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\(\frac{x-2}{18}-\frac{2x+5}{12}>\frac{x+6}{9}-\frac{x-3}{6}\)
\(\Leftrightarrow\frac{2\left(x-2\right)}{36}-\frac{3\left(2x+5\right)}{36}>\frac{4\left(x+6\right)}{36}-\frac{6\left(x-3\right)}{36}\)
\(\Leftrightarrow2x-4-6x-15>4x+24-6x+18\)
\(\Leftrightarrow2x-6x-4x+6x>24+18+4+15\)
\(\Leftrightarrow-2x>61\)
\(\Leftrightarrow x< -\frac{61}{2}\)
Vậy nghiệm của bất phương trình là \(x< -\frac{61}{2}\)
Bài b và c làm cách mình thì dễ hiểu hơn nhiều :3
\(\left(2x-2\right)\left(2x+3\right)\le0\)
TH1 : \(\hept{\begin{cases}2x-3\le0\\2x+3\ge0\end{cases}< =>\hept{\begin{cases}2x\le3\\2x\ge-3\end{cases}}}\)
\(< =>\hept{\begin{cases}x\le\frac{3}{2}\\x\ge-\frac{3}{2}\end{cases}}\)
TH2 : \(\hept{\begin{cases}2x-3\ge0\\2x+3\le0\end{cases}< =>\hept{\begin{cases}2x\ge3\\2x\le-3\end{cases}}}\)
\(< =>\hept{\begin{cases}x\ge\frac{3}{2}\\x\le-\frac{3}{2}\end{cases}}\)
Vậy ...
a, \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\left(ĐKXĐ:x\ne\pm2;\pm5\right)\)
\(\frac{x+9}{\left(x-5\right)\left(x+2\right)}-\frac{x+15}{\left(x+5\right)\left(x-5\right)}=\frac{1}{x+2}\)
\(\frac{\left(x+9\right)\left(x+5\right)}{\left(x-5\right)\left(x+2\right)\left(x+5\right)}-\frac{\left(x+15\right)\left(x+2\right)}{\left(x+5\right)\left(x-5\right)\left(x+2\right)}=\frac{\left(x+5\right)\left(x-5\right)}{\left(x+2\right)\left(x+5\right)\left(x-5\right)}\)
Khử mẫu : \(\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)=\left(x+5\right)\left(x-5\right)\)
\(x^2+14x+45-x^2-17x-30=x^2-25\)
\(-3x+15-x^2+25=0\)
\(-3x-x^2+40=0\)( giải delta ta đc )
\(x_1=-5;x_2=8\)
b, \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1ĐKXĐ\left(x\ne1;\frac{1}{3}\right)\)
\(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{\left(3x-1\right)\left(x-1\right)}=1\)
\(\frac{x-1}{\left(3x-1\right)\left(x-1\right)}+\frac{\left(2x+2\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}-\frac{3x^2+1}{\left(3x-1\right)\left(x-1\right)}=\frac{\left(3x-1\right)\left(x-1\right)}{\left(3x-1\right)\left(x-1\right)}\)
Khửi mẫu \(x-1+\left(2x+2\right)\left(3x-1\right)-3x^2-1=\left(3x-1\right)\left(x-1\right)\)( bn tự nốt nhé)
c, \(\left(x+3\right)^2-10\ge\left(x+3\right)\left(x+2\right)-4\)
\(x^2+6x+9-10\ge x^2+5x+6-4\)
\(x-3\ge0\Leftrightarrow x\ge3\)
a) \(\frac{x+9}{x^2-3x-10}-\frac{x+15}{x^2-25}=\frac{1}{x+2}\); ĐKXĐ: x # -2; x # +-5
<=> \(\frac{x+9}{\left(x+2\right)\left(x-5\right)}-\frac{x+15}{\left(x-5\right)\left(x+5\right)}=\frac{1}{x+2}\)
<=> \(\frac{\left(x+9\right)\left(x+5\right)-\left(x+15\right)\left(x+2\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)\left(x+5\right)}{\left(x+2\right)\left(x-5\right)\left(x+5\right)}\)
<=> (x + 9)(x + 5) - (x + 15)(x + 2) = (x - 5)(x + 5)
<=> -3x + 15 = x^2 - 25
<=> -3x + 15 - x^2 + 25 = 0
<=> -3x + 40 - x^2 = 0
<=> x^2 + 3x - 40 = 0
<=> (x - 5)(x + 8) = 0
<=> x - 5 = 0 hoặc x + 8 = 0
<=> x = 5 (ktm0 hoặc x = -8 (tm)
b) \(\frac{1}{3x-1}+\frac{2x+2}{x-1}-\frac{3x^2+1}{3x^2-4x+1}=1\); ĐKXĐ: x # 1/3; x # 1
<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{x\left(3x-1\right)-\left(3x-1\right)}=1\)
<=> \(\frac{1}{3x-1}+\frac{2\left(x+1\right)}{x-1}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=1\)
<=> \(\frac{x-1}{\left(x-1\right)\left(3x-1\right)}+\frac{2\left(x+1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}-\frac{3x^2+1}{\left(x-1\right)\left(3x-1\right)}=\frac{\left(x-1\right)\left(3x-1\right)}{\left(x-1\right)\left(3x-1\right)}\)
<=> x - 1 + 2(x + 1)(3x - 1) - 3x^2 + 1 = (x - 1)(3x - 1)
<=> 5x - 4 + 3x^2 = 3x^2 - 4x + 1
<=> 5x - 4 = -4x + 1
<=> 5x + 4x = 1 + 4
<=> 9x = 5
<=> x = 5/9 (tm)
c) (x + 3)^2 - 10 >= (x + 3)(x + 2) - 4
<=> x^2 + 3x + 3x + 9 - 10 >= x^2 + 2x + 3x + 6 - 4
<=> x^2 + 6x + 9 - 10 >= x^2 + 5x + 6 - 4
<=> x^2 + 6x - 1 >= x^2 + 5x + 2
<=> x^2 + 6x - 1 - x^2 - 5x - 2 >= 0
<=> x - 3 >= 0
<=> x >= 3
a)Ta có: \(\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)
\(\Leftrightarrow\dfrac{3x+9+x+1}{3\left(x+1\right)}\ge0\)
\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1>0\\4x+10\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le-\dfrac{5}{2}\end{matrix}\right.\)
b) Ta có: \(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)
\(\Leftrightarrow\dfrac{3x+6+x+3}{3\left(x+3\right)}\le0\)
\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+9>0\\4x+9\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow-3< x\le-\dfrac{9}{4}\)
a)\(\dfrac{x+3}{x+1}\ge-\dfrac{1}{3}\left(x\ne-1\right)\)
\(\Leftrightarrow\dfrac{x+3}{x+1}+\dfrac{1}{3}\ge0\)
\(\Leftrightarrow\dfrac{3x+9+x+1}{3x+3}\ge0\)
\(\Leftrightarrow\dfrac{4x+10}{3x+3}\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+10\ge0\\3x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+10\le0\\3x+3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{5}{2}\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{-5}{2}\\x< -1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>-1\\x\le\dfrac{-5}{2}\end{matrix}\right.\)
b) \(\dfrac{x+2}{x+3}\le-\dfrac{1}{3}\left(x\ne-3\right)\)
\(\dfrac{x+2}{x+3}+\dfrac{1}{3}\le0\)
\(\Leftrightarrow\dfrac{3x+6+x+3}{3x+9}\le0\)
\(\Leftrightarrow\dfrac{4x+9}{3x+9}\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x+9\ge0\\3x+9< 0\end{matrix}\right.\\\left\{{}\begin{matrix}4x+9\le0\\3x+9>0\end{matrix}\right.\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{9}{4}\\x< -3\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{9}{4}\\x>-3\end{matrix}\right.\end{matrix}\right.\)
TH1: loại
TH2: TM
Vậy no của BPT là :\(-\dfrac{9}{4}\ge x>-3\)
chúc bạn học tốt
a, Ta có\(\left(x+3\right)^2+3\left(x-1\right)\ge x^2-4\)
\(\Leftrightarrow x^2+6x+9+3x-3\ge x^2-4\)
\(\Leftrightarrow x^2+9x+6\ge x^2-4\)
\(\Leftrightarrow9x+10\ge0\Leftrightarrow x\ge-\frac{10}{9}\)
\(\left(x+3\right)^2+3\left(x-1\right)\ge x^2-4\)
\(\Leftrightarrow x^2+6x+9+3x-3\ge x^2-4\)
\(\Leftrightarrow x^2+6x+3x-x^2\ge-4-9+3\)
\(\Leftrightarrow9x\ge-10\)
\(\Leftrightarrow x\ge-\frac{10}{9}\)
\(1.A=x^2+3x-1=-\left(x^2-2.x.\frac{3}{2}+\frac{3}{2}^2-\frac{5}{4}\right)\)
\(A=-\left(x-\frac{3}{2}\right)^2+\frac{5}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0,x\in R\)
do đó \(-\left(x-\frac{3}{2}\right)^2\le0,x\in R\)
nên \(-\left(x-\frac{3}{2}\right)^2+\frac{5}{4}\le\frac{5}{4},x\in R\)
Vậy \(Max_A=\frac{5}{4},x=\frac{3}{2}\)
a)\(\left|x-2\right|\ge1\)
* x-2 \(\ge\)0 \(\Rightarrow\)x\(\ge\)2
x-2\(\ge\)1 \(\Leftrightarrow\)x\(\ge\)3 ( t/m )
*x-2<0\(\Rightarrow x< 2\)
-x+2 \(\ge1\)\(\Leftrightarrow\) -x\(\ge\)-1 \(\Leftrightarrow x\le1\)(t/m)
Vây bpt co nghiem la x\(\ge\)3;x\(\le1\)
b)\(\left|2-x\right|< 3\)
* \(2-x\ge0\Rightarrow x\le2\)
\(2-x< 3\Leftrightarrow-x< 1\Leftrightarrow x>-1\)(t/m)
*\(2-x< 0\Leftrightarrow-x< -2\Rightarrow x>2\)
\(-2+x< 3\Leftrightarrow x< 5\)(t/m)
Các ý còn lại tương tự nhé 
\(\left(8x-4x^2-1\right)\left(x^2+2x+1\right)=4\left(x^2+x+1\right)\)
\(\Leftrightarrow8x^3+16x^2+8x-4x^4-8x^3-4x^2-x^2-2x-1=4x^2+4x+4\)
\(\Leftrightarrow11x^2+6x-4x^4-1=4x^2+4x+4\)
\(\Leftrightarrow11x^2+6x-4x^2-1-4x^2-4x-4=0\)
\(\Leftrightarrow7x^2+2x-4x^4-4=0\)
\(\Leftrightarrow\left(-4x^3-4x^2+3x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(-4x^2-8x-5\right)\left(x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
câu a là nhân 2 hay mũ 2
* la the nao