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x^2+4x>0
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gọi 2+2căn 3 là a đi cho đỡ phải gõ nhiều
ủa? bạn xem lại đề đi: x+1/x hay x^2+1/x^2 đây???
`x - 1 / 2 = [-3] / 4 : 3 / 2`
`x - 1 / 2 = [-3] / 4 . 2 / 3`
`x - 1 / 2 = [-1] / 2`
`x = [-1] / 2 + 1 / 2`
`x = 0`
x - 4,6 = 8,3 - 3 x - 1/2 = -2/4
x - 4,6 = 5,3 x = -2/4 + 1/2
x = 5,3 + 4,6 x = 0
x = 9,9
\(\text{#}Irumaa:\)\(3\)
\(x - 4,6 = 8,3 - 2 × 1,5\)
\( x - 4,6 = 8,3 - 3\)
\( x - 4,6 = 5,3\)
\( x = 5,3 + 4,6\)
\( x = 11,1\)
\(--------------------------\)
\(x-\dfrac{1}{2}=-\dfrac{3}{4}:\dfrac{3}{2}\)
\(x-\dfrac{1}{2}=-\dfrac{3}{4\text{ }}\text{×}\dfrac{2}{3}\)
\(x-\dfrac{1}{2}=-\dfrac{1}{2}\)
\(x=-\dfrac{1}{2}+\dfrac{1}{2}\)
\(x=0\)
x + y = 0000000000000000000000000000000000000000000000000000000000000000
a) \(\dfrac{x+6}{x^2-4}+\dfrac{1}{x+2}=\dfrac{x+6}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+6+x-2}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{2x+4}{\left(x+2\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
b) \(x^2+xy-5\left(x+y\right)=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\)
a) Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow x^2+5x-2x-10=0\)
\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy: S={-5;2}
b) Ta có: \(3x^2-7x+1=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)
c) Ta có: \(3x^2-7x+8=0\)
\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)
mà 3>0
nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)
\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)
Vậy: \(x\in\varnothing\)
\(x^2+4x>0\Leftrightarrow x\left(x+4\right)>0\)
\(\Leftrightarrow\orbr{\begin{cases}x>0\\x+4>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x>0\\x>-4\end{cases}\Leftrightarrow}x>0}\)
\(\Leftrightarrow\orbr{\begin{cases}x< 0\\x+4< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 0\\x< -4\end{cases}\Leftrightarrow}x< -4}\)
vậy...........
Bài làm:
Ta có: \(x^2+4x>0\)
\(\Leftrightarrow x\left(x+4\right)>0\)
Ta thấy \(x< x+4\) nên => \(\orbr{\begin{cases}x>0\\x+4< 0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>0\\x< -4\end{cases}}\)
Vậy \(x>0\) hoặc \(x< -4\)