a) -25x⁶ - y⁸ + 10x³y⁴ = -25x⁶ + 10x³y⁴ - y⁸

= - ( 25x⁶ - 10x³y⁴ + y⁸ )

= - [ ( 5x³ )² - 2 . 5x³y⁴ + ( y⁴ )² ]

= - ( 5x³ - y⁴ )²

b) \(\dfrac{1}{4}\)x² - 5xy + 25y² = (\(\dfrac{1}{2}\)x)² - 2 . \(\dfrac{1}{2}\) x . 5y + ( 5y )²

= (\(\dfrac{1}{2}\) x - 5y )²

c) ( x - 5 )² - 16 = ( x - 5 )² - 4²

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

= ( x - 9 ) . ( x - 1 )

d) 25 - ( 3 - x )² = 5² - ( 3 - x )²

= ( 5 - 3 + x ) . ( 5 + 3 - x )

= ( x + 2 ) . ( 8 - x )

3x^2(5x^2-7x+4)

=15x^4-21x^3+12x^2

xy^2(2x^2y-5xy+y)

=2x^3y^3-5x^2y^3+xy^3

(2x^2-5x)(3x^2-2x+1)

=6x^4-4x^3+2x^2-15x^3+10x^2-5x

=6x^4-19x^3+12x^2-5x

(x-3y)(2xy+y^2+x)

=2x^2y+xy^2+x^2-6xy^2-3y^3-3xy

=-3y^3+2x^2y-5xy^2+x^2-3xy

1, \(25x^2-10xy+y^2=\left(5x-y\right)^2\)

2, \(8x^3+36x^2y+54xy^2+27y^3=\left(2x+3y\right)^3\)

4, \(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)-a^3-b^3-c^3\)

\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

5, \(2x^3+3x^2+2x+3\)

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

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

6, \(x^3z+x^2yz-x^2z^2-xyz^2\)

\(=x^3z-x^2z^2+x^2yz-xy^2\)

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

\(=xz\left[x\left(x-z\right)+y\left(x-z\right)\right]\)

\(=xz\left(x+y\right)\left(x-z\right)\)

8, \(x^3+3x^2y+3xy^2+y+y^3\)\(=\left(x+y\right)^3+y\)

9, \(x^2-6x+8\)

\(=x^2-4x-2x+8\)

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

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

10, \(x^2-8x+12\)

\(=x^2-6x-2x+12\)

\(=x\left(x-6\right)-2\left(x-6\right)\)

\(=\left(x-2\right)\left(x-6\right)\)

Chỗ còn lại mai làm nốt nha.

Gặp chút sự cố đăng nhập nên hơi muộn, xin lỗi nha

11, \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)

\(=a^2b-ab^2+abc-a^2c+b^2c-abc+ac^2-c^2b\)

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

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

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

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

12, \(x^3-7x-6\)

\(=x^3-3x^2+3x^2-9x+2x-6\)

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

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

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

\(=\left(x-3\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]\)

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

13, \(x^4+4\)

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

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

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

14, \(a^4+64\)

\(=a^4+16a^2+64-16a^2\)

\(=\left(a^2+8\right)^2-16a^2\)

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

15, \(x^5+x+1\)

\(=x^5-x^2+x^2+x+1\)

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

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

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

16, \(x^5+x-1\)

\(=x^5-x^4+x^3+x^4-x^3+x^2-x^2+x-1\)

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

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

17, \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

\(=\left(x^2+x\right)\left(x^2+x-2\right)-15\)

19, \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\) (*)

Đặt \(x^2+8x+7=a\) ta có:

(*) \(\Leftrightarrow a\left(a+8\right)+15\)

\(\Leftrightarrow a^2+8a+15\)

\(\Leftrightarrow a^2+3a+5a+15\)

\(\Leftrightarrow a\left(a+3\right)+5\left(a+3\right)\)

\(\Leftrightarrow\left(a+3\right)\left(a+5\right)\)

Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

20, \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\) (*)

Đặt \(x^2+3x+1=a\) ta có:

(*) \(\Leftrightarrow a\left(a+1\right)-6\)

\(\Leftrightarrow a^2+a-6\)

\(\Leftrightarrow a^2+3a-2a-6\)

\(\Leftrightarrow a\left(a+3\right)-2\left(a+3\right)\)

\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\)

Trả lại biến cũ ta có: (*) \(\Leftrightarrow\left(x^2+3x-1\right)\left(x^2+3x+5\right)\)

a)\(x^3y^3+x^2y^2+4\)

\(=x^3y^3-x^2y^2+2xy+2x^2y^2-2xy+4\)

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

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

b)\(x^4+x^3+6x^2+5x+5\)

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

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

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

c)\(x^4-2x^3-12x^2+12x+36\)

\(=x^4-2x^3-6x^2-6x^2+12x+36\)

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

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

d)\(x^8y^8+x^4y^4+1\)

\(=x^8y^8+2x^4y^4+1-x^4y^4\)

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

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

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

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

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

bạn lm pb = cách nhẩm nghiệm đc không

Bài 1:

a) 25x² - 10xy + y² = (5x - y)²

b) 81x² - 64y² = (9x)² - (8y)² = (9x - 8y)(9x + 8y)

c) 8x³ + 36x²y + 54xy² + 27y³

= 8x³ + 27y³ + 36x²y + 54xy²

= (2x + 3y)(4x² - 6xy + 9y²) + 18xy(2x + 3y)

= (2x + 3y)(4x² - 6xy + 18xy + 9y²)

= (2x + 3y)(4x² + 12xy + 9y²)

= (2x + 3y)(2x + 3y)² = (2x + 3y)³

c) (a² + b² - 5)² - 4(ab + 2)² = (a² + b² - 5)² - 2²(ab + 2)²

= (a² + b² - 5)² - (2ab + 4)²

= (a² + b² - 5 - 2ab - 4)(a² + b² - 5 + 2ab + 4)

= (a² - 2ab + b² - 9)(a² + 2ab + b² - 1)

= \(\left [ (a - b)^{2} - 3^{2} \right ]\)\(\left [ (a + b)^{2} - 1\right ]\)

= (a - b - 3)(a - b + 3)(a + b - 1)(a + b + 1)

pn đăng mỗi lần vài bài thôi chứ đăng nhìn ngán lắm

Bài 2:

a) 2x³ + 3x² + 2x + 3

= 2x³ + 2x + 3x² + 3

= 2x(x² + 1) + 3(x² + 1)

= (x² + 1)(2x + 3)

b)x³z + x²yz - x²z² - xyz²

= xz(x² + xy - xz - yz)

= \(xz\left [ x(x + y) - z(x + y) \right ]\)

= xz(x + y)(x - z)

c) x²y + xy² - x - y

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

= (x + y)(xy - 1)

d) 8xy³ - 5xyz - 24y² + 15z

= 8xy³ - 24y² - 5xyz + 15z

= 8y²(xy - 3) - 5z(xy - 3)

= (xy - 3)(8y² - 5z)

e) x³ + y(1 - 3x²) + x(3y² - 1) - y³

= x³ - y³ + y - 3x²y + 3xy² - x

= (x - y)(x² + xy + y²) - 3xy(x - y) - (x - y)

= (x - y)(x² + xy + y² - 3xy - 1)

= (x - y)(x² - 2xy + y² - 1)

= \((x - y)\left [ (x - y)^{2} - 1 \right ]\)

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

câu f tương tự

Câu 1:

a/ (-5x³)(2x²+3x-5)

=-10x⁵-15x⁴+25x³

b/(2x-1)x

=2x²-x

c/(x-y)(3x²+4xy)

=3x³+4x²y-3x²y-4xy²

=3x³ +x²y-4xy²

Câu 2:

a/ x³-2x²+x

=x(x²-2x+1)

=x(x-1)²

b/x²-x-12

=x² +3x-4x-12

=(x² +3x)+(-4x-12)

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

=(x+3)(x-4)

c/ 2x-6

=2(x-3)

e/ x²+4x+4-y²

=(x²+4x+4)-y²

=(x+2)²-y²

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

d/ x²-2xy+y²-16

=(x²-2xy+y²⁾-16

=(x-y)²-16

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

Câu 3:

a: \(=\dfrac{5xy-4+3xy+4}{2x^2y^3}=\dfrac{8xy}{2x^2y^3}=\dfrac{4}{xy^2}\)

b: \(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)

\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)

c: \(=\dfrac{3x+1-2x+3}{x+y}=\dfrac{x+4}{x+y}\)

d: \(=\dfrac{4x+7+5x+7}{9}=\dfrac{9x+14}{9}\)

e: \(=\dfrac{5\left(x+2\right)}{2\left(2x-1\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-5\left(x-2\right)}{2x-1}\)