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Ta có: \(\left(ax+by\right)^2=\left(a^2+b^2\right)\left(x^2+y^2\right)\)

\(\Leftrightarrow a^2x^2+2abxy+b^2y^2=a^2x^2+a^2y^2+x^2b^2+b^2y^2\)

\(\Leftrightarrow2abxy=a^2y^2+x^2b^2\)

\(\Leftrightarrow\left(ay-xb\right)^2=0\)

\(\Leftrightarrow ay=xb\)

hay \(\dfrac{a}{x}=\dfrac{b}{y}\)

13 tháng 8 2017

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)\(\left(a-b\right)\left(x^2+y\right)\)

10 tháng 9 2018

1 ) \(x\left(a-b\right)+a-b=\left(x+1\right)\left(a-b\right)\)

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

3 ) \(-2x-2y+ax+ay=-2\left(x+y\right)+a\left(x+y\right)=\left(a-2\right)\left(x+y\right)\)

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

5 ) \(5x^2y+5xy^2+a^2x+a^2y\)

\(=5xy\left(x+y\right)+a^2\left(x+y\right)\)

\(=\left(5xy+a^2\right)\left(x+y\right)\)

6 ) \(2x^2-6xy+5x-15y\)

\(=2x\left(x-3y\right)+5\left(x-3y\right)\)

\(=\left(2x+5\right)\left(x-3y\right)\)

7 ) \(ax^2-3axy+bx-3by\)

\(=\left(ax^2+bx\right)-\left(3axy+3by\right)\)

\(=x\left(ax+b\right)-3y\left(ax+b\right)\)

\(=\left(x-3y\right)\left(ax+b\right)\)

8 ) \(x^2+4x-5x-20=0\)

\(\Leftrightarrow x\left(x+4\right)-5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-4\end{matrix}\right.\)

9 ) \(x^2+10x-2x-20=0\)

\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-10\end{matrix}\right.\)

10 ) \(x^2-6x-4x+24=0\)

\(\Leftrightarrow x\left(x-6\right)-4\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)

:D

2 tháng 7 2018

1) x - y - a(x - y) = (x - y) - a(x - y) = (1 - x)(x - y)

2) a - b + x(a - b) = (a - b) + x(a - b) = (1 + x)(a - b)

3) a(x - y) - x + y = a(x - y) - (x - y) = (a - 1)(x - y)

4) x(a - b) - a + b = x(a - b) - (a - b) = (x - 1)(a - b)

5) ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)

6) ax + ay - bx - by = a(x + y) - b(x + y) = (a - b)(x + y)

7) - 2x - 2y + ax + ay = -2(x + y) + a(x + y) = (a - 2)(x + y)

8) x2 - xy - 2x + 2y = x(x - y) - 2(x - y) = (x - 2)(x - y)

Sorry nha, giờ mình chỉ rảnh làm 8 câu thôi

Bài 3:

a: =>(2x-7)(x-2)=0

=>x=7/2 hoặc x=2

b: =>(x-1)(x+2)=0

=>x=1 hoặc x=-2

d: =>2x+3=0

hay x=-3/2

1 tháng 8 2023

`ab(x-3) -a^2(x-3)`

`=(x-3)(a^2-ab)`

__

`ax+ay+bx+by`

`=a(x+y)+b(x+y)`

`=(x+y)(a+b)`

__

`ax+ay -2x-2y`

`=(ax+ay)-(2x+2y)`

`=a(x+y)-2(x+y)`

`=(x+y)(a-2_`

__

`2x-2y +ax-ay`

`=2(x-y)+a(x-y)`

`=(x-y)(2+a)`

__

`10ax -5ay -2x+y`

`= 5a(2x-y) -(2x-y)`

`=(2x-y)(5a-1)`

1: =(x-3)(ab-a^2)

=a(b-a)(x-3)

2: =a(x+y)+b(x+y)

=(x+y)(a+b)

3: =a(x+y)-2(x+y)

=(x+y)(a-2)

4: =2(x-y)+a(x-y)

=(x-y)(a+2)

5: =5a(2x-y)-(2x-y)

=(2x-y)(5a-1)

19 tháng 8 2017

\(a,7x^2-7xy-4x+4y\)

\(=7x\left(x-y\right)-4\left(x-y\right)\)

\(=\left(7x-4\right)\left(x-y\right)\)

\(b,2x-2y+ax-ay\)

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

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

\(c,x^2-x-y^2-y\)

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

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

\(=\left(x+y\right)\left(x-y-1\right)\)

\(d,ax+ay-2x-2y\)

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

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

\(e,x\left(a+b\right)-a-b=x\left(a+b\right)-\left(a+b\right)\)

\(=\left(x-1\right)\left(a+b\right)\)

16 tháng 7 2018

câu 2 nx