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Sửa đề: Cho \(x+y=a;x^2+y^2=b;x^3+y^3=c\)
Chứng minh: \(a^3-2ab+2c=0\)
Giải:
Ta có:
\(a^3-3ab+2c=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)
\(=x^3+y^3+3xy\left(x+y\right)-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)
\(=3\left(x^3+y^3\right)+3\left(x+y\right)\left(xy-x^2-y^2\right)=3\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(x+y\right)\left(xy-x^2-y^2\right)\)
\(=3\left(x+y\right)\left(x^2-xy+y^2+xy-x^2-y^2\right)=3\left(x+y\right).0\)
\(=0\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(a^3-3ab+2c=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)
\(=x^3+3x^2y+3xy^2+y^3-3\left(x^3+x^2y+xy^2+y^3\right)+2\left(x^3+y^3\right)\)
\(=x^3+3x^2y+3xy^2+y^3-3x^3-3xy^2-3x^2y-3y^3+2x^3+2y^3\)
\(=0\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^3=\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3\)
\(3ab=3\left(x+y\right)\left(x^2+y^2\right)=3\left(x^3+x^2y+xy^2+y^3\right)\)
\(2c=2x^3+2y^3\)
\(a^3-3ab+2c=\left(x^3+y^3-3x^2-3y^2+2x^3+2y^3\right)+3\left(x^2y-xy^2+xy^2-xy^2\right)=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a^3 - 3ab + 2c
= (x + y)^3 - 3(x + y)(x^2 + y^2) + 2(x^3 + x^3)
= x^3 + y^3 + 3xy(x + y) - 3(x + y)(x^2 + y^2) + 2(x^3 + y^3)
= [x^3 + y^3 + 2(x^3 + y^3)] + [3xy(x + y) - 3(x + y)(x^2 + y^2)]
= 3(x^3 + x^3) - 3(x + y)(x^2 - xy + y^2)
= 3(x^3 + x^3) - 3(x^3 + y^3)
= 0
a^3 - 3ab + 2c
= (x + y)^3 - 3(x + y)(x^2 + y^2) + 2(x^3 + x^3)
= x^3 + y^3 + 3xy(x + y) - 3(x + y)(x^2 + y^2) + 2(x^3 + y^3)
= [x^3 + y^3 + 2(x^3 + y^3)] + [3xy(x + y) - 3(x + y)(x^2 + y^2)]
= 3(x^3 + x^3) - 3(x + y)(x^2 - xy + y^2)
= 3(x^3 + x^3) - 3(x^3 + y^3)
= 0
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Dựa vào các hằng đẳng thức đáng nhớ ta có:
\(a^3-3ab+2c=(x+y)^3-3(x+y)(x^2+y^2)+2(x^3+y^3)\)
\(=x^3+y^3+3x^2y+3xy^2-3(x^3+xy^2+x^2y+y^3)+2(x^3+y^3)\)
\(=(x^3-3x^3+2x^3)+(y^3-3y^3+2y^3)+(3x^2y-3x^2y)+(3xy^2-3xy^2)\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^3-3ab+2c\)
\(=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)
\(=x^3+3x^2y+3xy^2+y^3-3x^3-3x^2y-3xy^2-3y^3+2x^3+2y^3\)
\(=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 2
Giải:x6+y6)-3(x4+y4)
2(x6+y6)−3(x4+y4)2(x6+y6)−3(x4+y4)
⇔2(x2+y2)(x4−x2y2+y4)−3x4−3y4⇔2(x2+y2)(x4−x2y2+y4)−3x4−3y4
⇔2(x4−x2y2+y4)−3x4−3y4⇔2(x4−x2y2+y4)−3x4−3y4
⇔2x4−2x2y2+2y4−3x4−3y4⇔2x4−2x2y2+2y4−3x4−3y4
⇔−2x2y2−x4−y4⇔−2x2y2−x4−y4
⇔−(x4+2x2y2+y4)⇔−(x4+2x2y2+y4)
⇔−(x2+y2)2⇔−(x2+y2)2
⇔−1
bài 1
bạn thay vào hết và tính ra là được
\(\Leftrightarrow\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)
\(\Leftrightarrow3x^3+3y^3+3xy\left(x+y\right)-3x^3-3y^3-3xy\left(x+y\right)=0\)(điều phải c/m)
\(a^3-3ab+2c=0\)
\(=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)\)
\(=\left(x+y\right)^3-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-3x^2-3y^2+2x^2-2xy+2y^2\right]\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-3x^2-3y^2+2x^2-2xy+2y^2\right)\)
\(=\left(x+y\right).0\)
\(=0\)