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9x - 7i > 3 . \((3x-7u)\)
=> 9x - 71 > 9x - 21u
=> -7i > -21u
=> 7i < 21u
=> i < 3u
9x - 7i > 3 (3x - 7u)
9x-7i>9x-21u
-7i>-21u
i<3u
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\(\left|x-7\right|+\left|3-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
Ta có: \(\left\{{}\begin{matrix}\left|x-7\right|+\left|3-x\right|\ge\left|x-7+3-x\right|=4\\\dfrac{12}{\left|y+1\right|+3}\le\dfrac{12}{3}=4\end{matrix}\right.\)
Mà theo đề bài: \(\left|x-7\right|+\left|3-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
\(\Rightarrow\left|x-7\right|+\left|3-x\right|=\dfrac{12}{\left|y+1\right|+3}=4\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}3\le x\le7\\y=-1\end{matrix}\right.\)
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a)\(M\left(x\right)=3x^4-x^3-2x^2+5x+7\)
\(N\left(x\right)=-3x^4+x^3+10x^2+x-7\)
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`@` `\text {Ans}`
`\downarrow`
`3x(x+3)`
`= 3x*x + 3x*3`
`= 3x^2 + 9x`
`=> 3x(x+3) = 3x^2 + 9x.`