a) \(\left(2x-7\right)^2-6\left(2x-7\right)\left(x-3\right)=0\)
\(\Leftrightarrow4x^2-28x+49-6\left(2x^2-13x+21\right)=0\)
\(\Leftrightarrow4a^2-28x+49-12x^2+78x-126=0\)
\(\Leftrightarrow-8x^2+50x-77=0\)
\(\Leftrightarrow-8x^2+28x+22x-77=0\)
\(\Leftrightarrow-4x\left(2x-7\right)+11\left(2x-7\right)=0\)
\(\Leftrightarrow\left(11-4x\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}11-4x=0\\2x-7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{11}{4}\\x=\frac{7}{2}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là : \(S=\left\{\frac{11}{4};\frac{7}{2}\right\}\)
b) Hình như sai đề ạ ! Sửa đề như này tớ thấy hợp lí hơn :33
\(x^3+1+\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)+\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2-x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(ktm\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là \(S=\left\{-2\right\}\)
Bài 2:
a) Ta có: \(\left(2x-7\right)^2-6\left(2x-7\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left[\left(2x-7\right)-6\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x-7-6x+18\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(11-4x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\11-4x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\4x=11\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{11}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{7}{2};\frac{11}{4}\right\}\)
a) \(\left(2x-7\right)^2-6\left(2x-7\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(1-6x+18\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(19-6x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\19-6x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{19}{6}\end{matrix}\right.\)
b) Đề sai. Như này thì phải :
\(x^3+1+\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)+\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(loại\right)\\x=-2\end{matrix}\right.\)
