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

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

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

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

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

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

(x+1/5) -4 = -2

x= -6-1/5

x= -31/5

Ta có : \(\left(2x+3\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=-3\\x=7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\)

Tìm x,biết :

\(a,\left(x+\frac{1}{5}\right)-4=-2\) 

\(\left(x+\frac{1}{5}\right)=2\)  

\(x+\frac{1}{5}=2\) 

\(x=\frac{9}{5}\) 

b,\(\left(2x+3\right)\left(x-7\right)=0\) 

\(\Rightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\) 

\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\) 

\(c,\frac{31}{9}x-\frac{5}{2}=\frac{8}{3}\) 

\(\frac{31}{9}x=\frac{8}{3}+\frac{5}{2}\) 

\(\frac{31}{9}x=\frac{31}{6}\) 

\(x=\frac{3}{2}\)

Đăng từng câu đio

,(8x-3).(3x+2)-(4x+7).(x+4)=(2x+1).(5x-1)-33 đúng không bạn

\(a,\left(2x-10\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-10=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)

Vậy .........

\(b,\left(x+5\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x^2-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

Vậy ......

\(a,\left(2x-10\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-10=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=10\\x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)

\(b,\left(x+5\right)\left(x^2-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x^2=9\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=3or-3\end{cases}}}\)

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

=>x^2+6x=0

=>x(x+6)=0

=>x=0 hoặc x=-6