( x+4) ( x2 -4x-4x+16)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Thực hiện rút gọn VT = -2x – 64
Giải phương trình -2x – 64 = 0 thu được x = -32.
b) Thực hiện rút gọn VT = -62 x +12
Giải phương trình -62x + 12 = -50 thu được x = 1.
![](https://rs.olm.vn/images/avt/0.png?1311)
b) (x – 4). (x2 + 4x + 16) – x. (x2 - 6) = 2
⇔ x3 + 4x2 + 16x – 4x2 – 16x – 64 – (x3 - 6x ) – 2= 0
⇔ x3 + 4x2 + 16x – 4x2 – 16x – 64 – x3 + 6x – 2= 0
⇔ 6x – 66 =0
⇔ 6x = 66
⇔ x = 66 : 6
⇔ x = 11
Vậy x = 11
![](https://rs.olm.vn/images/avt/0.png?1311)
1)(x2-4x+16)(x+4)-x(x+1)(x+2)+3x2=0
\(\Rightarrow\)(x3+64)-x(x2+2x+x+2)+3x2=0
\(\Rightarrow\)x3+64-x3-2x2-x2-2x+3x2=0
\(\Rightarrow\)-2x+64=0
\(\Rightarrow\)-2x=-64
\(\Rightarrow\)x=\(\dfrac{-64}{-2}\)
\(\Rightarrow x=32\)
2)(8x+2)(1-3x)+(6x-1)(4x-10)=-50
\(\Rightarrow\)8x-24x2+2-6x+24x2-60x-4x+10=50
\(\Rightarrow\)-62x+12=50
\(\Rightarrow\)-62x=50-12
\(\Rightarrow\)-62x=38
\(\Rightarrow\)x=\(-\dfrac{38}{62}=-\dfrac{19}{31}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x-3\right)\left(x+3\right)=8\)
\(\Rightarrow x^3-1-x\left(x^2-9\right)=8\)
\(\Rightarrow x^3-1-x^3+9x=8\)
\(\Rightarrow9x=9\Rightarrow x=1\)
c) \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)\left(x^2+4x+4\right)=-16\)
\(\Rightarrow x^3-4x^2+2x-8-\left(x+2\right)\left(x+2\right)^2=-16\)
\(\Rightarrow x^3-4x^2+2x-8-\left(x+2\right)^3=-16\)
\(\Rightarrow x^3-4x^2+2x-8-\left(x^3+6x^2+12x+8\right)=-16\)
\(\Rightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)
\(\Rightarrow-10x^2-10x-16=-16\)
\(\Rightarrow10x^2+10x=0\)
\(\Rightarrow10x\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải pt à bạn:P?
\(\left(x+4\right)\left(x^2-4x+16\right)-\left(x-2\right)^3=0\)
\(\Leftrightarrow x^3+4^3-\left(x^3-8-6x^2+12x\right)=0\)
\(\Leftrightarrow x^3+4^3-x^3+8+6x^2-12x=0\)
\(\Leftrightarrow72+6x^2-12x=0\Leftrightarrow6\left(x^2-2x+12\right)=0\Leftrightarrow x^2-2x+12=0\)
Ta lại có: \(x^2-2x+12=x^2-2x+1+11=\left(x-1\right)^2+11\ge11>0\ne0\)
=> Pt vô nghiệm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+4\right)\left(x^2-4x+16\right)=x^3+64\)
\(\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x 2 – x – 12 b) x 3 – 64.
c) m 3 n 3 – m 2 n + 5 mn 2 – 5 d) 16 x 4 – 1.
(x+4)(x2 -4x-4x+16)
=(x+4)(x2 -8x+16)
=(x-8x)(4+16)(x2)
=-7x.20.x2
=>sap xep x2.(-7x).20
chuc ban hoc tot