Câu 1 :

\(3\left(x-3\right)\left(x+7\right)+\left(1-4\right)\left(x+4\right)+18\)

\(=3\left(x^2+4x-21\right)-3\left(x+4\right)\)

\(=3x^2+12x-63-3x-12=3x^2+9x-75\)

Thay x = 1/2 vào ta được 

\(\dfrac{3.1}{4}+\dfrac{9}{2}-75=-\dfrac{279}{4}\)

Câu 2 : 

\(5x^2+5xy+5x=5x\left(x+y+1\right)\)

Thay x = 60 ; y = 50 ta được 

\(300\left(60+50+1\right)=33300\)

Câu 3 : 

\(4x^2y^2+2xy^2+6x^2y=2xy\left(2xy+y+3x\right)\)

Thay x = 10 ; y  = 1/2 ta được 

\(\dfrac{2.10.1}{2}\left(\dfrac{2.10.1}{2}+\dfrac{1}{2}+30\right)=405\)

1: \(=3\left(x^2+4x-21\right)+x^2-16+18\)

\(=3x^2+12x-63+x^2+2\)

\(=4x^2+12x-61\)

\(=4\cdot\dfrac{1}{4}+12\cdot\dfrac{1}{2}-61=1-61+6=-54\)

2: \(=5\cdot60^2+5\cdot60\cdot50+5\cdot60=33300\)

3: \(=4\cdot10^2\cdot\dfrac{1}{4}+2\cdot10\cdot\dfrac{1}{4}+6\cdot100\cdot\dfrac{1}{2}=405\)

\(a,M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\\ \Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\\ \Rightarrow M=x^2+11xy-y^2\\ b,\left(25x^2y-13xy^2+y^3\right)-M=11xy^2-2y^3\\ \Rightarrow M=25x^2y-13xy^2+y^3-11xy^2+2y^3\\ \Rightarrow M=25x^2y-24xy^2+3y^3\)

1) x²-y²-x-y=(x²-y²)-(x+y)

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

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

2)x⁴-5x²+4

=x⁴-4x²-x²+4

=(x⁴-x²)-(4x²-4)

=x²(x²-1)-4(x²-1)

=(x²-1)(x²-4)

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

3)25-x²+2xy-y²

=25-(x²-2xy+y²)

=25-(x-y)²

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

1, x²-y²-x-y=(x²-y²)-(x+y)=(x-y)(x+y)-(x+y)=(x+y)(x-y-1)

2, x⁴-5x²+4=x⁴-x²-4x²+4=(x⁴-x²)-(4x²-4)=x²(x²-1)-4(x²-1)=(x²-1)(x²-4)

3, 25-x²+2xy-y²=5²-(x²-2xy+y²)=5²-(x-y)²=(5-x+y)(5+x-y)

_{1, \(x^2-y^2-x-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)}

_{=\(\left(x+y\right)\left(x-y-1\right)\)}

2, \(x^4-5x^2+4=x^4-4x^2-x^2+4\)

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

\(=\left(x^2-4\right)\left(x^2-1\right)\)

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

3, \(25-x^2+2xy-y^2=5^2-\left(x^2-2xy+y^2\right)\)

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

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

Bài 1:

a) \(x^2-2xy-25+y^2\) (Sửa đề)

\(=x^2-2xy+y^2-25\)

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

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

Vậy ...

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

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

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

Vậy ...

c) \(7x+7y-\left(x+y\right)\) (Sửa đề)

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

\(=\left(x+y\right)\left(7-1\right)\)

\(=6\left(x+y\right)\)

Vậy ...

d) \(x^4+y^4\)

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

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

\(=\left(x^2+y^2-\sqrt{2}xy\right)\left(x^2+y^2+\sqrt{2}xy\right)\)

Vậy ...

Bạn xem lại 1 sô câu sai và bài 2 hộ mk

a,  x²+2xy+y²+2x+2y-15

<=> (x+y )²+2(x+y)+1-16

Đặt x+y =a

<=> a²+2a+1-4²

<=> (a+1)²-4²

<=> (a+5)(a-3) =>( x+y+5)(x+y-3)

b, x²-4xy+4y²-2x-4y-35

<=> (x-2y)²-2(x-2y)+1-36

Đặt (x-2y)  =b 

=> b²-2b+1-6²

<=> (b-1)²-6²

<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)

c, 

a,A= x^2+2xy+y^2+2x+2y-15

= (x+y)^2+(x+y)-15

Đặt x+y=a, ta có:

A=a^2+2a-15

  =a^2+2a+1-16

  =(a+1)^2-4^2

  =(a+1+4)(a+1-4)

  =(a+5)(a-3)

Thay a=x+y, ta có: A=(x+y+5)(x+y-3).

b,B= x^2 - 4xy+4y^2-2x-4y-35

   Hình như là sai đề đó bạn. Phải là x^2 - 4xy+4y^2-2x+4y-35 hoặc x^2 - 4xy+4y^2+2x-4y-35 hoặc x^2 + 4xy+4y^2-2x-4y-35 mới đúng đó bạn. Bạn xem lại đi nha.

c,C=6x^4 - 5x^3+8x^2-5x+6

C= x^2(6x^2-5x+8-5/x+6/x^2)

  =x^2(6(x^2+2+1/x^2)-5(x+1/x)-4)

  =x^2(6(x+1/x)^2-5(x+1/x)-4)

Đặt x+1/x=a, ta có:

C=x^2(6a^2-5a-4)

  =x^2(6a^2+3a-8a-4)

  =x^2(2a+1)(3a-4)

Thay a=x+1/x vào là được bạn nhé.

Giải:

a) x² - 2xy + y² - 25 = (x - y)² - 5² = (x - y - 5)(x - y + 5)

b) x² - 10x - y² + 25 = (x² - 10x + 25) - y² = (x - 5)² - y² = (x - 5 - y)(x - 5 + y)

c) 3x + 3y + zx + zy = (3x + 3y) + (zx + zy) = 3(x + y) + z(x + y) = (x + y)(3 + z)

d) x² - 4y² + 5x +10y = (x² - 4y²) + (5x +10y) = (x - 2y)(x + 2y) + 5(x + 2y) = (x + 2y)(x - 2y + 5)

\(\cdot x^2-4x-5=x^2-5x+x-5=\left(x^2-5x\right)+\left(x-5\right)=x\left(x-5\right)+\left(x-5\right)=\left(x-5\right)\left(x+1\right)\)

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