30ax - 34bx - 15a + 17b

=2x(15a-17b)-1(15a-17b)

=(15a-17b)(2x-1).

​

\(30ax-34bx-15a+17b.\)

\(=2\times\left(15a-17b\right)-1\times\left(15a-17b\right)\)

\(=\left(15a-17b\right)\times\left(2x-1\right)\)

Ta có: \(30ax-34bx-15a-17b\)

\(=2x\left(15a-17b\right)-1\left(15a-17b\right)\)

\(=\left(2x-1\right)\left(15a-17b\right)\)

c)

\(30ax-34bx-15a+17b\)

\(=(30ax-15a)-(34bx-17b)\)

\(=15a(2x-1)-17b(2x-1)\)

\(=(2x-1)(15a-17b)\)

d)

\(x^3-x^2y-x^2z-xyz\)

\(=x[x^2-xy-xz-yz]\)

a)

\(12x^3+4x^2-27x-9\)

\(=(12x^3+4x^2)-(27x+9)\)

\(=4x^2(3x+1)-9(3x+1)\)

\(=(3x+1)(4x^2-9)=(3x+1)[(2x)^2-3^2]\)

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

b)

\(x^6-x^4+2x^3+2x^2\)

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

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

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

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

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

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

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

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

Phân tích đa thcusw thành nhân tử (Phương pháp nhân hạng tử):

1. 3a - 3b + ax - bx

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

= (a - b)(3 + x)

2. x³ + 3x² + 3x + 9

= x³ + 3x² + 3x + 1 + 8

= (x + 1)³ + 8

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

= (x + 3)(x² + 2x + 1 - 2x - 2 + 4)

= (x + 3)(x² + 3)

3. 10ay - 5by + 2ax - bx

= 5y(2a - b) + x(2a - b)

= (2a - b)(5y + x)

4. 5a² - 5ax - 7a + 7x

= 5a(a - x) - 7(a - x)

= (a - x)(5a - 7)

5. x² - 3xy + x - 3x

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

= x(x - 3y - 2)

6. 7x² - 7xy - 4x + 4y

= 7x(x - y) - 4(x - y)

= (x - y)(7x - 4)

7. 3ax - 4by - 4ay + 3bx

= (3ax + 3bx) - (4ay + 4by)

= 3x(a + b) - 4y(a + b)

= (a + b)(3x - 4y)

8. 30ax - 34bx - 15a + 17b

= (30ax - 15a) - (34bx -17b )

= 15a(x - 1) - 17b(x - 1)

= (x - 1)(15a - 17b)

9. 3x² - 3xy + 3y² - 3xy

= 3(x² - xy + y² - xy)

= 3(x² - 2xy + y²)

= 3(x - y)²

10. 12a² - 6ab + 3b² - 6ab

= 3(4a² - 2ab + b² - 2ab)

= 3(4a² - 4ab + b²)

= 3(2a - b)²

Có : B=\(15a-17b+4a-2b\)

\(\Rightarrow B=15a+4a-17b-2b\)

\(\Rightarrow B=19a-19b\)

\(\Rightarrow B=19x\left(a-b\right)\)

Mà a-b=20

\(\Rightarrow B=19x20\)

\(\Rightarrow B=380\)

Thay b = a - 20

=> B = 15a - 17(a-20) - 4a - 2(a - 20)

        = 15a - 17a - 340 - 4a - 2a - 40

        = (15 - 17- 4 - 2)a - (340 - 40)

        = -8a - 300

Có : ( 16a + 17b ) ( 17a + 16b ) : 11 ( vì 11 là số nguyên tố )

= 16a + 17b : 11

    17a + 16b : 11

=G/s 16a + 17b : 11₍₁₎

Mà ( 16a + 17b ) + ( 17a + 16b ) = ( 33a + 33b ) = 11 ( 3a + 3b ) : 11

= 17a + 16b : 11₍₂₎

Từ ( 1 ) , ( 2 ) = ( 16a + 17b ) ( 17a  +16b ) : 121

Ta có: \(\left(16a+17b\right)\left(17a+16b\right)⋮11\)

\(\Rightarrow\orbr{\begin{cases}16a+17b⋮11\\17a+16b⋮11\end{cases}}\)

Giả sử \(16a+17b⋮11\)

\(\Rightarrow16a+17b+17a+16b=\left(16a+17a\right)+\left(17b+16b\right)=33a+33b=33\left(a+b\right)\)

Vì \(33⋮11\) nên \(33\left(a+b\right)⋮11\)

Mà \(16a+17b⋮11\)

\(\Rightarrow17a+16b⋮11\)

Lại có: 11 là số nguyên tố

\(\Rightarrow\left(16a+17b\right)\left(17a+16b\right)⋮11^2=121\)

Vậy \(\left(16a+17b\right)\left(17a+16b\right)⋮121\).

Cách làm tương tự: Câu hỏi của lekhanhhung - Toán lớp 7 - Học toán với OnlineMath