Tìm x a. 3.(x-2)-2x.(2-x)=0 b, (3-x) : ( $\frac{2}{3}$ .x +7) < 0 c, 3.x ² - $\frac{5}{3}$ =0 28/07/2021 Bởi Arianna Tìm x a. 3.(x-2)-2x.(2-x)=0 b, (3-x) : ( $\frac{2}{3}$ .x +7) < 0 c, 3.x ² - $\frac{5}{3}$ =0
Đáp án: a, Ta có : `3(x – 2) – 2x(2 – x) = 0` ` <=> 3(x – 2) + 2x(x – 2) = 0` ` <=> (3 + 2x)(x – 2) = 0` <=> \(\left[ \begin{array}{l}3 + 2x = 0\\x – 2 = 0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x = -3/2\\x=2\end{array} \right.\) Vậy `x = -3/2` và `x = 2` b, Ta có : `(3 – x) : (2/3 . x + 7) < 0` <=> \(\left[ \begin{array}{l}3 – x < 0 ; 2/3 . x + 7 > 0\\3 – x > 0 ; 2/3 . x + 7 < 0\end{array} \right.\) <=> \(\left[ \begin{array}{l}x > 3 ; x > -21/2\\x < 3 ; x < -21/2\end{array} \right.\) <=> \(\left[ \begin{array}{l}x > 3 \\x < -21/2\end{array} \right.\) c, Ta có : `3.x^2 – 5/3 = 0` ` <=> 3.x^2 = 5/3` ` <=> x^2 = 5/9` ` <=> x = ± \sqrt{5/9}` Giải thích các bước giải: Bình luận
$a$) $3.(x-2)-2x.(2-x)=0$ $⇔ (3+2x)(x-2) = 0$ $⇒$ \(\left[ \begin{array}{l}3+2x=0\\x-2=0\end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x = -\dfrac{3}{2}\\x=2\end{array} \right.\) Vậy $x$ $∈$ `{-3/2;2}`. $b$) `(3-x):(2/3 .x+7) < 0` `⇒` `3-x;2/3 .x+7` khác dấu $TH1$. $\left\{\begin{matrix} 3-x < 0& \\\dfrac{2}{3}x + 7 > 0& \end{matrix}\right.$ $⇒$ $x>3$ $TH2$. $\left\{\begin{matrix} 3-x > 0& \\\dfrac{2}{3}x + 7 < 0& \end{matrix}\right.$ $⇒$ $x< \dfrac{21}{2}$ Vậy $x>3;x < \dfrac{-21}{2}$. $c$) $3x^2 – \dfrac{5}{3} = 0$ $⇔ 3x^2 = \dfrac{5}{3}$ $⇔ x^2 = \dfrac{5}{9}$ $⇔ x = ± \sqrt{\dfrac{5}{9}}$ Vậy $x = ± \sqrt{\dfrac{5}{9}}$. Bình luận
Đáp án:
a, Ta có :
`3(x – 2) – 2x(2 – x) = 0`
` <=> 3(x – 2) + 2x(x – 2) = 0`
` <=> (3 + 2x)(x – 2) = 0`
<=> \(\left[ \begin{array}{l}3 + 2x = 0\\x – 2 = 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x = -3/2\\x=2\end{array} \right.\)
Vậy `x = -3/2` và `x = 2`
b, Ta có :
`(3 – x) : (2/3 . x + 7) < 0`
<=> \(\left[ \begin{array}{l}3 – x < 0 ; 2/3 . x + 7 > 0\\3 – x > 0 ; 2/3 . x + 7 < 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x > 3 ; x > -21/2\\x < 3 ; x < -21/2\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x > 3 \\x < -21/2\end{array} \right.\)
c, Ta có :
`3.x^2 – 5/3 = 0`
` <=> 3.x^2 = 5/3`
` <=> x^2 = 5/9`
` <=> x = ± \sqrt{5/9}`
Giải thích các bước giải:
$a$) $3.(x-2)-2x.(2-x)=0$
$⇔ (3+2x)(x-2) = 0$
$⇒$ \(\left[ \begin{array}{l}3+2x=0\\x-2=0\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x = -\dfrac{3}{2}\\x=2\end{array} \right.\)
Vậy $x$ $∈$ `{-3/2;2}`.
$b$) `(3-x):(2/3 .x+7) < 0`
`⇒` `3-x;2/3 .x+7` khác dấu
$TH1$. $\left\{\begin{matrix} 3-x < 0& \\\dfrac{2}{3}x + 7 > 0& \end{matrix}\right.$
$⇒$ $x>3$
$TH2$. $\left\{\begin{matrix} 3-x > 0& \\\dfrac{2}{3}x + 7 < 0& \end{matrix}\right.$
$⇒$ $x< \dfrac{21}{2}$
Vậy $x>3;x < \dfrac{-21}{2}$.
$c$) $3x^2 – \dfrac{5}{3} = 0$
$⇔ 3x^2 = \dfrac{5}{3}$
$⇔ x^2 = \dfrac{5}{9}$
$⇔ x = ± \sqrt{\dfrac{5}{9}}$
Vậy $x = ± \sqrt{\dfrac{5}{9}}$.