\(\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\)

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

\(\left(a^2+1\right)^2-6\left(a^2+1\right)+9=\left(a^2+1-3\right)^2=\left(a^2-2\right)^2\)

\(\left(x+y\right)^2+\left(x+y\right)x+\frac{1}{4}x^2=\left(x+y+\frac{1}{2}x\right)^2=\left(\frac{3}{2}x+y\right)^2\)

\(16x^4-9x^2=x^2\left(16x^2-9\right)=x^2\left(4x-4\right)\left(4x+3\right)\)

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

(x + 2y)² - 16

= (x + 2y)² - 4²

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

(x - 2y)² - 4.(x - 2y) + 4

= (x - 2y)² - 2.(x - 2y).2 + 2²

= (x - 2y - 2)²

(a² + 1)² - 6.(a² + 1) + 9

= (a² + 1)² - 2.(a² + 1).3 + 3²

= (a² + 1 - 3)²

= (a² - 2)²

(x + y)² + (x + y).x + 1/4.x²

= (x + y)² + 2.(x + y).1/2.x + (1/2.x)²

= (x + y + 1/2.x)²

= (3/2.x + y)²

16x⁴ - 9x²

= (4x²)² - (3x)²

= (4x² - 3x).(4x² + 3x)

a² - b⁴

= a² - (b²)²

= (a - b²).(a + b²)

a) 

áp dụng hằng đẳng thức hiệu 2 bình phương 

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

b) 

áp dụng HDT : bình phương của 1 hiệu

\(\left(x-2y\right)^2-2.2.\left(x-2y\right)+2^2=\left(x-2y-2\right)^2=\left(x-2y-2\right)\left(x-2y-2\right)\)

c) 

áp dụng HDT : bình phương của 1 hiệu

\(\left(a^2+1\right)^2-2.3.\left(a^2+1\right)+3^2=\left(a^2+1-3\right)^2=\left(a^2-2\right)^2=\left(a^2-2\right)\left(a^2-2\right)\)

d) áp dụng HDT : bình phương của 1 tồng

\(\left(x+y\right)^2+2.\frac{1}{2}.\left(x+y\right).x+\left(\frac{1}{2}x\right)^2=\left(x+y+\frac{1}{2}x\right)^2=\left(\frac{3}{2}x+y\right)\left(\frac{3}{2}x+y\right)\)

Chúc bạn học tốt nha!!! 

T I C K ủng hộ nha

hoa mắt chóng mặt

sao nhiều thế bạn

quá nhiều

Bài 2;

\(a)x^4-16x=0\Rightarrow x^4=16x\Leftrightarrow x^3=16\Leftrightarrow x=\sqrt[3]{16}\)

\(c)4x^2-\frac{1}{4}=0\Leftrightarrow4x^2=\frac{1}{4}\Leftrightarrow x^2=\frac{1}{16}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)

\(x.\left(x-3\right)-x+3=0\)

\(x.\left(x-3\right)-\left(x-3\right)=0\)

\(\left(x-3\right)\left(x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

\(x^3-3x^2+3x-1=0\)

\(\left(x-1\right)^3=0\)( hằng đẳng thức số 5 )

\(\Rightarrow x=1\)

Vậy \(x=1\)

1)\(21x^2y-12xy^2=xy.\left(21x-12y\right)\)

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

3)\(3x.\left(x-1\right)+7x^2\left(x-1\right)=\left(x-1\right).\left(3x+7x^2\right)=x.\left(x-1\right)\left(3+7x\right)\)

15)\(\left(2a+3\right)^2-\left(2a+1\right)^2=\left(2a+3-2a-1\right)\left(2a+3+2a+1\right)=2.\left(4a+4\right)=8\left(a+1\right)\)

14) \(-4y^2+4y-1=-\left[\left(2y\right)^2-2.2y.1+1^2\right]=-\left(2y-1\right)^2\)

13) \(x^6+1=\left(x^2\right)^3+1=\left(x^2+1\right)\left(x^4-x^2+1\right)\)

12) \(\left(x+1\right)^2-\left(y+6\right)^2=\left(x+1-y-6\right)\left(x+1+y+6\right)=\left(x-y-5\right)\left(x+y+7\right)\)

4) \(3x\left(x-a\right)+4a\left(a-x\right)=3x.\left(x-a\right)-4a\left(x-a\right)=\left(x-a\right)\left(3x-4a\right)\)

Sao nhiều thế!

Đúng là nhiều thật , dù sao cx cảm ơn bn nhìn nha!!!

a) 5xy ( x - y ) - 2x + 2y

= 5xy ( x - y ) - 2 ( x - y )

= ( x - y ) ( 5xy - 2 )

b) 6x-2y-x(y-3x)

= 2 ( y - 3x ) - x ( y - 3x )

= ( y - 3x ( ( 2 - x )

c)  x² + 4x - xy-4y

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

( x + 4 ) ( x - y )

d) 3xy + 2z - 6y - xz 

= ( 3xy - 6y ) + ( 2z - xz )

= 3y ( x - 2 ) + z ( x - 2 )

= ( x - 2 ) ( 3y + z )

a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)

b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)

c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)

d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)

11)

a,4-9x^2=0

(2-3x)(2+3x)=0

2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3

b,x^2 +x+1/4=0

(x+1/2)^2 =0

x+1/2=0

x=-1/2

c,2x(x-3)+(x-3)=0

(x-3)(2x+1)=0

x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2

d,3x(x-4)-x+4=0

3x(x-4)-(x-4)=0

(x-4)(3x-1)=0

x-4=0=>x=4 hoặc 3x-1=0=>x=1/3

e,x^3-1/9x=0

x(x^2-1/9)=0

x(x+1/3)(x-1/3)=0

x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3

f,(3x-y)^2-(x-y)^2 =0

(3x-y-x+y)(3x-y+x-y)=0

2x(4x-2y)=0

4x(2x-y)=0

x=0hoặc 2x-y=0=>x=y/2

thu thủy học lớp 9 chưa có long lone

1a/ z² - 6z + 5 - t² - 4t = z² - 2 . 3z + 3² - 4 - t² - 4t = (z² - 2 . 3z + 3²) - (2² + 2 . 2t + t²) = (z - 3)² - (2 + t)²

b/ x² - 2xy + 2y² + 2y² + 1 = x² - 2xy + y² + y² + 2y + 1 = (x² - 2xy + y²) + (y² + 2y + 1) = (x - y)² + (y + 1)²

c/ 4x² - 12x - y² + 2y + 8 = (2x)² - 12x - y² + 2y + 3² - 1 = [ (2x)² - 2 . 3 . 2x + 3² ] - (y² - 2y + 1) = (2x - 3)² - (y - 1)²

2a/ (x + y + 4)(x + y - 4) = x² + xy - 4x + xy + y² - 4y + 4x + 4y + 16 = x² + (xy + xy) + (-4x + 4x) + (-4y + 4y) + y² + 16

= x² + 2xy + y² + 4² = (x + y)² + 4²

b/ (x - y + 6)(x + y - 6) = x² + xy - 6x - xy - y² + 6y + 6x + 6y - 36 = x² + (xy - xy) + (-6x + 6x) + (6y + 6y) - y² - 36

= x² - y² + 12y - 6² = x² - (y² - 12y + 6²) = x² - (y² - 2 . 6y + 6²) = x² - (y - 6)²

c/ (y + 2z - 3)(y - 2z - 3) = y² -2yz - 3y + 2yz - 4z² - 6z - 3y + 6z + 9 = y² + (-2yz + 2yz) + (-3y - 3y) + (-6z + 6z) - 4z² + 9

= y² - 6y - 4z² + 9 = (y² - 6y + 9) - 4z² = (y - 3)² - (2z)²

d/ (x + 2y + 3z)(2y + 3z - x) = 2xy + 3xz - x² + 4y² + 6yz - 2xy + 6yz + 9z² - 3xz = (2xy - 2xy) + (3xz - 3xz) - x² + (6yz + 6yz) + 9z² + 4y²

= -x² + 4y² + 12yz + 9z² = (4y² + 12yz + 9z²) - x² = [ (2y)² + 2 . 2 . 3yz + (3z)² ] - x² = (2y + 3z)² - x²

:v dễ mà có trong nâng cao mới hc qua :3

a, x2+10x+26+y2+2y

=(x2+2.x.5+52)+(12+2.1.y+y2)

=(x+5)2+(y+1)2

b, x2−2xy+2y2+2y+1

=x2−2xy+y2+y2+2y+1

=(x2−2.x.y+y2)+(y2+2.y.1+12)

=(x−y)2+(y+1)2

c,z2−6z+5−t2−4t

=−(t2+4t−z2+6z−5)

=−(t2+2.t.2+22−z2+2.z.3−32)

=−((t2+2.t.2+22)−(z2−2.z.3+32))

=−((t+2)2−(z−3)2)

=(z−3)2−(t+2)2