A) 7X² - 7Y² - 14X + 14Y

= ( 7X² - 7Y² ) - ( 14X - 14Y )

= 7( X² - Y² ) - 14( X - Y )

= 7( X - Y )( X + Y ) - 14( X - Y )

= 7( X - Y )( X + Y - 14 )

B) X² - Y² + 14X + 49

= ( X² + 14X + 49 ) - Y²

= ( X + 7 )² - Y²

= ( X - Y + 7 )( X + Y + 7 )

C) X² - Y² - X + Y

= ( X² - Y² ) - ( X - Y )

= ( X - Y )( X + Y ) - ( X - Y )

= ( X - Y )( X + Y - 1 )

D) X² + 12Y - Y² - 36

= X² - ( Y² - 12Y + 36 )

= X² - ( Y - 6 )²

= ( X - Y + 6 )( X + Y - 6 )

E) X³ + X² - 9X - 9

= ( X³ + X² ) - ( 9X + 9 )

= X²( X + 1 ) - 9( X + 1 )

= ( X + 1 )( X² - 9 )

= ( X + 1 )( X - 3 )( X + 3 )

sửa ý a) thành 7( x - y )( x + y - 2 ) nhé ;-;

Ta có a^5-a luôn chia hết cho 6

suy ra a^5+...+d^5 -2016 chia hết cho 6

dpcm

=2xy(2y-y)

\(4xy^2-2xy^2\)

\(=2xy\left(2y-y\right)\)

giải nhanh hộ mik bài trên mới mình hứa sẽ tích nhaaa

bn viết phần đề bài có dấu đc ko. Mk ko dịch đc

Sử dụng delta thôi!

Xét \(4x^2+\sqrt{2}x-\sqrt{2}=0\) có \(4\cdot\left(-\sqrt{2}\right)=-4\sqrt{2}< 0\) nên PT có 2 nghiệm phân biệt

Mà a là nghiệm nguyên dương của PT nên ta có: \(4a^2+\sqrt{2}a-\sqrt{2}=0\)

Vì a > 0 \(\Rightarrow4a^2=-\sqrt{2}a+\sqrt{2}\)

\(\Rightarrow a^2=\frac{\sqrt{2}-\sqrt{2}a}{4}=\frac{\left(1-a\right)\sqrt{2}}{4}=\frac{1-a}{2\sqrt{2}}\)

\(\Rightarrow a^4=\left(\frac{1-a}{2\sqrt{2}}\right)^2=\frac{1-2a+a^2}{8}\)

Thay vào ta được:

\(B=\frac{a+1}{\sqrt{a^4+a+1}-a^2}=\frac{\left(a+1\right)\left(\sqrt{a^4+a+1}+a^2\right)}{\left(\sqrt{a^4+a+1}\right)^2-a^4}\)

\(=\frac{\left(a+1\right)\left(\sqrt{a^4+a+1}+a^2\right)}{a^4+a+1-a^4}=\frac{\left(a+1\right)\left(\sqrt{a^4+a+1}+a^2\right)}{a+1}=\sqrt{a^4+a+1}+a^2\)

\(=\sqrt{\frac{1-2a+a^2}{8}+a+1}+\frac{1-a}{2\sqrt{2}}=\sqrt{\frac{a^2+6a+9}{8}}+\frac{1-a}{2\sqrt{2}}\)

\(=\frac{a+3}{2\sqrt{2}}+\frac{1-a}{2\sqrt{2}}=\frac{4}{2\sqrt{2}}=\sqrt{2}\)

Vậy \(B=\sqrt{2}\)

a) ( 3 - x )( x² + 2x - 7 ) + ( x - 3 )( x² + x - 5 )

= ( 3 - x )( x² + 2x - 7 ) - ( 3 - x )( x² + x - 5 )

= ( 3 - x )( x² + 2x - 7 - x² - x + 5 )

= ( 3 - x )( x - 2 )

b) ( x - 5 )² + 3( 5 - x )

= ( x - 5 )² - 3( x - 5 )

= ( x - 5 )( x - 5 - 3 ) = ( x - 5 )( x - 8 )

c) 2x( x - 1 )² - ( 1 - x )³

= 2x( 1 - x )² - ( 1 - x )³

= ( 1 - x )²( 2x - 1 + x ) = ( 1 - x )²( 3x - 1 )

d) x² + 8x + 16 = ( x + 4 )²

e) x² - 4xy + 4y² = ( x - 2y )²

g) 4x² - 25y² = ( 2x )² - ( 5y )² = ( 2x - 5y )( 2x + 5y )

h) 25( x + 1 )² - 4( x - 3 )²

= 5²( x + 1 )² - 2²( x - 3 )²

= ( 5x + 5 )² - ( 2x - 6 )²

= ( 5x + 5 - 2x + 6 )( 5x + 5 + 2x - 6 )

= ( 3x + 11 )( 7x - 1 )

i) x³ + 27 = ( x + 3 )( x² - 3x + 9 )

k) 8x³ - 125 = ( 2x )³ - 5³ = ( 2x - 5 )( 4x² + 10x + 25 )

l) x³ + 6x² + 12x + 8 = ( x + 2 )³

m) -x³ + 9x² - 27x + 27 = -( x³ - 9x² + 27x - 27 ) = -( x - 3 )³

Đề:........

<=> (x - 2y)² - z²

<=> [(x - 2y) - z].[(x - 2y) + z]

<=> (x - 2y - z).(x - 2y + z)

x² - 4xy + 4y² - z²

= ( x - 2y )² - z²

= ( x - 2y - z )( x - 2y + z )

\(4\left(6-x\right)+x^2-12x+36=0\)

\(24-4x+x^2-12x+36=0\)

\(x^2-16x+60=0\)

\(x^2-2x8+8^2-8^2+60=0\)

\(\left(x-8\right)^2-4=0\)

\(\left(x-8\right)^2=4\)

\(\left(x-8\right)^2=\left(\pm2\right)^2\)

\(\orbr{\begin{cases}x-8=2\Rightarrow x=10\\x-8=-2\Rightarrow x=6\end{cases}}\)

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

\(=2yx+2y^2+3x^2-3xy+5\)

\(=2y^2+3x^2-xy+5\)