<=>x^3-6x^2+12x-8-x^3+6x^2=7

<=>12x=15

<=> x=5/4

chúc bạn hk tốt

a)\(2x^4-6x^3+x^2+6x-3=0\)

\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)

\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)

b)\(x^3+9x^2+26x+24=0\)

\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)

\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)

\(\Leftrightarrow\left(2x+3-2x+5\right)^2=x^2+6x+64\)

=>x^2+6x=0

=>x(x+6)=0

=>x=0 hoặc x=-6

\(6x^2-\left(2x+5\right)\left(3x-2\right)=7.\)

\(6x^2-\left(6x^2-4x+15x-10\right)=7\)

\(6x^2-6x^2+4x-15x+10=7\)

\(-11x+10=7\)

\(-11x=-3\)

\(x=\frac{-3}{-11}=\frac{3}{11}\)

\(6x^2-\left(2x+5\right)\left(3x-2\right)=7\)

\(6x^2-\left(6x^2-4x+15x-10\right)=7\)

+ Áp dụng quy tắc bở ngoặc ta có :

\(6x^2-6x^2+4x-15x+10=7\)

\(-11x+10=7\)

        \(-11x=7-10\)

            \(-11x=-3\)

                        \(x=-3:-11\)

                         \(x=\frac{-3}{-11}\)

                       \(\Rightarrow x=\frac{3}{11}\)

     Vậy \(x=\frac{3}{11}\)

Chúc bạn học tốt !!!

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

\(x^3-2x^2-3x+2x^2-5x=7\)

\(x^3-8x-7=0\)

lam the nao nua

X = 0 

nha

kb nha