a) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-5\right)\left(x+5\right)=-12\)
\(\Leftrightarrow x^3-3x^2+9x+3x^2-9x+27-x^3-5x^2+5x^2+25x=-12\)
\(\Leftrightarrow25x+27=-12\)
\(\Leftrightarrow25x=-12-27\)
\(\Leftrightarrow25x=-39\)
\(\Leftrightarrow x=-\dfrac{39}{25}\)
b) \(6\left(x+1\right)^2+2\left(x-1\right)\left(x^2+x+1\right)-2\left(x+1\right)^3=32\)
\(\Leftrightarrow6x^2+12x+6+2x^3+2x^2+2x-2x^2-2x-2-2x^3-4x^2-2x-2x^2-4x-2=32\)
\(\Leftrightarrow6x+2=32\)
\(\Leftrightarrow6x=32-2\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
c) \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x+10=30\)
\(\Leftrightarrow10x=30-10\)
\(\Leftrightarrow10x=20\)
\(\Leftrightarrow x=2\)
c: Ta có: \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
a: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-5\right)\left(x+5\right)=-12\)
\(\Leftrightarrow x^3+27-x\left(x^2-25\right)=-12\)
\(\Leftrightarrow x^3+27-x^3+25x=-12\)
\(\Leftrightarrow25x=-39\)
hay \(x=-\dfrac{39}{25}\)
b: Ta có: \(6\left(x+1\right)^2+2\left(x-1\right)\left(x^2+x+1\right)-2\left(x+1\right)^3=32\)
\(\Leftrightarrow6\left(x^2+2x+1\right)+2\left(x^3-1\right)-2\left(x^3+3x^2+3x+1\right)=32\)
\(\Leftrightarrow6x^2+12x+6+2x^3-2-2x^3-6x^2-6x-2=32\)
\(\Leftrightarrow6x=30\)
hay x=5
