Question

Giải bất phương trình : $\frac{\left( 3x-2 \right)\left( 5-x \right)}{\left( 2-7x \right)}\ge 0$.

Answer 1

\(\dfrac{\left(3x-2\right)\left(5-x\right)}{2-7x}\)=\(\dfrac{3x^2-17x+10}{7x-2}\)≥0

TH1: \(\left\{{}\begin{matrix}3x^2-17x-10\ge0\\7x-2>0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{\left(17-\sqrt{409}\right)}{6}\\x\ge\dfrac{\left(17+\sqrt{409}\right)}{6}\end{matrix}\right.\\x>\dfrac{2}{7}\end{matrix}\right.\)=> x≥\(\dfrac{\left(17+\sqrt{409}\right)}{6}\)

TH2:\(\left\{{}\begin{matrix}3x^2-17x-10\le0\\7x-20< 0\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}\dfrac{17-\sqrt{409}}{6}\le x\le\dfrac{17+\sqrt{409}}{6}\\x< \dfrac{2}{7}\end{matrix}\right.\)=>\(\dfrac{17-\sqrt{409}}{6}\le x< \dfrac{2}{7}\)

Answer 2

678: 6

 

 

Answer 3

ok

(3x−2)(5−x)2−7x=

17−4096≤x≤17+4096x<27=>

Answer 4

(3x−2)(5−x)2−7x=

17−4096≤x≤17+4096x<27=>

Answer 5

 

BPT ⇔ 3 x 2 − 6 x + 3 > 0 ⇔3x2−6x+3>0 ⇔ 3 ( x − 1 ) 2 > 0 ⇔3(x−1)2>0 Có ( x − 1 ) 2 ≥ 0 ∀ x (x−1)2≥0∀x . Dấu ''='' xảy ra khi x = 1 => Để 3 ( x − 1 ) 2 > 0 3(x−1)2>0 thì ( x − 1 ) 2 ≠ 0 ⇔ x ≠ 1 (x−1)2≠0⇔x≠1 Vậy 3 x 2 − 5 x − x + 3 > 0 3x2−5x−x+3>0 ⇔ x ≠ 1