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a: (x-3)(x-2)<0
=>x-2>0 và x-3<0
=>2<x<3
b: \(\left(x+3\right)\left(x+4\right)\left(x^2+2\right)\ge0\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\ge0\)
=>x>=-3 hoặc x<=-4
c: \(\dfrac{x-1}{x-2}\ge0\)
nên \(\left[{}\begin{matrix}x-2>0\\x-1\le0\end{matrix}\right.\Leftrightarrow x\in(-\infty;1]\cup\left(2;+\infty\right)\)
d: \(\dfrac{x+3}{2-x}\ge0\)
\(\Leftrightarrow\dfrac{x+3}{x-2}\le0\)
hay \(x\in[-3;2)\)
a)
\(4x-10< 0\\ 4x< 10\\ x< \dfrac{10}{4}=\dfrac{5}{2}\)
b)
\(2x+x+12\ge0\\ 3x\ge-12\\ x\ge-\dfrac{12}{3}=-4\)
c)
\(x-5\ge3-x\\ 2x\ge8\\ x\ge4\)
d)
\(7-3x>9-x\\ -2>2x\\ x< -1\)
đ)
\(2x-\left(3-5x\right)\le4\left(x+3\right)\\ 2x-3+5x\le4x+12\\ 3x\le15\\ x\le5\)
e)
\(3x-6+x< 9-x\\ 5x< 15\\ x< 3\)
f)
\(2t-3+5t\ge4t+12\\ 3t\ge15\\ t\ge5\)
g)
\(3y-2\le2y-3\\ y\le-1\)
h)
\(3-4x+24+6x\ge x+27+3x\\ 0\ge2x\\ 0\ge x\)
i)
\(5-\left(6-x\right)\le4\left(3-2x\right)\\ 5-6+x\le12-8x\\ \\ 9x\le13\\ x\le\dfrac{13}{9}\)
k)
\(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\\ 10x-15-20x+28\ge19-2x-22\\ 13-10x\ge-2x-3\\ -8x\ge-16\\ x\le\dfrac{-16}{-8}=2\)
l)
\(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\\ \dfrac{40x-100}{60}-\dfrac{90x-30}{2}< \dfrac{36-12x}{60}-\dfrac{30x-15}{60}\\ \Rightarrow40x-100-90x+30< 36-12x-30x+15\\ 130-50x< 51-42x\\ 92x< -79\\ x< -\dfrac{79}{92}\)
m)
\(5x-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+x\\ \dfrac{10x}{2}-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+\dfrac{2x}{2}\\ \Rightarrow10x-3+2x>7x-5+2x\\ 12x-3>9x-5\\ 3x>-2\\ x>-\dfrac{2}{3}\)
n)
\(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\\ \dfrac{28x-8}{12}-\dfrac{24x}{12}< \dfrac{60}{12}-\dfrac{3x-6}{12}\\ \Rightarrow28x-8-24x< 60-3x+6\\ 4x-8< -3x+66\\ 7x< 74\\ x< \dfrac{74}{7}\)
a) \(4x-10< 0\)
\(\Leftrightarrow4x< 10\)
\(\Leftrightarrow x< \dfrac{5}{2}\)
b) ???
c) \(x-5\ge3-x\)
\(\Leftrightarrow2x-5\ge3\)
\(\Leftrightarrow2x\ge8\)
\(\Leftrightarrow x\ge4\)
d) \(7-3x>9-x\)
\(\Leftrightarrow7-2x>9\)
\(\Leftrightarrow-2x>2\)
\(\Leftrightarrow x< -1\)
đ) ???
e) \(3x-6+x< 9-x\)
\(\Leftrightarrow4x-6< 9-x\)
\(\Leftrightarrow5x-6< 9\)
\(\Leftrightarrow5x< 15\)
\(\Leftrightarrow x< 3\)
f) ???
g) ???
h) \(3-4x+24+6x\ge x+27+3x\)
\(\Leftrightarrow2x+27\ge4x+27\)
\(\Leftrightarrow-2x\ge0\)
\(\Leftrightarrow x\le0\)
i) \(5-\left(6-x\right)\le4\left(3-2x\right)\)
\(\Leftrightarrow5-6+x\le12-8x\)
\(\Leftrightarrow x-1\le12-8x\)
\(\Leftrightarrow9x-1\le12\)
\(\Leftrightarrow9x\le13\)
\(\Leftrightarrow x\le\dfrac{13}{9}\)
k) \(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\)
\(\Leftrightarrow10x-15-20x+28\ge19-2x-22\)
\(\Leftrightarrow-10x+23\ge-3-2x\)
\(\Leftrightarrow-8x+13\ge-3\)
\(\Leftrightarrow-8x\ge-16\)
\(\Leftrightarrow x\ge2\)
l) \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\)
\(\Leftrightarrow-\dfrac{5}{6}x-\dfrac{7}{6}< -\dfrac{7}{10}x+\dfrac{17}{20}\)
\(\Leftrightarrow-\dfrac{2}{15}x-\dfrac{7}{6}< \dfrac{17}{20}\)
\(\Leftrightarrow-\dfrac{2}{15}x< \dfrac{121}{60}\)
\(\Leftrightarrow x>-\dfrac{121}{8}\)
m, n) làm tương tự:
đáp án: m. \(x>-\dfrac{2}{3}\); n. \(x< \dfrac{74}{7}\)
Mk thấy mấy cái này dễ mà, toàn trong sách giáo khoa hết á. Bạn cố gắng đọc và lm đi. Sắp lên lớp 9 rồi đó 
a)\(\dfrac{2x^2+10}{1-x}\le0\Rightarrow1-x< 0\Leftrightarrow x>1\)
b) \(\dfrac{3x-4}{x+2}\ge4\Leftrightarrow\dfrac{3x-4}{x+2}-\dfrac{4\left(x+2\right)}{x+2}\ge0\Leftrightarrow\dfrac{-x-12}{x+2}\ge0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x-12\le0\\x+2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-12\\x< -2\end{matrix}\right.\Leftrightarrow-12\le x< -2}}\\\left\{{}\begin{matrix}-x-12\ge0\\x+2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-12\\x>-2\end{matrix}\right.\end{matrix}\right.\)\(S=\left\{x|-12\le x< -2\right\}\)
c) \(\dfrac{1}{x+4}\le\dfrac{1}{x-2}\Leftrightarrow\dfrac{6}{\left(x+4\right)\left(x-2\right)}\le0\Rightarrow\left(x+4\right)\left(x-2\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+4>0\\x-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-4\\x< 2\end{matrix}\right.\Leftrightarrow-4< x< 2}}\\\left\{{}\begin{matrix}x+4< 0\\x-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -4\\x>2\end{matrix}\right.\end{matrix}\right.\)
\(S=\left\{x|-4< x< 2\right\}\)
a. \(5x-10=0\)
\(\Leftrightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy \(S=\left\{2\right\}\)
b. \(\dfrac{2}{x+1}-\dfrac{3}{x-2}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
ĐKXĐ: \(x\ne-1;x\ne2\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)}{x+1\left(x-2\right)}-\dfrac{3\left(x+1\right)}{x-2\left(x+1\right)}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow2\left(x-2\right)-3\left(x+1\right)=2x-6\)
\(\Leftrightarrow2x-4-3x-3=2x-6\)
\(\Leftrightarrow2x-4-3x-3-2x+6=0\)
\(\Leftrightarrow-3x-1=0\)
\(\Leftrightarrow-3x=1\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy \(S=\left\{-\dfrac{1}{3}\right\}\)
c) \(3x-5\ge-7\) (3)
\(\Leftrightarrow3x\ge-7+5\)
\(\Leftrightarrow3x\ge-2\)
\(\Leftrightarrow x\ge-\dfrac{2}{3}\)
Vậy tập nghiệm của BPT (3) là \(x\ge-\dfrac{2}{3}\)
d) \(3x-1=0\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy \(S=\left\{\dfrac{1}{3}\right\}\)
a. 5x - 10 = 0
⇔ 5x = 10
⇔ x = 2
Vậy S ={2}.
b.\(\dfrac{2}{x+1}\) - \(\dfrac{3}{x-2}\) = \(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇔\(\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\) - \(\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\) = \(\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
⇔ 2(x - 2) - 3(x+1) = 2x - 6
⇔ 2x - 4 - 3x -3 = 2x - 6
⇔ 2x - 2x - 3x = -6 + 4 + 3
⇔ -3x = 1
⇔ x = \(-\dfrac{1}{3}\)
a) \(\left|x-1\right|+\left|x-2\right|>x+3\)
ta có các trường hợp
trường hợp 1:\(\left|x-1\right|< 0\Leftrightarrow\left|x-2\right|< 0\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=-x+1\\\left|x-2\right|=-x+2\end{matrix}\right.\Leftrightarrow x< 1\)
trường hợp 2: \(\left|x-1\right|\ge0và\left|x-2\right|< 0\Leftrightarrow\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|x-2\right|=-x+2\end{matrix}\right.\Leftrightarrow1\le x< 2\)
trường hợp 3:\(\left|x-2\right|\ge0\Leftrightarrow\left|x-1\right|>0\Leftrightarrow\left\{{}\begin{matrix}\left|x-2\right|=x-2\\\left|x-1\right|=x-1\end{matrix}\right.\Leftrightarrow x\ge2\)\(\) \(\)
xét trường hợp 1:ta có BPT:
\(-x+1-x+2>x+3\Leftrightarrow-x-x-x>-1-2+3\\ \Leftrightarrow-3x>0\Leftrightarrow x< 0\)
vì điều kiện là x<1 nên mọi giá trị của x<0 đều thỏa mãn
trường hợp 2:
\(x-1-x+2>x+3\Leftrightarrow x-x-x>1-2+3\\ \Leftrightarrow-x>2\Leftrightarrow x< -2\)
vì điều kiện là \(1\le x< 2\) nên không có giá trị nào của x TM
trường hợp 3:
\(x-1+x-2>x+3\Leftrightarrow x+x-x>1+2+3\\ \Leftrightarrow x>6\)
vì điều kiện là x>=2 nên với mọi giá trị x>6 đều TM
Vậy nghiệm BPT là: x<0 hoặc x>6
c)
\(\left(x+5\right)\left(7-2x\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+5>0\\7-2x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-5\\-2x>-7\Leftrightarrow x< \dfrac{7}{2}\end{matrix}\right.\Leftrightarrow-5< x< \dfrac{7}{2}\\\left\{{}\begin{matrix}x+5< 0\\7-2x< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< -5\\-2x< -7\Leftrightarrow x>\dfrac{7}{2}\end{matrix}\right.\end{matrix}\right.\)
Vì trường hợp 2 không có giá trị nào của x TM nên ta loại
Vậy tập nghiệm của BPT là {x/5<x<7/2}
a)
\(\left(a\right)\Leftrightarrow\dfrac{x+1}{x-1}\le0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\x-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\x-1\ge0\end{matrix}\right.\end{matrix}\right.\)
(I) \(\Rightarrow\left\{{}\begin{matrix}x\ge-1\\x< 1\end{matrix}\right.\) \(\Rightarrow-1\le x< 1\)
(II)\(\Rightarrow\left\{{}\begin{matrix}x\le-1\\x>1\end{matrix}\right.\) vô nghiệm
Kết luận ;\(-1\le x< 1\)
\(\left(b\right)\Leftrightarrow\dfrac{2x+3}{5x-2}\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+3\ge0\\5x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+3\le0\\5x-2< 0\end{matrix}\right.\end{matrix}\right.\)
(I)\(\Rightarrow x\le-\dfrac{3}{2}\)
(II)\(\Rightarrow x>\dfrac{2}{5}\)
Kết luận nghiệm \(\left[{}\begin{matrix}x\le-\dfrac{3}{2}\\x>\dfrac{2}{5}\end{matrix}\right.\)
\(a,2x+7\ge0\Leftrightarrow2x\ge-7\Rightarrow x\ge\dfrac{-7}{2}\)
\(b,5-2x\le0\Leftrightarrow-2x\le-5\Leftrightarrow x\ge\dfrac{5}{2}\)
\(c,\dfrac{x+2}{x^2+1}\ge0\Leftrightarrow x+2\ge x^2+1\Leftrightarrow x+2-x^2-1\ge0\Leftrightarrow x-x^2+1\ge0\)\(\Leftrightarrow-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{5}{4}\ge0\Leftrightarrow-\left(x-\dfrac{1}{2}\right)^2\ge-\dfrac{5}{4}\Rightarrow\left(x-\dfrac{1}{2}\right)^2\ge\dfrac{5}{4}\)\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}\ge\sqrt{\dfrac{5}{4}}\\x-\dfrac{1}{2}\ge-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ge\sqrt{\dfrac{5}{4}}+\dfrac{1}{2}\\x\ge-\sqrt{\dfrac{5}{4}}+\dfrac{1}{2}\end{matrix}\right.\)
\(d,\dfrac{x^2+3}{2-x}< 0\Leftrightarrow x^2+3< 2-x\Leftrightarrow x^2+3-2+x\ge0\Leftrightarrow\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{3}{4}\ge0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2\ge\dfrac{-3}{4}\)( vô lí )
Vậy : BPT trên vô nghiệm