\(^{\left(x+2\right)^2}-3x+1< =x\left(x-1\right)\)
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a)\(\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow6x=36\Leftrightarrow x=6\)
A = \(\dfrac{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)+1}{x^2+5x+5}=\dfrac{\left(x^2+5x+5\right)^2}{x^2+5x+5}=x^2+5x+5\)B = \(\dfrac{\left|x-1\right|+\left|x\right|+x}{3x^2-4x+1}\)với x < 0
Với x < 0 thì |x-1| = 1-x, |x| = -x, ta có:
\(\dfrac{1-x-x+x}{\left(x-1\right)\left(3x-1\right)}=\dfrac{1-x}{\left(x-1\right)\left(3x-1\right)}=\dfrac{x-1}{\left(x-1\right)\left(1-3x\right)}=\dfrac{1}{1-3x}\)
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\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)
\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)
\(\Leftrightarrow4x+4x>-1\)
\(\Leftrightarrow8x>-1\)
\(\Leftrightarrow x>-\frac{1}{8}\)
\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)
\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-6x^2< 1+3\)
\(\Leftrightarrow-2x^2< 4\)
\(\Leftrightarrow x^2>2\)
\(\Leftrightarrow x>\pm\sqrt{2}\)
a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
\(\left(x+2\right)^2-3x+1\le x\left(x-1\right)\)
\(\Leftrightarrow x^2+4x+4-3x+1\le x^2-x\)
\(\Leftrightarrow x^2+x+5-x^2+x\le0\)
\(\Leftrightarrow2x+5\le0\)
\(\Leftrightarrow x\le\frac{-5}{2}\)
( x + 2 )2 - 3x + 1 < x ( x - 1 )
= x2 + 4x + 4 - 3x + 1 < x2 - x
= x2 + x + 5 - x2 + x < - 0
= 2x + 5 < 0
= x < -5/2