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1) (x + 1)2 + (x - 1)(x2 + x + 1) + (x - 1)3
= x2 + 2x + 1 + x3 - 1 + x3 - 3x2 + 3x - 1
= 2x3 - 2x2 + 5x + 1
2) (x - 2)2 + (2x + 1)2 + (x + 1)3
= x2 - 4x + 4 + 4x2 + 4x + 1 + x3 + 3x2 + 3x + 1
= x3 + 8x2 + 3x + 6
3) (x + 1)(x2 - x + 1) - (x - 3)2
= x3 + 1 - x2 + 6x - 9
= x3 - x2 + 6x - 8
4) (3x + 2)2 + (2x - 1)2 - (x + 3)2
= 9x2 + 12x + 4 + 4x2 - 4x + 1 - x2 - 6x - 9
= 12x2 + 2x - 4
a) Ta có: \(\left(x-2\right).\left(x^2+2x+4\right)+\left(x-2\right)^3-\left(x-2\right).\left(x+2\right)\)
\(=\left(x^3-8\right)+\left(x-2\right)^3-\left(x^2-4\right)\)
\(=x^3-8+x^3-6x^2+12x-8-x^2+4\)
\(=2x^3-7x^2+12x-12\)
b) Ta có: \(\left(3-2x\right)^2-\left(x+3\right)^2-\left(2x+1\right)\left(2x-1\right)\)
\(=9-12x+4x^2-x^2-6x-9-4x^2+1\)
\(=3x^2-18x+1\)
1) đặt 2x+1 = a => \(a^4-3a^2+2=\left(a^2-1\right)\left(a^2-2\right)=\)\(\left(a-1\right)\left(a+1\right)\left(a-\sqrt{2}\right)\left(a+\sqrt{2}\right)\)
=(2x+1-1)(2x+1+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\)) = 4x(x+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\))
2) =(x2-x)(x2-x-2)-3
đặt x2-x = b => b(b-2)-3 = b2-2b-3 = (b+1)(b-3) = (x2-x+1)(x2-x-3)
3) đặt x2+2x-1 = c => c2-3xc+2x2 = (c-x)(c-2x) = (x2+2x-1-x)(x2+2x-1-2x) = (x2+x-1)(x2-1) = (x2+x-1)(x-1)(x+1)
tìm x
x3-8 +(x-2)(x+1)=0 <=> (x-2)(x2+2x+4)+(x-2)(x+1)=0 <=>(x-2)(x2+2x+4+x+1)=0 <=> x=2 (vì x2+3x+5= (x+\(\frac{3}{2}\))2 +\(\frac{11}{4}\)>0)
vậy x=2
2) \(x\left(x-1\right)\left(x+1\right)\left(x-2\right)-3\)
\(=\left(x^2-x\right)\left(x^2-x-2\right)-3\)(1)
Đặt \(x^2-x=t\)
\(\Rightarrow\left(1\right)=t\left(t-2\right)-3=t^2-2t+1-4\)
\(=\left(t-1\right)^2-4\)
\(=\left(t+3\right)\left(t-5\right)\)
Thay \(x^2-x=t\), ta được:
\(BTDNT=\left(x^2-x+3\right)\left(x^2-x-5\right)\)
\(d,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=15\\ \Leftrightarrow24x=-10\Leftrightarrow x=-\dfrac{5}{12}\\ e,\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=10\Leftrightarrow x=\dfrac{10}{9}\\ f,\Leftrightarrow9x^2+18x+9-18x=36+x^3-27\\ \Leftrightarrow x^3-9x^2=0\Leftrightarrow x^2\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\)
a) Ta có :
\(3x=3\left(x+2\right)\)
\(\Leftrightarrow3x=3x+2\)
\(\Leftrightarrow0=2\) ( vô lí )
Do đó pt đã cho vô nghiệm
b) Ta có \(\left|x\right|=-x^2-2\) (1)
Nhân xét : VT (1) : \(\left|x\right|\ge0\forall x\)
VP (1) : \(-x^2\le0\Leftrightarrow-x^2-2\le-2\forall x\)
Do đó : \(VT\ne VP\)
Vì vậy pt đã cho vô nghiệm