\(\left(4x-1\right)\cdot\sqrt{x^3+1}=2x^3+2x+1\)
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15 tháng 6 2018
1> 3x(x-2)-2x(2x-1)=(1-x)(1+x)
⇔\(3x^2\)-6x-\(4x^2\)+2x=1-\(x^2\)
⇔-1\(x^2\) - 4x= 1- \(x^2\)
⇔ -1\(x^2\) -4x+ \(x^2\) = 1
⇔-4x=1
⇔ x = \(\dfrac{-1}{4}\)
21 tháng 7 2017
1) \(\left(x+1\right)^3-\left(x-1\right)^3=6.\left(x+2\right)^2-9\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)=6\left(x^2+4x+4\right)-9\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+24x+24-9\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2-24x-24+9=0\)
\(\Leftrightarrow-24x-13=0\Leftrightarrow-24x=13\Leftrightarrow x=\dfrac{-13}{24}\) vậy \(x=\dfrac{-13}{24}\)
21 tháng 7 2017
2) \(\left(2x-1\right).\left(4x^2+2x+1\right)+\left(1-2x\right)^3=3.\left(2x+3\right)^2\)
\(\Leftrightarrow8x^3+4x^2+2x-4x^2-2x-1+1-6x+12x^2-8x^3=3\left(.4x^2+12x+9\right)\)
\(\Leftrightarrow8x^3+4x^2+2x-4x^2-2x-1+1-6x+12x^2-8x^3=12x^2+36x+27\)
\(\Leftrightarrow8x^3+4x^2+2x-4x^2-2x-1+1-6x+12x^2-8x^3-12x^2-36x-27=0\)
\(\Leftrightarrow-42x-27=0\Leftrightarrow-42x=27\Leftrightarrow x=\dfrac{-27}{42}\) vậy \(x=\dfrac{-27}{42}\)
22 tháng 5 2022
1: \(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(4x+3\right)^2+8\left(x+3\right)^2\)
\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)
\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)
=56
2: \(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)
\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)
\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)
\(=6\)
TN
13 tháng 7 2017
\(a,4x^2-\left(3x+1\right)\left(2x-1\right)=2\left(x-3\right)^2\)
\(\Leftrightarrow4x^2-\left(6x^2-3x+2x-1\right)=2\left(x^2-6x+9\right)\)
\(\Leftrightarrow4x^2-6x^2+x+1-2x^2+12x-18=0\)
\(\Leftrightarrow-4x^2+13x-17=0\)
\(\Leftrightarrow-4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)-\dfrac{103}{16}=0\)
\(\Leftrightarrow-4\left(x-\dfrac{13}{8}\right)^2=\dfrac{103}{16}\)
\(\Leftrightarrow\left(x-\dfrac{13}{8}\right)^2=\dfrac{-103}{64}\Rightarrow\) pt vô nghiệm
\(b,\left(5x-1\right)\left(x+1\right)-\left(2x-1\right)\left(2x+1\right)=x.\left(x+1\right)\)\(\Leftrightarrow5x^2+5x-x-1-\left(4x^2-1\right)=x^2+x\)
\(\Leftrightarrow5x^2+5x-x-1-4x^2+1-x^2-x=0\) \(\Leftrightarrow3x=0\Rightarrow x=0\)
\(c,7x^2-\left(2x-3\right)^2=1+3\left(x+2\right)^2\)
\(\Leftrightarrow7x^2-\left(4x^2-12x+9\right)=1+3\left(x^2+4x+4\right)\)
\(\Leftrightarrow7x^2-4x^2+12x-9=1+3x^2+12x+12\)\(\Leftrightarrow7x^2-4x^2+12x-9-1-3x^2-12x-12=0\)\(\Leftrightarrow-22=0\) ( vô lí)
Vậy phương trình vô nghiệm
22 tháng 5 2022
1: \(=x^3-6x^2+12x-8-8x^3-36x^2-54x-27+7\left(x-1\right)^3\)
\(=-7x^3-42x^2-42x-35+7x^3-21x^2+21x-7\)
\(=-63x^2-21x-42\)
2: \(=x^3+125-\left(x^3-8\right)=125+8=133\)
3: \(=8x^3-27-8x^3-12x^2-6x-1=-12x^2-6x-28\)
21 tháng 5 2022
Sửa đề: \(\left(2x^2+1\right)\left(3x^2-1\right)-\left(4x^2-3\right)\left(x^2+1\right)=x^2\left(2x^2+1\right)\)
\(\Leftrightarrow6x^4-2x^2+3x^2-1-\left(4x^4+4x^2-3x^2-3\right)=2x^4+x^2\)
\(\Leftrightarrow6x^4+x^2-1-4x^4-x^2+3-2x^4-x^2=0\)
\(\Leftrightarrow-x^2+2=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)