Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1a,P=\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right).\)
\(=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24=0\)
\(b,Q=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)
\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6\left(x^2-1\right)\)
\(=-6x^2-2+6x^2-6=-8\)
dấu <=> thứ 4 em làm nhầm rồi, 4x - 6x = - 2x chứ! Rồi tiếp theo em nên đưa về hằng đẳng thức chứ giải vậy ko đc đâu.
a) Mình không hiểu đề cho lắm 
b) \(3x\left(x-1\right)^2-2x\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)
\(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x\left(x-4\right)\)
\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)
\(=x^3-2x^2+5x\)
c) \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(2x+5\right)^2+3\left(4x+1\right)\left(4x-1\right)\)
\(=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)
\(=8x^2+40x+50+48x^2-3\)
\(=56x^2+40x+47\)
d) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
e) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)
\(=y^4-81-y^4+4\)
\(=-77\)
b) \(3x\left(x-1\right)^2-2x\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)
\(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x^2-16x=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)\(=x^3-2x^2+5x\)
c) \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)
\(=8x^2+40x+50+48x^2-3=56x^2+40x+47\)
d) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)=x\left(x^2-16\right)-x^4+1=x^3-x^4-16x+1\)
e) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)=\left(y^2-9\right)\left(y^2+9\right)-y^4+4=y^4-81-y^2+4=-77\)
b)Ta có:\(B=\left(0,5x^2+x\right)^2-3\left|0,5x^2+x\right|\)
\(B=\left|0,5x^2+x\right|^2-3\left|0,5x^2+x\right|+\dfrac{9}{4}-\dfrac{9}{4}\)
\(B=\left(\left|0,5x^2+x\right|-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\)
"="<=>\(\left|0,5x^2+x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
g)Ta có:\(G=\left(x^2+x-6\right)\left(x^2+x+2\right)\)
Đặt \(x^2+x-2=t\)
\(\Rightarrow G=\left(t-4\right)\left(t+4\right)\)
\(G=t^2-16\ge-16\)
"="<=>\(x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
E=\(x^4-6x^3+9x^2+x^2-6x+9\)
\(=x^2\left(x^2-6x+9\right)+x^2-6x+9\\ =x^2\left(x-3\right)^2+\left(x-3\right)^2\ge0\forall x\\ E_{min}=0\Leftrightarrow x=3\)
a: \(=\left(x^2-4\right)\left(x^2+4\right)-x^2+3\)
\(=x^4-16-x^2+3\)
\(=x^4-x^2-13\)
b: \(=x^3-6x^2+12x-8-x^3-1+6x^2-12x+6\)
\(=-3\)
c: \(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2-b^3-6a^2b\)
\(=2b^2\)