Đè phân tích hả 

a) 10x(x-y) - 8y(y-x)

= 10x(x-y) + 8y(x-y)

= (10x-8y)(x-y)

b) x^2 - 2xy + y^2 - 25 

 =( x- y)^2 - 25

= ( x- y- 5 )( x-  y+ 5 )

c) 5x^2 + 10x^2y + 5xy^2 

= 5( x^2 + 2x^2y + xy^2)

= 5x ( x + 2xy + y^2)

( 5x^3 hay 5x^2) 

Giải thích vì sao câu a có đáp án như vậy?

\(10x\left(x-y\right)-8y\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=2\left(x-y\right)\left(5x+4y\right)\)

 1/x-1/y=1/6 
<=> 6(y-x) = xy 
<=>y(6-x) = 6x (1) 
<=>y= 6x/(6-x) (2) 
(1) => x<6 => x=1,2,3,4,5 
Thay vào (2) ta có các cặp số nguyên thỏa đề bài là : 
(x;y)= (2;3);(3;6)

10x (x-y) -8y(y-x)

=10x(x-y) + 8y(x-y)

=(x-y)(10x+8y)

=2(x-y)(5x+4y)

\(A=\dfrac{7yz}{x}+\dfrac{8zx}{y}+\dfrac{9xy}{z}=\dfrac{3yz}{x}+\dfrac{3zx}{y}+\dfrac{4yz}{x}+\dfrac{4xy}{z}+\dfrac{5zx}{y}+\dfrac{5xy}{z}\)

\(bddt:\) \(AM-GM\Rightarrow\left\{{}\begin{matrix}\dfrac{3yz}{x}+\dfrac{3zx}{y}\ge2\sqrt{\dfrac{9yzxz}{xy}}\ge6z\\\dfrac{4yz}{x}+\dfrac{4xy}{z}\ge2\sqrt{\dfrac{16yzxy}{xz}}\ge8y\\\dfrac{5zx}{y}+\dfrac{5xy}{z}\ge2\sqrt{\dfrac{25zxxy}{yz}}\ge10x\\\end{matrix}\right.\)

\(\Rightarrow A\ge10x+8y+6z\)

a) \(x^2-3x=x\left(x-3\right)\)

b) \(10x\left(x-y\right)-8y\left(x-y\right)=2\left(x-y\right)\left(5x-4y\right)\)

c) \(x^2-9=\left(x-3\right)\left(x+3\right)\)

a) x²-3x=x(x-3)

c) x²-9=x²-3²=(x+3)(x-3)

a: \(x^2-3x=\left(x-3\right)\cdot x\)

c: \(x^2-9=\left(x-3\right)\left(x+3\right)\)

a) \(5x^2\)\(\left(x-2y\right)\)\(-\)\(15x\)\(\left(x-2y\right)\)

\(=\left(x-2y\right)\left(5x^2-15x\right)\)

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

b)  \(3\left(x-y\right)\)\(-\)\(5x\left(y-x\right)\)

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

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

c)  \(10x\left(x-y\right)\)\(-\)\(8y\left(y-x\right)\)

\(=\)\(10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=2\left(5x+4\right)\left(x-y\right)\)

d)  \(x^2\)\(\left(x-5\right)\)\(+\)\(4\)\(\left(5-x\right)\)

\(=x^2\)\(\left(x-5\right)\)\(-\)\(4\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2-4\right)\)

\(=\left(x-5\right)\left(x-2\right)\left(x-2\right)\)

a) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)\)

\(=\left(x-2y\right)\left(5x^2-15x\right)\)

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

b)\(3\left(x-y\right)-5x\left(y-x\right)\)

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

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

c)\(10x\left(x-y\right)-8y\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(10x+8y\right)\left(x-y\right)\)

\(=2\left(5x+4y\right)\left(x-y\right)\)

d)\(x^2\left(x-5\right)+4\left(5-x\right)\)

\(=x^2\left(x-5\right)-4\left(x-5\right)\)

\(=\left(x^2-4\right)\left(x-5\right)\)

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

Ta có 7x = 9y \(\Rightarrow\frac{x}{9}=\frac{y}{7}\Rightarrow\frac{10.x}{10.9}=\frac{8.y}{8.7}\Rightarrow\frac{10x}{90}=\frac{8y}{56}=\frac{10x-8y}{90-56}=\frac{68}{34}=2\)

                  Vậy \(\frac{10x}{90}=2\Rightarrow\frac{x}{9}=2\Rightarrow x=18\)

                          \(\frac{8y}{56}=2\Rightarrow\frac{y}{7}=2\Rightarrow y=14\)

7x=9y

=>x/9=y/7

=>10/10.x/9=8/8.y/7

=>10x/90=8y/56

áp dụng t/c dãy tỉ số = nhau:

10x/90=8y/56=10x-8y/90-56=68/34=2

10x/90=2=>10x=180=>x=18

8y/56=2=>8y=112=>y=14

vậy x=18

y=14

TL

y=14

Hok tốt

\(x^2-y^2+10x+8y+9\)

\(=x^2-y^2+9x+x+9y-y+9\)

\(=\left(x^2-y^2\right)+\left(x-y\right)+\left(9x+9y+9\right)\)

\(=\left(x-y\right)\left(x+y\right)+\left(x-y\right)+9\left(x+y+1\right)\)

\(=\left(x-y\right)\left(x+y+1\right)+9\left(x+y+1\right)\)

\(=\left(x+y+1\right)\left(x-y+9\right)\)

Phân tích đa thức thành nhân tử à bn

ta co : 

7x=9y => 7x/63=9y/63 => x/9=y/7 => 10x/90=8y/56 va 10x-8y = 68

Ap dung tinh chat day ti so bang nhau ta co : 

10x/90=8y/56 = 10x-8y/90-56=68/34=2

suy ra : 

10x/90=2=>10x=2.90=180=>x=18

8y/56=2=>8y=2.56=112=>y=112:8=14

Vay : x=18 va y= 14