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\(1,ax+ay+bx+by\)
\(=\left(ax+ay\right)+\left(bx+by\right)\)
\(=a\left(x+y\right)+b\left(x+y\right)\)
\(=\left(x+y\right)\left(a+b\right)\)
\(2,ax+ay+2x+2y\)
\(=\left(ax+ay\right)+\left(2x+2y\right)\)
\(=a\left(x+y\right)+2\left(x+y\right)\)
\(=\left(x+y\right)\left(a+2\right)\)
\(3,ax+ay-bx-by\)
\(=\left(ax+ay\right)-\left(bx+by\right)\)
\(=a\left(x+y\right)-b\left(x+y\right)\)
\(=\left(x+y\right)\left(a-b\right)\)
\(4,ax+ay-2x-2y\)
\(=\left(ax+ay\right)-\left(2x+2y\right)\)
\(=a\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(a-2\right)\)
ax + ay + bx + by
= a.(x+y) + b.(x+y)
= (x+y).(a+b)
ax + ay + 2x + 2y
= a.(x+y) + 2.(x+y)
= (x+y).(a+2)
ax + ay - bx - by
= a.(x+y) - b.(x+y)
= (x+y).(a-b)
ax + ay - 2x - 2y
= a.(x+y) - 2.(x+y)
= (x+y).(a-2)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)
\(=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\)
\(=\left(ax-ay-bx+by\right)\left(ax+ay+bx+by\right)\)
\(=\left[a\left(x-y\right)-b\left(x-y\right)\right]\left[a\left(x+y\right)+b\left(x+y\right)\right]\)
\(=\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)
(ax+ay+bx+by)(ax−ay+by−bx) \(=\left(ax+ay+bx+by\right)\left(ax-ay+by-bx\right)\)
\(=\left(a+b\right)\left(x+y\right)\left(a-b\right)\left(x-y\right)\)
\(\left(ax+by\right)^2-\left(ay+bx\right)^2=\left(ax+by-ay-bx\right)\left(ax+by+ay+bx\right)\)
\(=\left[a\left(x-y\right)-b\left(x-y\right)\right].\left[a\left(x+y\right)+b\left(x+y\right)\right]\)
\(=\left(a-b\right)\left(x-y\right)\left(a+b\right)\left(x+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{ax+ay-bx-by}{ax-ay-bx+by}=\dfrac{a\left(x+y\right)-b\left(x+y\right)}{a\left(x-y\right)-b\left(x-y\right)}=\dfrac{\left(a-b\right)\left(x+y\right)}{\left(a-b\right)\left(x-y\right)}=\dfrac{x+y}{x-y}\)
b) \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\dfrac{a+b-c}{a+c-b}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. (-2x - 1)(x2 - x - 3) - (x + 2)(x + 1)2
= -2x3 + 2x2 + 6x - x2 + x + 3 - (x + 2)(x2 + 2x + 1)
= -2x3 + x2 + 7x + 3 - x3 - 2x2 - x - 2x2 - 2x - 2
= -3x3 - 3x2 + 4x + 1
2. (x + 2)(x - 1) - (x - 3)(x + 2) = 3
=> (x + 2)(x - 1 - x + 3) = 3
=> (x + 2).0 = 3
...(xem lại đề)
\(\left(x+2\right)\left(x-1\right)-\left(x-3\right)\left(x+2\right)=3\)
\(\Leftrightarrow\left(x+2\right)\left(x-1-x+3\right)=3\)
\(\Leftrightarrow2\left(x+2\right)=3\)
\(\Leftrightarrow x+2=\frac{3}{2}\)
\(\Leftrightarrow x=\frac{3}{2}-2\)
\(\Leftrightarrow x=-\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ax^2+by^2-ay^2-bx^2=x^2\left(a-b\right)-y^2\left(a-b\right)=\left(a-b\right)\left(x^2-y^2\right)=\left(a-b\right)\left(x-y\right)\left(x+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bn post nhiều nên mình ghi đáp án thôi nhé phần nào sai đề mình cho qua
b)\(\left(x+1\right)\left(xy+1\right)\)
c)\(\left(a+b\right)\left(x+y\right)\)
d)\(\left(x-a\right)\left(x-b\right)\)
e)\(\left(x+y\right)\left(xy-1\right)\)
f)\(\left(a-b\right)\left(x^2+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ax-ay-bx+by+(y-x)2
= a(x-y)-b(x-y)-(x-y)2
= (a-b-1)(x-y)
chả bít dúng hay sai ,hình như là đúng