Đáp án:
Giải thích các bước giải:
b) `(2/3+x)(1/5-2x)=0`
`⇔` \(\left[ \begin{array}{l}\dfrac{2}{3}+x=0\\\dfrac{1}{5}-2x=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{2}{3}\\x=\dfrac{1}{10}\end{array} \right.\)
Vậy `x \in {2/3.;1/10}`
d) `22/9 - (x+1/2)^2 =7/3`
`⇔ (x+1/2)^2=1/9`
`⇔ (x+1/2)^2=(1/3)^2`
`⇔` \(\left[ \begin{array}{l}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{array} \right.\)
Vậy `x \in {- 1/6;- 5/6}`
