\(x^3-11x^2+30x=0\)
\(\left(x-6\right).\left(x-5\right).x=0\)
\(=>\orbr{\begin{cases}x-6=0\\x-5=0,x=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=6\\x=5,x=0\end{cases}}\)
P/S: mk mới lớp 7 sai sót mong bỏ qua
\(8x^2+30x+7=0\)
\(8x^2+28x+2x+7=0\)
\(2x.\left(4x+1\right)+7.\left(4x+1\right)=0\)
\(\left(2x+7\right).\left(4x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=-7\\4x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{2}\\x=-\frac{1}{4}\end{cases}}\)
vậy ....
P/S sorry mk làm hơi lâu :)__chờ tí làm câu a cho
\(x^2+3x-18=0\)
\(x^2-3x+6x-18=0\)
\(x.\left(x-3\right)+6.\left(x-3\right)=0\)
\(\left(x+6\right).\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-6\\x=3\end{cases}}}\)
P/S:nếu cách làm sai sót hay dài thì bn thồn cảm nha mk kiểm tra kết quả thấy ko sai
a) \(x^2+3x-18=0\)
\(x^2+6x-3x-18=0\)
\(x\left(x+6\right)-3\left(x+6\right)=0\)
\(\left(x+6\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-6\\x=3\end{cases}}}\)
b) \(8x^2+30x+7=0\)
\(8x^2+2x+28x+7=0\)
\(2x\left(4x+1\right)+7\left(4x+1\right)=0\)
\(\left(4x+1\right)\left(2x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x+1=0\\2x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-7}{2}\end{cases}}}\)
c) \(x^3-11x^2+30x=0\)
\(x\left(x^2-11x+30\right)=0\)
\(x\left(x^2-6x-5x+30\right)=0\)
\(x\left[x\left(x-6\right)-5\left(x-6\right)\right]=0\)
\(x\left(x-6\right)\left(x-5\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-6=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=6\\x=5\end{cases}}}\)
P.s: câu c) dùng ngoặc vuông
a) \(x^2+3x-18=0\)
\(\Leftrightarrow\left(x^2+6x\right)-\left(3x+18\right)=0\)
\(\Leftrightarrow x\left(x+6\right)-3\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-3\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+6=0\\x-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-6\\x=3\end{cases}}\)
b) \(8x^2+30x+7=0\)
\(\Leftrightarrow\left(8x^2+2x\right)+\left(28x+7\right)=0\)
\(\Leftrightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\)
\(\Leftrightarrow\left(4x+1\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+1=0\\2x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=-1\\2x=-7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{7}{2}\end{cases}}\)
Easy mà!
c) \(\(x^3-11x^2+30x=0\)\)
\(\(\Leftrightarrow x\left(x^2-11x+30\right)=0\)\)
\(\(\Leftrightarrow x\left(x^2-5x-6x+30\right)=0\)\)
\(\(\Leftrightarrow x\left[\left(x^2-5x\right)-\left(6x-30\right)\right]=0\)\)
\(\(\Leftrightarrow x\left[x\left(x-5\right)-6\left(x-5\right)\right]=0\)\)
\(\(\Leftrightarrow x\left(x-5\right)\left(x-6\right)=0\)\)
\(\(\Leftrightarrow\hept{\begin{cases}x=0\\x-5=0\\x-6=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=0\\x=5\\x=6\end{cases}}\)\)
