(x2 - 36)(x2 - 81) ≤ 0
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\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-36\ge0\\x^2-81\le0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-36\le0\\x^2-81\ge0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow36\le x^2\le81\\ \Leftrightarrow-6\le x\le9\)
`#3107.101107`
\(\left(3^x-81\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3^x-81=0\\x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3^x=81\\x^2=-1\left(\text{vô lý}\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3^x=3^3\\x\in\varnothing\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x\in\varnothing\end{matrix}\right.\)
Vậy, `x = 3.`
\(x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(4x^2+4x+1-4x^2-12x-9=0\)
\(-8x-8=0\Leftrightarrow x=-1\)
\(\left(x-6\right)^2=0\)
\(x-6=0\Leftrightarrow x=6\)
c)\(x^2-12x=-36\)
\(x^2-12x+36=0\)
\(\left(x-6\right)^2=0\)
\(\Rightarrow x-6=0\)
........
\(\Leftrightarrow\left(x+6\right)^2\left(5-3x\right)\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x\ge\dfrac{5}{3}\end{matrix}\right.\)
Để pt: \(x^2-3x+m-2=0\) có hai nghiệm : \(x_1;x_2\) điều kiện là:
\(\Delta=9-4\left(m-2\right)\ge0\)
<=> \(m\le\frac{17}{4}\)( @@)
Áp dụng định lí viet ta có:
\(\hept{\begin{cases}x_1+x_2=3\\x_1.x_2=m-2\end{cases}}\)=> \(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=9-4\left(m-2\right)=17-4m\ge0\)
=> \(x_1-x_2=\sqrt{17-4m}\)
Ta có:
\(x_1^3-x_2^3+9x_1x_2=\left(x_1-x_2\right)^3+3\left(x_1-x_2\right)x_1x_2+9x_1x_2\)
\(=\sqrt{\left(17-4m\right)^3}+3\sqrt{17-4m}\left(m-2\right)+9\left(m-2\right)\)
Theo bài ra ta có phương trình:
\(\sqrt{\left(17-4m\right)^3}+3\sqrt{17-4m}\left(m-2\right)+9\left(m-2\right)=81\)
<=> \(\left(\sqrt{17-4m}\right)^3-3^3+3\left(m-2\right)\left(\sqrt{17-4m}-3\right)=0\)
<=> \(\left(\sqrt{17-4m}-3\right)\left(17-4m+3\sqrt{17-4m}+9+3\left(m-2\right)\right)=0\)
<=> \(\left(\sqrt{17-4m}-3\right)\left(20-m+3\sqrt{17-4m}\right)=0\)
TH1: \(\sqrt{17-4m}-3=0\Leftrightarrow17-4m=9\Leftrightarrow m=2\left(tm@@\right)\)
TH2: \(20-m+3\sqrt{17-4m}=0\)
<=> \(3\sqrt{17-4m}=m-20\)=> \(m-20\ge0\)=> \(m\ge20\) vô lí với (@@)
Vậy m = 2.