a) 11x + 11y + x² + xy

= 11.(x+y) + x.(x+y)

= (x+y).(11+x)

b) 255 + x² - 4xy + y²

= 255 + 2xy + x² -2xy + y²

= 255 + 2xy + (x-y)²

...

Nhấn vào câu hỏi tương tự 

:))) 

a, \(11x+11y+x^2+xy=\left(11x+11y\right)+\left(x^2+xy\right)=11\left(x+y\right)+x\left(x+y\right)=\left(x+y\right)\left(x+11\right)\)

b. \(255-4x^2-4xy-y^2=255-\left(4x^2+4xy+y^2\right)=255-\left(2x+y\right)^2=\left(15+2x+y\right)\left(15-2x-y\right)\)

Bài 2:

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

\(=\left(72-2\right)\left(102-2\right)=70.100=7000\) ( x+y=102, x-y=72 )

Bài 1: 

a, 11x+11y+x²+xy=11(x+y)+x(x+y)=(x+y)(11+x)

b,\(b,255-4x^2-4xy-y^2=255-\left(2x+y\right)^2\)

\(=\left(\sqrt{255}-2x-y\right)\left(\sqrt{255}+2x+y\right)\)

Bài 2:

A=x²-y²-4x+4=(x²-4x+4)-y²=(x-2)²-y²=(x-2-y)(x-2+y)    (1)

Ta có: x+y=102 và x-y=72

\(\Rightarrow\hept{\begin{cases}x=\frac{102+72}{2}=87\\y=\frac{102-72}{2}=15\end{cases}}\)Thay vào (1) ta có:

A=(87-2-15)(87-2+15)=70.100=7000

Vậy A=7000

Bài 1:

\(a,=11\left(x+y\right)+x\left(x+y\right)=\left(x+11\right)\left(x+y\right)\\ b,=225-\left(2x+y\right)^2=\left(15-2x-y\right)\left(15+2x+y\right)\)

Bài 2:

\(A=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\\ A=\left(72-2\right)\left(120-2\right)=70\cdot118=8260\)

Bài 3:

\(a,\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\\ \Leftrightarrow24x+25=49\\ \Leftrightarrow24x=24\Leftrightarrow x=1\)

thk you very much UwU

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

câu a đê đúng ko vậy?

A/\(4x^2-12+9\)

\(=\left(2x\right)^2-2.2.3+3^2\)

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

B/\(11x+11y-x^2-xy\)

\(=\left(11x-x^2\right)+\left(11y-xy\right)\)

\(=x\left(11-x\right)+y\left(11-x\right)\)

\(=\left(11-x\right)\left(x+y\right)\)

C/\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

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

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

a, \(4x^2-12x+9=4x^2-6x-6x+9\)

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

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

b, \(11x+11y-x^2-xy=x\left(11-x\right)+y\left(11-x\right)\)

                                                 \(=\left(11-x\right)\left(x+y\right)\)

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

                                                        \(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)

                                                         \(=\left(a+b-c\right)\left(a+b+c\right)\left(c-a+b\right)\left(c+a-b\right)\)

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

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

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

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

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

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

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn.

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

b) \(x^2-2x-y^2+1=\left(x^2-2x+1\right)-y^2=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\)

c) \(4x^2-4xy+y^2=\left(2x\right)^2-2.2x.y+y^2=\left(2x-y\right)^2\)

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

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

Câu 1: (2x+y)(y-2x)+4x²=y²-4x²+4x²=y²

Với y=10 giá trị biểu thức trên là 10²=100

Câu 2: 

a. xy-11x=x.(y-11)

b. x²+4y²+4xy-16=(x²+4xy+4y²)-16

=(x+2y)²-16=(x+2y+4)(x+2y-4)

I don know