\(\Leftrightarrow3x-12=0\)

\(\Rightarrow x=4\)

Vậy \(x=4\)

\(3x^2-15x=0\)

\(\Leftrightarrow3x\left(x-5\right)=0\Leftrightarrow x=0;5\)

\(\frac{2}{x}=\frac{y}{15}=\frac{\left(-25\right)}{75}\)

\(\Rightarrow\frac{y}{15}=\frac{\left(-1\right)}{3}\)

\(\frac{y}{15}=\frac{\left(-5\right)}{15}\)

\(\Rightarrow15y=15\times\left(-5\right)\)

\(15y=\left(-75\right)\)

\(y=\left(-75\right)\div15\)

\(y=\left(-5\right)\)

\(\Rightarrow\frac{2}{x}=\frac{\left(-5\right)}{15}\)

\(\frac{2}{x}=\frac{\left(-1\right)}{3}\)

\(\Rightarrow-x=3\times2\)

\(-x=6\)

\(x=\left(-6\right)\)

\(V\text{ậy}x=\left(-6\right);y=\left(-5\right)\)

thif 1 trong 2 so = 0