Question

[Lê Thanh Dương]

Tìm x biết:

a,2x(x+1)-3-2x=5

b,2x(3x+1)+(4-2x)=7\

c,(x-3)^3-(x-3)(x^2+3x+9)+6(x-1)^2=6

Answer 1

[Trịnh Thành Công]

a)\(2x\left(x+1\right)-3-2x=5\)

\(\Leftrightarrow2x^2+2x-3-2x=5\)

\(\Leftrightarrow2x^2=8\)

\(\Leftrightarrow x^2=4=\left(-2\right)^2=2^2\)

              \(\Rightarrow x=2;-2\)

b)\(2x\left(3x+1\right)+\left(4-2x\right)=7\)

\(\Leftrightarrow6x^2+2x+4-2x=7\)

\(\Leftrightarrow6x^2+4=7\)

\(\Leftrightarrow6x^2=3\)

\(\Leftrightarrow x^2=\frac{1}{2}=-\sqrt{\frac{1}{2}}=\sqrt{\frac{1}{2}}\)

c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x-1\right)^2=6\)

\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x^2-2x+1\right)=6\)

\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)

​\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)​​

​\(\Leftrightarrow3x^2+15x=0\)

\(\Leftrightarrow3x\left(x+5\right)=0\)

         \(\Rightarrow\orbr{\begin{cases}3x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)