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a, \(\left(x-3\right)\left(x+2\right)>0\)
th1 : \(\hept{\begin{cases}x-3>0\\x+2>0\end{cases}\Rightarrow\hept{\begin{cases}x>3\\x>-2\end{cases}\Rightarrow}x>3}\)
th2 : \(\hept{\begin{cases}x-3< 0\\x+2< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 3\\x< -3\end{cases}\Rightarrow}x< -3}\)
vậy x > 3 hoặc x < -3
b, \(\left(x+5\right)\left(x+1\right)< 0\)
th1 : \(\hept{\begin{cases}x+5>0\\x+1< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-5\\x< -1\end{cases}\Rightarrow x\in\left\{-4;-3;-2\right\}}}\)
th2 : \(\hept{\begin{cases}x+5< 0\\x+1>0\end{cases}\Rightarrow\hept{\begin{cases}x< -5\\x>-1\end{cases}\Rightarrow}x\in\varnothing}\)
vậy x = -4; -3; -2
c, \(\frac{x-4}{x+6}\le0\)
xét \(\frac{x-4}{x+6}=0\)
\(\Rightarrow x-4=0;x\ne-6\)
\(\Rightarrow x=4\ne-6\)
xét \(\frac{x-4}{x+5}< 0\)
th1 : \(\hept{\begin{cases}x-4< 0\\x+5>0\end{cases}\Rightarrow\hept{\begin{cases}x< 4\\x>-5\end{cases}\Rightarrow}x\in\left\{3;2;1;0;-1;-2;-3;-4\right\}}\)
th2 : \(\hept{\begin{cases}x-4>0\\x+5< 0\end{cases}\Rightarrow\hept{\begin{cases}x>4\\x< -5\end{cases}\Rightarrow x\in\varnothing}}\)
d tương tự c
\(\frac{\left(x-6\right)}{x-7}\ge0\)
Th1: x - 6 < 0
<=> x - 6 + 6 < 0 + 6
<=> x - 6 + 6 > 0 + 6
=> x < 6
Th2: x - 7
<=> x - 7 + 7 < 0 + 7
<=> x - 7 + 7 > 0 + 7
=> x > 7
=> x < 6 hoặc x > 7
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a)\(\frac{x+1}{5}+\frac{x+3}{4}=\frac{x+5}{3}+\frac{x+7}{2}\)
\(\Leftrightarrow\frac{12\left(x+1\right)}{60}+\frac{15\left(x+3\right)}{60}=\frac{20\left(x+5\right)}{60}+\frac{30\left(x+7\right)}{60}\)
\(\Leftrightarrow12x+12+15x+45=20x+100+30x+210\)
\(\Leftrightarrow27x+57=50x+310\)
\(\Leftrightarrow27x+57-50x-310=0\)
\(\Leftrightarrow-23x-253=0\)
\(\Leftrightarrow x=-\frac{253}{23}\)
b)Tự làm
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1) \(\left|x\right|< 4\Leftrightarrow-4< x< 4\)
2) \(\left|x+21\right|>7\Leftrightarrow\orbr{\begin{cases}x+21>7\\x+21< -7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-14\\x< -28\end{cases}}\)
3) \(\left|x-1\right|< 3\Leftrightarrow-3< x-1< 3\Leftrightarrow-2< x< 4\)
4) \(\left|x+1\right|>2\Leftrightarrow\orbr{\begin{cases}x+1>2\\x+1< -2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -3\end{cases}}\)
\(\left|x+\frac{1}{2}\right|+\left|3-y\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|3-y\right|\ge0\end{cases}}\Rightarrow\)\(\left|x+\frac{1}{2}\right|+\left|3-y\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|3-y\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=3\end{cases}}\)
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Ta có : \(\frac{x+1}{x-4}>0\)
Thì sảy ra 2 trường hợp
Th1 : x + 1 > 0 và x - 4 > 0 => x > -1 ; x > 4
Vậy x > 4
Th2 : x + 1 < 0 và x - 4 < 0 => x < -1 ; x < 4
Vậy x < (-1) .
Ta có : \(\left(x+2\right)\left(x-3\right)< 0\)
Th1 : \(\hept{\begin{cases}x+2< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>3\end{cases}}\left(\text{Vô lý }\right)}\)
Th2 : \(\hept{\begin{cases}x+2>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x< 3\end{cases}\Rightarrow}-2< x< 3}\)
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a ) \(\left(\frac{2}{5}-x\right):1\frac{1}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}+\frac{1}{2}=-4\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-4-\frac{1}{2}\)
\(\left(\frac{2}{5}-x\right):\frac{4}{3}=-\frac{9}{2}\)
\(\frac{2}{5}-x=-\frac{9}{2}.\frac{4}{3}\)
\(\frac{2}{5}-x=-3\)
\(x=\frac{2}{5}-\left(-3\right)\)
\(x=\frac{2}{5}+3\)
\(x=\frac{3}{5}-\frac{15}{5}\)
\(x=-\frac{12}{5}\)
Vay \(x=-\frac{12}{5}\)
b ) \(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(1+\frac{2}{5}+\frac{2}{3}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15}{15}+\frac{6}{15}+\frac{10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\left(\frac{15+6+10}{15}\right)=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right):\frac{31}{15}=-\frac{5}{4}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{5}{4}.\frac{31}{15}\)
\(\left(-3+\frac{3}{x}-\frac{1}{3}\right)=-\frac{1}{4}.\frac{31}{3}\)
\(-3+\frac{3}{x}-\frac{1}{3}=-\frac{31}{12}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{1}{2}\)
\(-3+\frac{3}{x}=-\frac{31}{12}+\frac{6}{12}\)
\(-3+\frac{3}{x}=\frac{-25}{12}\)
\(\frac{3}{x}=\frac{-25}{12}+3\)
\(\frac{3}{x}=\frac{-25}{12}+\frac{36}{12}\)
\(\frac{3}{x}=\frac{5}{6}\)
\(\frac{18}{6x}=\frac{5x}{6x}\)
Đèn dây , bạn tự làm tiếp nhé , de rồi chứ
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\orbr{\begin{cases}x=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
b)\(\orbr{\begin{cases}3x=0\\2x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
c)\(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}}\)
d)\(\orbr{\begin{cases}x^2\\x+4=0\end{cases}=0\Rightarrow\orbr{\begin{cases}x=0\\x=-4\end{cases}}}\)
e)\(\orbr{\begin{cases}\left(x+1\right)^2\\3x-5=0\end{cases}=0}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)
g)\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varphi\)
h)Tương tự các câu trên
i) x = 0
k)\(\left(\frac{3}{4}\right)^x=1=\left(\frac{3}{4}\right)^0\Rightarrow x=0\)
l)\(\left(\frac{2}{5}\right)^{x+1}=\frac{8}{125}=\left(\frac{2}{5}\right)^3\)
=> x + 1 = 3 => x = 2
x.(x+1)=0
suy ra x=0 hoac x+1=0
x=0-1
x=-1
vay x=0 hoac x=-1
mấy câu sau cũng làm tương tự