Đáp án:
Giải thích các bước giải:
a, Ta có: 2x-9=0 <=> 2x=9 <=> x=9/2 . Vậy nghiệm của P(x) là 9/2
b, Ta có: (2-$\frac{2}{3}$ x ).( -2x+1) =0
⇔\(\left[ \begin{array}{l}2-2/3x=0\\-2x+1=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}2/3x=2\\2x=1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=3\\x=0,5\end{array} \right.\)
Vậy ....
c, Ta có: 7-2x=0 ⇔ 2x=7 ⇔x=$\frac{7}{2}$
Vậy....
d, Ta có: $x^{3}$ -4x =0 ⇔ x(x^2-4) =0
⇔\(\left[ \begin{array}{l}x=0\\x^2-4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x^2=4\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x =2 hoặc x=-2\end{array} \right.\)
Vậy....
e, Ta có: 3x-5=0 ⇔ 3x=5⇔ x=$\frac{5}{3}$
Vậy ...
f, Ta có: (x-$\frac{4}{7}$ ).( 2-$\frac{1}{3}$x )
⇔\(\left[ \begin{array}{l}x-4/7=0\\2-1/3x=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4/7\\1/3x=2\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4/7\\x=6\end{array} \right.\)
Vậy ....
g, Ta có: 6x-2x=0 ⇔ 4x=0⇔x=0
Vậy ...
h, Ta có: 3x²-x=0 ⇔ x(3x-1)=0
⇔\(\left[ \begin{array}{l}x=0\\3x-1=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\3x=1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=0\\x=1/3\end{array} \right.\)
Vậy....
i, Ta có: 3x-$\frac{5}{2}$ =0 ⇔3x =$\frac{5}{2}$ ⇔x=$\frac{5}{6}$
Vậy...
j, Ta có: 2/5x(x - 1)+3(x - 1)=0 ⇔ (x-1)(2/5x+3)=0
⇔\(\left[ \begin{array}{l}x-1=0\\2/5x+3=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\2/5x=-3\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1\\x=-15/2\end{array} \right.\)
Vậy ...
