a) (x - 7)(2x + 8) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\2x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy: S = {7; -4}
b) Tương tự câu a
c) (x - 1)(2x + 7)(x2 + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\\x^2+2=0\end{matrix}\right.\)
Mà: x2 + 2 > 0 với mọi x
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{2}\right\}\)
d) (2x - 1)(x + 8)(x - 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-8\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-8;5\right\}\)
a/ Pt \(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy \(S=\left\{7;-4\right\}\)
b/ pt \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\5x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)
c/ pt \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\) (\(x^2+2>0\forall x\))\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
d/ pt \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
a)(x-7)(2x+8)=0
⇔x-7=0 hoặc 2x+8=0
1.x-7=0⇔x=7
2.2x+8=0⇔2x=-8⇔x=-4
phương trình có 1 nghiệm x=7 và x=-4
b)(3x+1)(5x-2)=0
⇔3x+1=0 hoặc 5x-2=0
1.3x+1=0⇔3x=-1⇔x=-1/3
2.5x-2=0⇔5x=2⇔x=5/2
phương trình có 2 nghiệm x=-1/3 và x=5/2
c)(x-1)(2x+7)(x2+2)=0
⇔x-1=0 hoặc 2x+7=0 hoặc x2+2=0
1.x-1=0⇔x=1
2.2x+7=0⇔2x=-7⇔x=-7/2
3.x2+2=0⇔x2=-2⇔x=√(-2)2
phương trình có 3 nghiệm x=1 ; x=-7/2 và x=√(-2)2
d)(2x-1)(x+8)(x-5)=0
⇔2x-1=0 hoặc x+8=0 hoặc x-5=0
1.2x-1=0⇔2x=1⇔x=1/2
2.x+8=0⇔x=-8
3.x-5=0⇔x=5
phương trình có 3 nghiệm x=1/2 ;x=-8 và x=5
\(\left(x-7\right)\left(2x+8\right)=0\)
\(< =>\orbr{\begin{cases}x-7=0\\2x+8=0\end{cases}}\)
\(< =>\orbr{\begin{cases}x=7\\2x=-8\end{cases}}\)
\(< =>\orbr{\begin{cases}x=7\\x=-\frac{8}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=7\\x=-4\end{cases}}\)
\(\left(3x+1\right)\left(5x-2\right)=0\)
\(< =>\orbr{\begin{cases}3x+1=0\\5x-2=0\end{cases}}\)
\(< =>\orbr{\begin{cases}3x=-1\\5x=2\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{2}{5}\end{cases}}\)
\(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)
\(< =>\left(x-1\right)\left(2x+7\right)=0\)
\(< =>\orbr{\begin{cases}x-1=0\\2x+7=0\end{cases}}\)
\(< =>\orbr{\begin{cases}x=1\\2x=-7\end{cases}}\)
\(< =>\orbr{\begin{cases}x=1\\x=-\frac{7}{2}\end{cases}}\)