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\(Giải:\)
\(ĐK:x\ne\left(-2\right);x\ne\left(-1\right)\)
\(\frac{x^2+2x+2}{x+1}>\frac{x^2+4x+5}{x+2}-1\Leftrightarrow\frac{x^2+2x+2}{x+1}>\frac{x^2+3x+3}{x+2}\)
\(\Leftrightarrow\frac{x^2+2x+1}{x+1}+\frac{1}{x+1}-\frac{x^2+3x+2+1}{x+2}>0\)
\(\Leftrightarrow\frac{\left(x+1\right)^2}{x+1}-\frac{\left(x+1\right)\left(x+2\right)}{x+2}+\frac{1}{x+1}-\frac{1}{x+2}>0\)
\(\Leftrightarrow x+1-x-1+\frac{1}{x+1}-\frac{1}{x+2}>0\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}>0\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}=\frac{1}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x+1>0\\x+2>0\end{cases}}hoặc\hept{\begin{cases}x+1< 0\\x+2< 0\end{cases}}\)
\(+,\hept{\begin{cases}x+1>0\\x+2>0\end{cases}}\Rightarrow x>\left(-2\right)\)
\(+,\hept{\begin{cases}x+1< 0\\x+2< 0\end{cases}}\Rightarrow x< \left(-2\right)\)
BPT đã được giải quyết
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2+2x+2\right)-\left(x^2+4x+5\right)\left(x+1\right)+\left(x^2+3x+2\right)}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{x^3+4x^2+6x+4-\left(x^3+5x^2+9x+5\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)}>0\)
=>\(\dfrac{x^3+5x^2+9x+6-x^3-5x^2-9x-5}{\left(x+1\right)\left(x+2\right)}>0\)
=>(x+1)(x+2)>0
=>x>-1 hoặc x<-2
a: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)
\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)
=>3x+5<10x-30
=>-7x<-35
hay x>5
b: \(\Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-80-12x^2-6x>4x-12x^2-15x\)
=>14x-80>-11x
=>25x>80
hay x>16/5
2.\(\dfrac{x}{x-2}+\dfrac{x+2}{x}>2\) (ĐKXĐ: \(x\ne0;2\))
\(\dfrac{x}{x-2}+\dfrac{x+2}{x}>2\\ \Leftrightarrow\dfrac{x^2}{\left(x-2\right)x}+\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)x}>\dfrac{2x\left(x-2\right)}{\left(x-2\right)x}\\ \Rightarrow x^2+x^2-4>2x^2-4x\)
\(\Leftrightarrow-4>-4x\Leftrightarrow x>1\)
vậy bất phương trình có tập nghiệm là \(S=\left\{x|x>1;x\ne2\right\}\)
a) \(3\left(4x-1\right)-2x\left(5x+2\right)>8x-2\)
\(\Leftrightarrow12x-3-10x^2-4x>8x-2\)
\(\Leftrightarrow-10x^2>5\)
\(\Leftrightarrow x^2< \dfrac{-1}{2}\)(vô lí)
Vậy bất phương trình đã cho vô nghiệm.
h)
\(\dfrac{x+5}{x+7}-1>0\)
\(\Leftrightarrow\dfrac{x+5}{x+7}-\dfrac{x+7}{x+7}>0\)
\(\Leftrightarrow\dfrac{x+5-x-7}{x+7}>0\)
\(\Leftrightarrow\dfrac{-2}{x+7}>0\)
\(\Leftrightarrow x+7< 0\)
\(\Leftrightarrow x< -7\)
g)
\(\dfrac{4-x}{3x+5}\ge0\)
* TH1:
\(4-x\ge0\) và \(3x+5>0\)
\(\Leftrightarrow x\le4\) và \(x>\dfrac{-5}{3}\)
* TH2:
\(4-x\le0\) và \(3x+5< 0\)
\(\Leftrightarrow x\ge4\) và \(x< \dfrac{-5}{3}\) ( loại)
Vậy: \(-\dfrac{5}{3}< x\le4\)
a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)
\(\Leftrightarrow6x+2-20+8x>8x-6-6\)
\(\Leftrightarrow14x-18-8x+12>0\)
\(\Leftrightarrow6x-6>0\)
\(\Leftrightarrow6x>6\)
hay x>1
Vậy: S={x|x>1}
b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)
\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)
\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)
\(\Leftrightarrow-1< 0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
\(\dfrac{x^2+2x+2}{x+1}>\dfrac{x^2+4x+5}{x+2}-1\left(x\ne-1,-2\right)\)
\(\Leftrightarrow\dfrac{x^2+2x+1+1}{x+1}>\dfrac{x^2+4x+4+1}{x+2}-1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}>\dfrac{\left(x+2\right)^2+1}{x+2}-1\)\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x+1}+\dfrac{1}{x+1}>\dfrac{\left(x+2\right)^2}{x+2}+\dfrac{1}{x+2}-1\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}>x+2+\dfrac{1}{x+2}-1\)
\(\Leftrightarrow\dfrac{1}{x+1}>\dfrac{1}{x+2}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}>0\)
\(\Leftrightarrow\dfrac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}>0\)mà 1 > 0 \(\Rightarrow\left(x+1\right)\left(x+2\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x< -2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-1\\x< -2\end{matrix}\right.\)
x2+2x+2x+1>x2+4x+5x+2−1(x≠−1,−2)x2+2x+2x+1>x2+4x+5x+2−1(x≠−1,−2)
⇔x2+2x+1+1x+1>x2+4x+4+1x+2−1⇔x2+2x+1+1x+1>x2+4x+4+1x+2−1
⇔(x+1)2+1x+1>(x+2)2+1x+2−1⇔(x+1)2+1x+1>(x+2)2+1x+2−1⇔(x+1)2x+1+1x+1>(x+2)2x+2+1x+2−1⇔(x+1)2x+1+1x+1>(x+2)2x+2+1x+2−1
⇔x+1+1x+1>x+2+1x+2−1⇔x+1+1x+1>x+2+1x+2−1
⇔1x+1>1x+2⇔1x+1>1x+2
⇔1x+1−1x+2>0⇔1x+1−1x+2>0
⇔x+2−x−1(x+1)(x+2)>0⇔x+2−x−1(x+1)(x+2)>0
⇔1(x+1)(x+2)>0⇔1(x+1)(x+2)>0mà 1 > 0 ⇒(x+1)(x+2)>0⇒(x+1)(x+2)>0
⇔⎡⎢ ⎢ ⎢ ⎢⎣{x+1>0x+2>0{x+1<0x+2<0⇔⎡⎢ ⎢ ⎢ ⎢⎣{x>−1x>−2{x<−1x<−2⇔[x>−1x<−2