(x + y)3 = 12x + 3
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\(A=3x^2y^3-5x^2+3x^3y^2\)
bậc 5, hệ số 3
bạn xem lại đề B nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(12x^4y^3+12x^3y^3+3x^2y^3\)
\(=3x^2y^3\cdot4x^2+3x^2y^3\cdot4x+3x^2y^3\cdot1\)
\(=3x^2y^3\left(4x^2+4x+1\right)\)
\(=3x^2y^3\left(2x+1\right)^2\)
b: \(x^4+xy^3-x^3y-y^4\)
\(=\left(x^4+xy^3\right)-\left(x^3y+y^4\right)\)
\(=x\left(x^3+y^3\right)-y\left(x^3+y^3\right)\)
\(=\left(x^3+y^3\right)\left(x-y\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2-xy+y^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(2\left(x-y\right)+\left(x-y\right)^2+\left(y-x\right)^2\)
\(=2\left(-3-1000\right)+\left(-3-1000\right)^2+\left(3+1000\right)^2\)
\(=-2006+1006009+1006009\)
\(=2010012\)
2) \(x^3+12x^2+48x+64\)
\(=x^3+3.x^2.4+3.x.4^2+4^3\)
\(=\left(x+4\right)^3=\left(6+4\right)^3=10^3=1000\)
3) \(x^3-6x^2+12x-8\)
\(=x^3-3.x^2.2+3.x.2^2-2^3\)
\(=\left(x-2\right)^3=\left(22-2\right)^3=20^3=8000\)
\(2\left(x-y\right)+\left(x-y\right)^2+\left(y-x\right)^2\)
=\(2\left(x-y\right)+\left(x-y+y-x\right)\left(x-y-\left(y-x\right)\right)\)
= \(2\left(x-y\right)+\left(x-y+y-x\right)\left(x-y-y+x\right)\)
= \(2\left(x-y\right)\)
Thay x = -3,y = 1000 vào ta có : 2(x - y) = 2(-3 - 1000) = 2.(-1003) = -2006
\(x^3+12x^2+48x+64\)
\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3=\left(x+4\right)^3\)
Thay x = 6 vào ta có : (6 + 4)3 = 103 = 10000
\(x^3-6x^2+12x-8=x^3-3x^2\cdot2+3x\cdot2^2-2^3\)
\(=\left(x-2\right)^3\)
Thay x = 22 vào ta có : (22 - 2)3 = 203 = 8000
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(x^3+12x^2+48x+64\)
\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3\)
\(=\left(x+4\right)^3\)
b) Ta có: \(x^3-12x^2+48x-64\)
\(=x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3\)
\(=\left(x-4\right)^3\)
c) Ta có: \(8x^3+12x^2y+6xy^2+y^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)
\(=\left(2x+y\right)^3\)
d)Sửa đề: \(x^3-3x^2+3x-1\)
Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)
\(=\left(x-1\right)^3\)
e) Ta có: \(8-12x+6x^2-x^3\)
\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)
\(=\left(2-x\right)^3\)
f) Ta có: \(-27y^3+9y^2-y+\frac{1}{27}\)
\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)
\(=\left(\frac{1}{3}-3y\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(12x^3-12x^2+3x\)
\(=12x^3-9x+12x-12x^2\)
\(=3x.\left(4x^2-3\right)+3x.\left(4-4x^2\right)\)
\(=3x.\left(4x^2-3+4-4x^2\right)\)
\(=3x.\left(-1\right)=-3x\)
p/s: ko chắc =]
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 4:
\(x^3-2x^2+x=x\left(x-1\right)^2\)
\(5\left(x-y\right)-y\left(x-y\right)=\left(x-y\right)\left(5-y\right)\)
\(x^2-12x+36=\left(x-6\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{12x-15y+20z-12x+15y-20z}{7+9+11}=0\)
⇒\(\left\{{}\begin{matrix}12x=15y\\20z=12x\\15y=20z\end{matrix}\right.\)
⇔\(12x=15y=20z\)⇒\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{x+y+z}{5+4+3}=\dfrac{48}{12}=4\)
⇒\(\left\{{}\begin{matrix}x=5.4=20\\y=4.4=16\\z=3.4=12\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(x+1\right)^2+\left(x+3\right)\left(x-2\right)-4x\)
\(=x^2+2x+1+x^2+x-6-4x\)
\(=2x^2-x-6\)
\(\left(x+y\right)^3=12x+3\)lẻ\(\Rightarrow x+y\)lẻ
\(\left(x+y\right)^3=12x+3< 12x+12y\Rightarrow\left(x+y\right)^2< 12< 4^2\)
\(\Rightarrow x+y< 4\). Do đó: x+y=3
Ta có: 12x+3=27=> x=2 nên y=1