Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(\left(2-x\right)\left(x+3\right)>0\Leftrightarrow\left(x-2\right)\left(x+3\right)< 0\)
Vì \(x+3>x-2\)
nên \(\hept{\begin{cases}x+3>0\\x-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>-3\\x< 2\end{cases}\Leftrightarrow-3< x< 2}\)
c, \(\left(5-2x\right)\left(x+4\right)>0\)
TH1 : \(\hept{\begin{cases}5-2x>0\\x+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{5}{2}\\x>-4\end{cases}}\Leftrightarrow-4< x< \frac{5}{2}\)
TH2 : \(\hept{\begin{cases}5-2x< 0\\x+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{5}{2}\\x< -4\end{cases}}\)( vô lí )
bạn làm tương tự nhé
Tìm x biết :a) ( 2x - 3 ).( x +1 ) > 0b) ( x + 5 ).(x-7) < 0c) | 2x - 3 | + 8 = 10d) ( 2x + 5 ) . | x -8 | . ( x2 + 1 ) = 0
a) (2x+4) . (x-3) > 0
\(\Rightarrow\orbr{\begin{cases}2x+4< 0;x-3< 0\\2x+4>0;x-3>0\end{cases}}\Rightarrow\orbr{\begin{cases}2x< -4;x< 3\\2x>-4;x>3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< -2;x< 3\\x>-2;x>3\end{cases}}\Rightarrow\orbr{\begin{cases}x< -2\\x>3\end{cases}}\)thì (2x+4).(x-3) > 0
b) \(\frac{x+5}{x-1}< 0\)
\(\Rightarrow\orbr{\begin{cases}x+5< 0;x-1>0\\x+5>0;x-1< 0\end{cases}}\Rightarrow\orbr{\begin{cases}x< -5;x>1\\x>-5;x< 1\end{cases}}\Rightarrow-5< x< 1\)thì \(\frac{x+5}{x-1}< 0\)
c)\(\left(x-2\right)\left(x+5\right)< 0\)
\(\Rightarrow\orbr{\begin{cases}x-2< 0;x+5>0\\x-2>0;x+5< 0\end{cases}}\Rightarrow\orbr{\begin{cases}x< 2;x>-5\\x>2;x< -5\end{cases}}\Rightarrow-5< x< 2\)thì (x-2).(x+5) <0
a) Nhận xét: \(x-1< x+4\)
=> \(\hept{\begin{cases}x-1< 0\\x+4>0\end{cases}}\Rightarrow-4< x< 1\)
b) Nếu: \(\hept{\begin{cases}x>0\\4-x>0\end{cases}}\Rightarrow0< x< 4\)
Nếu: \(\hept{\begin{cases}x< 0\\4-x< 0\end{cases}}\Rightarrow∄x\)
c) Nếu: \(\hept{\begin{cases}1-3x>0\\8+x< 0\end{cases}}\Rightarrow x< -8\)
Nếu: \(\hept{\begin{cases}1-3x< 0\\8+x>0\end{cases}\Rightarrow}x>\frac{1}{3}\)
d) Nếu: \(\hept{\begin{cases}2x+6>0\\4-x>0\end{cases}}\Rightarrow-3< x< 4\)
Nếu: \(\hept{\begin{cases}2x+6< 0\\4-x< 0\end{cases}}\Rightarrow∄x\)
a: (2x-3)(3x+6)>0
=>(2x-3)(x+2)>0
=>x<-2 hoặc x>3/2
b: (3x+4)(2x-6)<0
=>(3x+4)(x-3)<0
=>-4/3<x<3
c: (3x+5)(2x+4)>4
\(\Leftrightarrow6x^2+12x+10x+20-4>0\)
\(\Leftrightarrow6x^2+22x+16>0\)
=>\(6x^2+6x+16x+16>0\)
=>(x+1)(3x+8)>0
=>x>-1 hoặc x<-8/3
f: (4x-8)(2x+5)<0
=>(x-2)(2x+5)<0
=>-5/2<x<2
h: (3x-7)(x+1)<=0
=>x+1>=0 và 3x-7<=0
=>-1<=x<=7/3
a) TH1:
\(\left\{{}\begin{matrix}x< 0\\8-x>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 0\\x< 8\end{matrix}\right.\) \(\Rightarrow x< 0\)
TH2:
\(\left\{{}\begin{matrix}x>0\\8-x< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>0\\x>8\end{matrix}\right.\)\(\Rightarrow x>8\)
Vậy \(\left[{}\begin{matrix}x< 0\\x>8\end{matrix}\right.\)
b) TH1:
\(\left\{{}\begin{matrix}2-x>0\\x+3>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x< 2\\x>-3\end{matrix}\right.\)\(\Rightarrow-3< x< 2\)
TH2:
\(\left\{{}\begin{matrix}2-x< 0\\x+3< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\)(vô nghiệm)
Vậy \(-3< x< 2\)
c) TH1:
\(\left\{{}\begin{matrix}2x-4>0\\5-x< 0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2x>4\\x>5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>2\\x>5\end{matrix}\right.\Rightarrow x>5\)
TH2:
\(\left\{{}\begin{matrix}2x-4< 0\\5-x>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x< 4\\x< 5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 2\\x< 5\end{matrix}\right.\Rightarrow x< 2\)
Vậy \(\left[{}\begin{matrix}x>5\\x< 2\end{matrix}\right.\)
a, \(x\left(8-x\right)< 0\\ 8x-x^2< 0\)
Có x2 ≥ 0 ∀ x (1)
\(\Rightarrow\) - x2 ≤ 0 ∀ x
Mà 8x - x2 < 0
\(\Rightarrow\) 8x < x2 (2)
Thay (1) vào (2) \(\Rightarrow\) 8x < 0
\(\Rightarrow\) x < 0
Vậy x < 0
b, \(\left(2-x\right)\left(x+3\right)>0\\ 2x+6-x^2-3x>0\\ \Rightarrow\left(6-x\right)-x^2>0\)
Có x2 ≥ 0 ∀ x (1)
⇒ - x2 ≤ 0 ∀ x
Mà (6 - x) - x2 > 0
\(\Rightarrow6-x>x^2\left(2\right)\)
Thay (1) vào (2) \(\Rightarrow6-x>0\\ \Rightarrow x< 0\)
Vậy x < 0
c, \(\left(2x-4\right)\left(5-x\right)< 0\\ 10x-2x^2-20+4x< 0\\ \Rightarrow\left(-20+14x\right)-2x^2< 0\)
Có x2 ≥ 0 ∀ x
⇒ 2x2 ≥ 0 ∀ x (1)
⇒ - 2x2 ≤ 0 ∀ x
Mà (-20 + 14x) - 2x2 < 0
\(\left(-20+14x\right)< 2x^2\left(2\right)\)
Thay (1) vào (2) \(\Rightarrow-20+14x< 0\\ \Rightarrow14x< 20\\ \Rightarrow x< \frac{10}{7}\)
Vậy \(x< \frac{10}{7}\)