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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
\(\dfrac{x^2+2x+2}{x+1}>\dfrac{x^2+4x+5}{x+2}-1\left(x\ne-1,-2\right)\)
\(\Leftrightarrow\dfrac{x^2+2x+1+1}{x+1}>\dfrac{x^2+4x+4+1}{x+2}-1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}>\dfrac{\left(x+2\right)^2+1}{x+2}-1\)\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{x+1}+\dfrac{1}{x+1}>\dfrac{\left(x+2\right)^2}{x+2}+\dfrac{1}{x+2}-1\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}>x+2+\dfrac{1}{x+2}-1\)
\(\Leftrightarrow\dfrac{1}{x+1}>\dfrac{1}{x+2}\)
\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}>0\)
\(\Leftrightarrow\dfrac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}>0\)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
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x^2+2x+2\right)-\left(x^2+4x+5\right)\left(x+1\right)+\left(x^2+3x+2\right)}{\left(x+1\right)\left(x+2\right)}>0\)
\(\Leftrightarrow\dfrac{x^3+4x^2+6x+4-\left(x^3+5x^2+9x+5\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)}>0\)
=>\(\dfrac{x^3+5x^2+9x+6-x^3-5x^2-9x-5}{\left(x+1\right)\left(x+2\right)}>0\)
=>(x+1)(x+2)>0
=>x>-1 hoặc x<-2
c: \(\Leftrightarrow2x-8>=2x+1\)
=>-8>=1(vô lý)
d: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)
\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)
=>10x-30>3x+5
=>7x>35
hay x>5
2.a)
\(2x\left(6x-1\right)>\left(3x-2\right)\left(4x+3\right)\)
\(\Leftrightarrow12x^2-2x>12x^2+9x-8x-6\)
\(\Leftrightarrow12x^2-2x-12x^2-9x+8x>6\)
\(\Leftrightarrow-3x>6\)
\(\Leftrightarrow3>\dfrac{6}{-3}\)
\(\Leftrightarrow x< -2\)
Vậy nghiệm của bpt \(S=\left\{-2\right\}\)
2.b)
\(\dfrac{2\left(x+1\right)}{3}-2\ge\dfrac{x-2}{2}\)
\(\Leftrightarrow4\left(x+1\right)-2.6\ge3x-6\)
\(\Leftrightarrow4x+4-12\ge3x-6\)
\(\Leftrightarrow4x-3x\ge-6-4+12\)
\(\Leftrightarrow x\ge2\)
vậy nghiệm của bpt x\(\ge\)2
a) 4x -8 ≥ 3(3x-1)-2x +1
⇒4x -8 ≥7x -2
⇒4x -7x ≥ -2 +8
⇒-3x ≥ 6
⇒x≤-2
Vậy bpt có nghiệm là:{x|x≤-2}
b) (x-3)(x+2)+(x+4)2≤ 2x (x+5)+4
⇔ x2+2x - 3x - 6 +x2 + 8x +16≤ 2x2 + 10x +4
⇔ x2 +2x - 3x + x2 + 8x - 2x2- 10x ≤ 4+6-16
⇔ -3x ≤ -6
⇔ x≥ 2
Vậy bpt có tập nghiệm là: {x|x≥2}
a/ ĐKXĐ: x khác 1; x khác - 2
pt <=> \(\dfrac{x-1}{\left(x+2\right)\left(x-1\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-1\right)}=\dfrac{4x-8}{\left(x+2\right)\left(x-1\right)}\)
\(\Leftrightarrow x-1-2x-4=4x-8\Leftrightarrow-5x=-3\Leftrightarrow x=\dfrac{3}{5}\left(tm\right)\)
Vậy........
b/ \(2x-3\ge5\Leftrightarrow2x\ge8\Leftrightarrow x\ge4\)
Vậy......
c,d tt
a. \(\dfrac{1}{x+2}-\dfrac{2}{x-1}=\dfrac{4x-8}{\left(x+2\right)\left(x-1\right)}\)
ĐKXĐ: \(x\ne-2;x\ne1\)
\(\Leftrightarrow\dfrac{1\left(x-1\right)}{x+2\left(x-1\right)}-\dfrac{2\left(x+2\right)}{x-1\left(x+2\right)}=\dfrac{4x-8}{\left(x+2\right)\left(x-1\right)}\)
\(\Rightarrow1\left(x-1\right)-2\left(x+2\right)=4x-8\)
\(\Leftrightarrow x-1-2x-4=4x-8\)
\(\Leftrightarrow x-2x-4x=-8+1+4\)
\(\Leftrightarrow-5x=-3\)
\(\Leftrightarrow x=\dfrac{3}{5}\)
Vậy \(S=\left\{\dfrac{3}{5}\right\}\)
b) \(2x-3\ge5\left(2\right)\)
\(\Leftrightarrow2x\ge8\)
\(\Leftrightarrow x\ge4\)
Vậy tập nghiệm của BPT (2) là \(x\ge4\)
c) \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
ĐKXĐ: \(x\ne2;x\ne-1\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)}{x+1\left(x-2\right)}-\dfrac{1\left(x+1\right)}{x-2\left(x+1\right)}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow2x-3-1-x=2x-6\)
\(\Leftrightarrow2x-x-2x=-6+3+1\)
\(\Leftrightarrow x=2\) (KTM)
Vậy pt vô \(n_o\)
d) \(3x-5\ge7\left(4\right)\)
\(\Leftrightarrow3x\ge12\)
\(\Leftrightarrow x\ge4\)
Vậy tập nghiệm của BPT (4) là \(x\ge4\)
a) \(\dfrac{4\left(x-4\right)}{12}\)-\(\dfrac{3x}{12}\)-\(\dfrac{12}{12}\) = 0
\(\dfrac{4x-16-3x-12}{12}=0\)
\(\dfrac{x-28}{12}\)\(=0\)
x - 28 = 0
x = 28
Vậy x = 28
ĐKXĐ : x khác -1
\(\dfrac{x^2+2x+2}{x+1}\ge\dfrac{x^2+3x+4}{x+1}\\ \Leftrightarrow\dfrac{x^2+2x+2}{x+1}\ge\dfrac{x^2+2x+2}{x+1}+\dfrac{x+2}{x+1}\\ \Leftrightarrow\dfrac{x+2}{x+1}\le0\\ \Leftrightarrow x+2\ge0;x+1< 0\Leftrightarrow-1>x\ge-2\)