1a)\(\left(x+2\right)^2-6\left(x+2\right)\le x^2-4\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\ge\left(x+2\right)\left(x+2-6\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x+2\right)\left(x-4\right)\ge0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2-x+4\right)\ge0\)
\(\Leftrightarrow\left(x+2\right)\cdot2\ge0\)
\(\Leftrightarrow x+2\ge0\)
\(\Leftrightarrow x\ge-2\)
b)\(-\dfrac{2}{x-1}>0\left(đkxđ:x\ne1\right)\)
\(\Leftrightarrow\dfrac{2}{x-1}< 0\)
\(\Leftrightarrow x-1< 0\)
\(\Leftrightarrow x< 1\)
1)
a) (x+2)2 - 6(x+2) \(\le\) x2 - 4
<=> (x+2).(x+2 - 6) \(\le\) (x+2)(x-2)
<=> x2 + 2x - 6x + 2x + 4 - 12 \(\le\) x2 - 2x +2x - 4
<=> x2 - x2 + 2x + 2x - 6x + 2x - 2x \(\le\) 12 - 4 - 4
<=> -2x \(\le\) 4
<=> x \(\ge\) -2
Vậy bpt có nghiệm x \(\ge\) -2
b) Để \(\dfrac{-2}{x-1}\) nhận giá trị không âm
=>\(\dfrac{-2}{x-1}\) \(\ge\) 0
<=> -2 . (x-1) \(\ge\) 0
<=> -2x + 2 \(\ge\) 0
<=> -2x \(\ge\) -2
<=> x \(\le\) 1
Vậy với x \(\le\) 1 thì biểu thức \(\dfrac{-2}{x-1}\) nhận giá trị không âm.
2)
a) \(|x+1|\) = 2x - 1
+) Nếu x+1 \(\ge\) 0 => x \(\ge\) -1 thì phương trình :
x + 1 = 2x -1
<=> -x = -2
<=> x = 2 (thỏa mãn)
+) Nếu x+1 < 0 => x < -1 thì phương trình :
-(x + 1) = 2x - 1
<=> -x -1 = 2x -1
<=> -3x = 0
<=> x = 0 (không thỏa mãn)
Vậy phương trình có nghiệm x = 2.
c) \(\dfrac{x+1}{x-2}\) - \(\dfrac{5}{x+2}\) = \(\dfrac{12}{x^2-4}\)+ 1 (ĐKXĐ: x \(\ne\) \(\pm\) 2)
<=>\(\dfrac{\left(x+1\right).\left(x+2\right)-5.\left(x-2\right)}{x^2-4}\) = \(\dfrac{12+x^2-4}{x^2-4}\)
<=> \(\dfrac{x^2-2x+12}{x^2-4}\) = \(\dfrac{x^2+8}{x^2-4}\)
<=> x2 - 2x +12 = x2 +8
<=> -2x = 8-12
<=> x = 2 ( không thỏa mãn ĐKXĐ)
Vậy phương trình vô nghiệm.
1/
a) \(\left(x+2\right)^2-6\left(x+2\right)\le x^2-4\)
\(\Leftrightarrow x^2+4x+4-6x-12-x^2+4\le0\)
\(\Leftrightarrow-2x\le4\Leftrightarrow x\ge-2\)
Vậy bpt có tập nghiệm: S = {x|\(x\ge2\)}
b) \(\dfrac{-2}{x-1}< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)
Vậy để biểu thức âm thì x < 1
2/
a) \(\left|x+1\right|=2x-1\) (1)
+) Nếu x + 1 \(\ge\) 0 <=> \(x\ge-1\)
thì |x + 1| = x + 1
(1) => x + 1 = 2x - 1
<=> x - 2x = -1 - 1
<=> -x = - 2 <=> x = 2 (nhận)
+) Nếu x + 1 < 0 <=> x < - 1
thì |x + 1| = -x - 1
(1) => -x - 1 = 2x - 1
<=> -x - 2x = -1 + 1
<=> -3x = 0
<=> x = 0 (loại)
Vậy pt (1) có 1 nghiệm x = 2
b/ \(x^3-1-x\left(x-1\right)=0\)
\(\Leftrightarrow x^3-x^2+x-1=0\)
\(\Leftrightarrow x^2\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=-1\left(loai\right)\end{matrix}\right.\)
Vậy pt có 1 nghiệm x = 1
c/ \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\) ĐKXĐ: \(x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-5\left(x-2\right)=12+\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2+3x-5x-x^2=12-4-10-2\)
\(\Leftrightarrow-2x=-4\Leftrightarrow x=2\)(không t/m đkxđ)
Vậy pt vô nghiệm
2a)\(\left|x+1\right|=2x-1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2x-1\left(x\ge-1\right)\\x+1=1-2x\left(x< -1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\left(loai\right)\end{matrix}\right.\)
b)\(x^3-1-x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x=1\)(vì \(x^2+1>0\))
