Là 2 con mắt

thiếu, làm tiếp nek

=>(x-3)(x³-x²-22x+40)=0

=>x-3=0
    x³-x²-22x+40=0

=>x=3
    x=4;x=-5
    

Đặt: \(x+2=t\)

\(pt\Leftrightarrow\left(t+1\right)^3-\left(t-1\right)^3=56\)

\(\Rightarrow\left(t^3+3t^2+3t+1\right)-\left(t^3-3t^23t-1\right)=56\)

\(\Rightarrow t^3+3t^2+3t+1-t^3+3t^2-3t+1=56\)

\(\Rightarrow\left(t^3-t^3\right)+\left(3t^2+3t^2\right)+\left(3t-3t\right)+\left(1+1\right)=56\)

\(\Rightarrow6t^2+2=56\Leftrightarrow6t^2=54\Leftrightarrow t^2=9\)

\(\Rightarrow\left[{}\begin{matrix}t=3\\t=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

\(A=\left|x\right|+3\ge3\)

Dấu "=" xảy ra khi: \(x=0\)

Bạn Trần Thùy Dung trả lời sai rồi nếu như thế thì sẽ ra kết quả là 2001 và dư 3

a) \(\left(x^2-5\right)\left(x^2-25\right)< 0\)

Vì \(x^2-5>x^2-25\) nên \(\left\{{}\begin{matrix}x^2-5>0\\x^2-25< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2>5\\x^2< 25\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{5}< x< -\sqrt{5}\left(vl\right)\\-5< x< 5\end{matrix}\right.\)

b) \(\left(x+5\right)\left(9+x^2\right)< 0\)

Vì \(9+x^2>0\) nên \(x+5< 0\Leftrightarrow x< -5\)

c) \(\left(x+3\right)\left(x^2+1\right)=0\)

Vì \(x^2+1>0\) nên \(x+3=0\Leftrightarrow x=-3\)

d) \(\left(x+5\right)\left(x^2-4\right)=0\)

\(\Rightarrow\left(x+5\right)\left(x+2\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-2\\x=2\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-28a=b\\35c=d\end{matrix}\right.\)

Vì \(b;d\in Z^-\) nên \(\left\{{}\begin{matrix}-28a< 0\\35c< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>0\\c< 0\end{matrix}\right.\)

Vậy \(a>c\)

\(3x-2⋮2x+1\)

\(\Rightarrow2\left(3x-2\right)⋮2x+1\)

\(\Rightarrow6x-4⋮2x+1\)

\(\Rightarrow6x+3-7⋮2x+1\)

\(\Rightarrow3\left(2x+1\right)-7⋮2x+1\)

\(\Rightarrow7⋮2x+1\)

\(\Rightarrow2x+1\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=1\\2x+1=-1\\2x+1=7\\2x+1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\\x=-4\end{matrix}\right.\)

\(xy-y-2x-2=5\)

\(\Rightarrow xy-y-2x+2=9\)

\(\Rightarrow y\left(x-1\right)-2\left(x-1\right)=9\)

\(\Rightarrow\left(y-2\right)\left(x-1\right)=9\)

Áp dụng bất đẳng thức Cauchy-Schwarz:

\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\left(đpcm\right)\)