Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

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

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

c)4x+4y-x^2(x+y)

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

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

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

d) y^2+y-x^2+x

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

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

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

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

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

f)9x^2-25y^2-6x+10y

= 

ko biết làm

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

Bài 1:

\(a,=3x\left(3xy+5y-1\right)\\ b,=\left(z-2\right)\left(3z-5\right)\\ c,=\left(x+2y\right)^2-4z^2=\left(x+2y+2z\right)\left(x+2y-2z\right)\\ d,=x^2-3x+5x-15=\left(x-3\right)\left(x+5\right)\)

Bài 2:

\(a,\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x+2-4x^2-12x=9\\ \Leftrightarrow4x^2+10x+7=0\\ \Leftrightarrow4\left(x^2+\dfrac{5}{2}x+\dfrac{25}{16}\right)+\dfrac{3}{4}=0\\ \Leftrightarrow4\left(x+\dfrac{5}{6}\right)^2+\dfrac{3}{4}=0\left(vô.lí\right)\\ \Leftrightarrow x\in\varnothing\\ c,\Leftrightarrow x^2-12x+36=0\\ \Leftrightarrow\left(x-6\right)^2=0\\ \Leftrightarrow x=6\)

a>(8x^2y+10xy6^2-6xy):2xy=4xy+5y-3

b>(3x^2-4x).(2x-6)=6x^3-26x^2+24x

kết quả thôi nha

bài 1 : a, sai đề . đề đúng là x³ + x² + 2² = x³ + 2x²- x²+ 4

= x² ( x + 2 ) - ( x - 2 ) ( x + 2 ) = ( x + 2 ) ( x² - x + 2 )

b, xy -y² -x -y = ( xy -x ) + ( y² - y ) = x ( y -1) + y ( y -1) = ( y -1 ) ( x +y )

sai đề thì sửa dùm mik nhé

giúp mik bài này với

CẦN GẤP

1.a) (4x - 6y)² - (8xy - 5)² = (4x - 6y - 8xy + 5)(4x - 6y + 8xy - 5)

   b) 16x² - 49y² = (4x)² - (7y)² = (4x - 7y)(4x + 7y)

   c) 36x² + 60x + 25 = (6x)² + 2.6x.5 + 5² = (6x + 5)²

   d) (2x - y)(x - y) - (3y - 4x)² + (y - 2x)(2y - 3x) = (y - 2x)(y - x) + (y - 2x)(2y - 3x) - (3y - 4x)²

     = (y - 2x)[(y - x) + (2y - 3x)] - (3y - 4x)² = (y - 2x)(3y - 4x) - (3y - 4x)² = (3y - 4x)[(y - 2x) - (3y - 4x)] = 2(3y - 4x)(x - y)

2.M = (3x - 4)(9x² - 12x + 16) + (6x - 8)² = (3x - 4)[(3x)² - 2.3x.4 + 4²] + [2(3x - 4)]² = (3x - 4)(3x - 4)² + 4(3x - 4)²

       = (3x - 4)²(3x - 4 + 4) = 3x(3x - 4)²

a) =(4x-6y-8xy+3)(4x-6y+8xy-3)

=[4x(1-2y)+3(1-2y)][4x(1+2y)-3(1+2y)]

=(4x+3)(4x-3)(1-2y)(1+2y)

câu a )Phan Thanh Tịnh sai, tui sửa lại