[(x^2)-1].[(x^2-4)].[(x^2-7)].[(x^2-10)]<10
Tìm X
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\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a, -5/7+ 1+ 30/-7< x < -1/6+ 1/3 +5/6
<=> -4< x <1
<=> x = -3; -2; -1; 0
a, \(\dfrac{-5}{7}+1+\dfrac{30}{-7}\le x\le\dfrac{-1}{6}+\dfrac{1}{3}+\dfrac{5}{6}\)
<=> -4 \(\le x\le1\)
Do x \(\in Z\Rightarrow x=-4;-3;-2;-1;0;1\)
b, \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< x< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)
<=> -\(\dfrac{1}{12}< x< \dfrac{1}{8}\)
Do x \(\in Z\Rightarrow x=0;1\)
@Mai Tran