a) x²y+xy+x+1= (x²y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)

b) x²-(a+b)x+ab=x²-ax-bx+ab=(x²-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)

c) ax²+ay-bx²-by=(ax²+ay)-(bx²+by)=a(x²+y)-b(x²+y)=(a-b)(x²+y)

d) ax-2x-a²+2a=(ax-2x)-(a²-2a)=x(a-2)-a(a-2)=(a-2)(x-a)

e) 2x²+4ax+x+2a=(2x²+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)

f) x³+ax²+x+a=(x³+ax²)+(x+a)=x²(x+a)+(x+a)=(x²+1)(x+a)

g: Ta có: \(x^4+2x^3-4x-4\)

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

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

TK
 

a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).

b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2

= (a + 4 – b)(a + 4 + b).

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

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

b: \(=\left(x-4\right)\left(x+3\right)\)

e: =(x+3)(x-2)

a) \(=x^2\left(2x+3\right)+\left(2x+3\right)=\left(2x+3\right)\left(x^2+1\right)\)

b) \(=x\left(x-4\right)+3\left(x-4\right)=\left(x-4\right)\left(x+3\right)\)

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

d) \(=4xy\left(y-3x+2\right)\)

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

f) \(=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-4z^2\right]=x\left(x+y-2z\right)\left(x+y+2z\right)\)

g) \(=x\left(x^2-2xy+y^2-25\right)=x\left[\left(x-y\right)^2-25\right]=x\left(x-y-5\right)\left(x-y+5\right)\)

h) \(=x\left(x+1\right)-3\left(x+1\right)=\left(x+1\right)\left(x-3\right)\)

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

từng câu 1 thôi:v

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

h: \(=\left(a-1-b\right)\left(a-1+b\right)\)

a: \(x^4-4x^3-8x^2+8x\)

\(=x\left(x^3-4x^2-8x+8\right)\)

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

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

b: \(x^2-1-xy+y\)

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

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

c: Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)^2\cdot\left(x-2\right)\)

\(=\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3-x-1\right)\)

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

Câu 2:

a: \(\left(x-1\right)\left(x+1\right)-x\left(x+3\right)+7=0\)

=>\(x^2-1-x^2-3x+7=0\)

=>-3x+6=0

=>-3x=-6

=>\(x=\dfrac{-6}{-3}=2\)

b: \(2x^3-22x^2+36x=0\)

=>\(2x\left(x^2-11x+18\right)=0\)

=>\(x\left(x^2-11x+18\right)=0\)

=>\(x\left(x^2-2x-9x+18\right)=0\)

=>\(x\left[x\left(x-2\right)-9\left(x-2\right)\right]=0\)

=>\(x\left(x-2\right)\left(x-9\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x-2=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=9\end{matrix}\right.\)

Câu 4:

1: Diện tích cỏ cần thay là:

\(105\cdot68=7140\left(m^2\right)\)

Số tiền BQL sân cần trả là:

\(7140\cdot120000=856800000\left(đồng\right)\)

2:

a: Xét tứ giác ABDC có

M là trung điểm chung của AD và BC

=>ABDC là hình bình hành

Hình bình hành ABDC có \(\widehat{BAC}=90^0\)

nên ABDC là hình chữ nhật

b: Xét ΔADE có

H,M lần lượt là trung điểm của AE,AD

=>HM là đường trung bình của ΔADE
=>HM//DE

=>BC//DE

=>\(\widehat{EDB}=\widehat{DBM}\)(hai góc so le trong)(1)

Ta có: ABDC là hình chữ nhật

=>AD=BC

mà \(MD=\dfrac{AD}{2};MB=\dfrac{BC}{2}\)

nên MD=MB

=>ΔMBD cân tại M

=>\(\widehat{MDB}=\widehat{MBD}\left(2\right)\)

Từ (1) và (2) suy ra \(\widehat{MDB}=\widehat{EDB}\)

=>\(\widehat{ADB}=\widehat{EDB}\)

=>DB là phân giác của góc ADE

a) x³y + x - y - 1

= (x³y - y) + (x - 1)

= y(x³ - 1) + (x - 1)

= y(x - 1)(x² + x + 1) + (x - 1)

= (x - 1)[y(x² + x + 1) + 1]

= (x - 1)(x²y + xy + y + 1)

b) x²(x - 2) + 4(2 - x)

= x²(x - 2) - 4(x - 2)

= (x - 2)(x² - 4)

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

= (x - 2)²(x + 2)

c) x³ - x² - 20x

= x(x² - x - 20)

= x(x² + 4x - 5x - 20)

= x[(x² + 4x) - (5x + 20)]

= x[x(x + 4) - 5(x + 4)]

= x(x + 4)(x - 5)

d) (x² + 1)² - (x + 1)²

= (x² + 1 - x - 1)(x² + 1 + x + 1)

= (x² - x)(x² + x + 2)

= x(x - 1)(x² + x + 2)

e) 6x² - 7x + 2

= 6x² - 3x - 4x + 2

= (6x² - 3x) - (4x - 2)

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

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

f) x⁴ + 8x² + 12

= x⁴ + 2x² + 6x² + 12

= (x⁴ + 2x²) + (6x² + 12)

= x²(x² + 2) + 6(x² + 2)

= (x² + 2)(x² + 6)

g) (x³ + x + 1)(x³ + x) - 2

Đặt u = x³ + x

x³ + x + 1 = u + 1

(u + 1).u - 2

= u² + u - 2

= u² - u + 2u - 2

= (u² - u) + (2u - 2)

= u(u - 1) + 2(u - 1)

= (u - 1)(u + 2)

= (x³ + x - 1)(x³ + x + 2)

= (x³ + x - 1)(x³ + x² - x² - x + 2x + 2)

= (x³ + x - 1)[(x³ + x²) - (x² + x) + (2x + 2)]

= (x³ + x - 1)[x²(x + 1) - x(x + 1) + 2(x + 1)]

= (x³ + x - 1)(x - 1)(x² - x + 2)

h) (x + 1)(x + 2)(x + 3)(x + 4) - 1

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

= (x² + 5x + 4)(x² + 5x + 6) - 1 (1)

Đặt u = x² + 5x + 4

u + 2 = x² + 5x + 6

(1) u.(u + 2) - 1

= u² + 2u - 1

= u² + 2u + 1 - 2

= (u² + 2u + 1) - 2

= (u + 1)² - 2

= (u + 1 + √2)(u + 1 - √2)

= (x² + 5x + 4 + 1 + √2)(x² + 5x + 4 + 1 - √2)

= (x² + 5x + 5 + √2)(x² + 5x + 5 - √2)

i: \(-\left(x^2+2\right)^2+4x\left(x^2+2\right)-3x^2\)

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

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

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

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

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

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

l: \(81x^4+4y^4\)

\(=81x^4+36x^2y^2+4y^4-36x^2y^2\)

\(=\left(81x^4+36x^2y^2+4y^4\right)-\left(6xy\right)^2\)

\(=\left[\left(9x^2\right)^2+2\cdot9x^2\cdot2y^2+\left(2y^2\right)^2\right]-\left(6xy\right)^2\)

\(=\left(9x^2+2y^2\right)^2-\left(6xy\right)^2\)

\(=\left(9x^2+2y^2+6xy\right)\left(9x^2+2y^2-6xy\right)\)

\(a,=x\left(x-8\right)\\ b,=x\left(x-y\right)-6\left(x-y\right)=\left(x-6\right)\left(x-y\right)\\ c,=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\\ d,=\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2+2\right)\)

a: \(x^2-8x=x\left(x-8\right)\)

c: \(x^2-6x+9-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

\(a,4x^2-4x+1\\ =\left(2x\right)^2-2.2x+1^2=\left(2x-1\right)^2\\ c,x^2-6xy-25z^2+9y^2\\ =\left(x^2-2.x.3y+9y^2\right)-\left(5z\right)^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left(x-3y-5z\right)\left(x-3y+5z\right)\)

Xem lại đề ý b

\(a,=x\left(x+y\right)+5\left(x+y\right)=\left(x+5\right)\left(x+y\right)\\ b,=x\left(y-x\right)-3\left(y-x\right)=\left(x-3\right)\left(y-x\right)\\ c,=18x-4x^3=2x\left(9-2x^2\right)\\ d,=\left(x-2\right)^2-4y^2=\left(x-2y-2\right)\left(x+2y-2\right)\\ e,=x^2-x-9x+9=\left(x-1\right)\left(x-9\right)\\ f,=4x^2-6x+2x-3=\left(2x-3\right)\left(2x+1\right)\)