Question

Giup vs.mih cần trong chiều nay nha mn.làm nhanh mih tik

1.4a²b²-(a²+b²-1)²

2.(xy+4)²-(2x+2y)²

3.(a²+b²+ab)²- a²b² -b²c²-c²a²

4.x²+x²y-4x-4y

5.(a+b+c)²+ (a+b-c)²-4c²

6.4a²b²-(a²+b²-c²)²

7.a⁴+b⁴+c⁴-2a²b²-2b²c²-2a²c²

8.a.(b³-c³)+b.(c³-a³)+c.(a³-b³)

Answer 1

1.\(4a^2b^2-\left(a^2+b^2-1\right)^2\)

\(=\left(2ab\right)^2-\left(a^2+b^2-1\right)^2\)

\(=\left(2ab+a^2+b^2-1\right)\left(2ab-a^2-b^2+1\right)\)

\(=\left[\left(a+b\right)^2-1\right]\left[1-\left(a-b\right)^2\right]\)

Answer 2

2.\(\left(xy+4\right)^2-\left(2x+2y\right)^2\)

\(=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)

\(=\left[x\left(y-2\right)-2\left(y-2\right)\right]\left[x\left(y+2\right)+2\left(y+2\right)\right]\)

\(=\left(x-2\right)\left(y-2\right)\left(x+2\right)\left(y+2\right)\)

Answer 3

3/\(\left(a^2+b^2+ab\right)^2-a^2b^2-b^2c^2-c^2a^2\)

\(=\left(a^2+b^2+ab-ab\right)\left(a^2+b^2+2ab\right)-\left[\left(bc\right)^2+\left(ca\right)^2\right]\)

\(=\left(a^2+b^2\right)\left(a+b\right)^2-\left[c^2\left(a^2+b^2\right)\right]\)

\(=\left(a^2+b^2\right)\left[\left(a+b\right)^2-c^2\right]\)

\(=\left(a^2+b^2\right)\left(a+b+c\right)\left(a+b-c\right)\)

Answer 4

4

.\(=x^2-4x+x^2y-4y\)

\(=x\left(x-4\right)+y\left(x-4\right)\left(x+4\right)\)

\(=\left(x-4\right)\left(x+xy+4y\right)\)

Answer 5

5.

\(=\left(a+b+c\right)^2+\left(a+b-c+2c\right)\left(a+b-c-2c\right)\)

\(=\left(a+b+c\right)^2+\left(a+b+c\right)\left(a+b-3c\right)\)

\(=\left(a+b+c\right)\left(a+v+c+a+b-3c\right)\)

\(=2\left(a+b+c\right)\left(a+b-c\right)\)

Answer 6

6/ \(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)

\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)

\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)

\(=\left(a+b-c\right)\left(a+b+c\right)\left(c-a+b\right)\left(c-a-b\right)\)

Answer 7

7.

=\(4a^2b^2-\left(a^4+2a^2b^2+b^4\right)+\left(2b^2c^2+2a^2c^2\right)-c^4\)

\(=\left(2ab\right)^2-\left[\left(a^2+b^2\right)^2-2c^2\left(a^2+b^2\right)+c^4\right]\)

\(=\left(2ab\right)^2-\left[\left(a^2+b^2\right)-c^2\right]^2\)

\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)

\(=\left(a+b+c\right)\left(a+b-c\right)\left(c-a+b\right)\left(c+a-b\right)\)

Answer 8

8.Ta có :\(\left(c^3-a^3\right)=-\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]\)

\(=a.\left(b^3-c^3\right)-b\left[\left(b^3-c^3\right)+\left(a^3-b^3\right)\right]+c\left(a^3-b^3\right)\)

\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)+c\left(a^3-b^3\right)-b\left(b^3-c^3\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)-\left(b-c\right)\left(a-b\right)\left(a^2+ab+c^2\right)\)

=\(\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)