c: \(x^3-8x^2+x+42\)

\(=x^3+2x^2-10x^2-20x+21x+42\)

\(=\left(x+2\right)\left(x^2-10x+21\right)\)

\(=\left(x+2\right)\left(x-3\right)\left(x-7\right)\)

a: \(x^3+6x^2+11x+6\)

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

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

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

a) Ta có: \(x^4+3x^3-7x^2-27x-18\)

\(=x^4-3x^3+6x^3-18x^2+11x^2-33x+6x-18\)

\(=x^3\left(x-3\right)+6x^2\left(x-3\right)+11x\left(x-3\right)+6\left(x-3\right)\)

\(=\left(x-3\right)\left(x^3+6x^2+11x+6\right)\)

\(=\left(x-3\right)\left(x^3+x^2+5x^2+5x+6x+6\right)\)

\(=\left(x-3\right)\left[x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)\right]\)

\(=\left(x-3\right)\left(x+1\right)\left(x^2+5x+6\right)\)

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

b) Ta có: \(x^3-8x^2+x+42\)

\(=x^3-7x^2-x^2+7x-6x+42\)

\(=x^2\left(x-7\right)-x\left(x-7\right)-6\left(x-7\right)\)

\(=\left(x-7\right)\left(x^2-x-6\right)\)

\(=\left(x-7\right)\left(x-3\right)\left(x+2\right)\)

c) Ta có: \(x^4+5x^3-7x^2-41x-30\)

\(=x^4+5x^3-7x^2-35x-6x-30\)

\(=x^3\left(x+5\right)-7x\left(x+5\right)-6\left(x+5\right)\)

\(=\left(x+5\right)\left(x^3-7x-6\right)\)

\(=\left(x+5\right)\left(x^3-x-6x-6\right)\)

\(=\left(x+5\right)\left[x\left(x^2-1\right)-6\left(x+1\right)\right]\)

\(=\left(x+5\right)\left[x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\right]\)

\(=\left(x+5\right)\left(x+1\right)\left(x^2-x-6\right)\)

\(=\left(x+5\right)\left(x+1\right)\left(x-3\right)\left(x+2\right)\)

a ) \(==>x^3.\left(x+3\right)-\left(7x^2+27x+18\right)\)

ko xét phần x^3.( x+3 ) nữa mà mik phân tích trong ngoặc nha zo thi ko lm như vậy mà ghi lại phần đó nha

\(7x^2+21x+6x+18\)

\(7x\left(x+3\right)+6\left(x+3\right)\)

\(\left(x+3\right)\left(7x+6\right)\)

==> \(x^3.\left(x+3\right)-\left(x+3\right)\left(7x+6\right)\)

==>\(\left(x+3\right)\left(x^3-7x-6\right)\)

1, x³+ 6x²+11x+6

= x³ + 2x² + 4x² + 8x + 3x + 6 

= x²(x + 2) + 4x(x + 2) + 3(x + 2)

= (x + 2)(x² + 4x + 3)

2, x⁴+3x³-7x²-27x-18

= x⁴ + 3x³ - 9x² + 2x² - 27x -18

= (x⁴ - 9x²) + (3x³ - 27x) + (2x² - 18)

= x²(x² - 9) + 3x(x² - 9) + 2(x² - 9)

= (x² - 9)(x² + 3x + 2)

= (x + 3)(x - 3)(x² + 3x + 2)

3, x³-8x²+x+42

= x³ - 7x² - x² + 7x - 6x + 42

= (x³ - 7x²) - (x² - 7x) - (6x - 42)

= x²(x - 7) - x(x - 7) - 6(x - 7)

= (x - 7)(x² - x - 6) 

4, x⁴+5x³-7x²-41x-30 

= x⁴ + x³ + 4x³ - 4x² - 11x² - 11x - 30x - 30

= (x⁴ + x³) + (4x³ - 4x²) - (11x² + 11x) - (30x + 30)

= x³(x + 1) + 4x²(x + 1) - 11x(x + 1) - 30(x + 1)

= (x³ + 4x² - 11x - 30)(x + 1)

5, x⁵+x-1

= x⁵ - x⁴ + x³ + x⁴ - x³ + x² - x²+ x -1 

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

= (x² - x + 1)(x³ + x² - 1)

6, x⁵-x⁴-1

= x⁵ - x³ - x² - x⁴ + x² + x + x³ - x - 1 

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

= (x² - x + 1)(x³ - x - 1)

1, x 3+ 6x 2+11x+6

= x 3 + 2x 2 + 4x 2 + 8x + 3x + 6

= x 2 ﴾x + 2﴿ + 4x﴾x + 2﴿ + 3﴾x + 2﴿

= ﴾x + 2﴿﴾x 2 + 4x + 3﴿

2, x 4+3x 3‐7x 2‐27x‐18

= x 4 + 3x 3 ‐ 9x 2 + 2x 2 ‐ 27x ‐18

= ﴾x 4 ‐ 9x 2 ﴿ + ﴾3x 3 ‐ 27x﴿ + ﴾2x 2 ‐ 18﴿

= x 2 ﴾x 2 ‐ 9﴿ + 3x﴾x 2 ‐ 9﴿ + 2﴾x 2 ‐ 9﴿

= ﴾x 2 ‐ 9﴿﴾x 2 + 3x + 2﴿

=﴾x + 3﴿﴾x ‐ 3﴿﴾x 2 + 3x + 2﴿

3, x 3‐8x 2+x+42

= x 3 ‐ 7x 2 ‐ x 2 + 7x ‐ 6x + 42

= ﴾x 3 ‐ 7x 2 ﴿ ‐ ﴾x 2 ‐ 7x﴿ ‐ ﴾6x ‐ 42﴿

= x 2 ﴾x ‐ 7﴿ ‐ x﴾x ‐ 7﴿ ‐ 6﴾x ‐ 7﴿

= ﴾x ‐ 7﴿﴾x 2 ‐ x ‐ 6﴿

4, x 4+5x 3‐7x 2‐41x‐30

= x 4 + x 3 + 4x 3 ‐ 4x 2 ‐ 11x 2 ‐ 11x ‐ 30x ‐ 30

= ﴾x 4 + x 3 ﴿ + ﴾4x 3 ‐ 4x 2 ﴿ ‐ ﴾11x 2 + 11x﴿ ‐ ﴾30x + 30﴿

= x 3 ﴾x + 1﴿ + 4x 2 ﴾x + 1﴿ ‐ 11x﴾x + 1﴿ ‐ 30﴾x + 1﴿

= ﴾x 3 + 4x 2 ‐ 11x ‐ 30﴿﴾x + 1﴿

5, x 5+x‐1

= x 5 ‐ x 4 + x 3 + x 4 ‐ x 3 + x 2 ‐ x 2+ x ‐1

= x 3 ﴾x 2 ‐ x + 1﴿+ x 2 ﴾x 2 ‐ x + 1﴿‐ ﴾x 2 ‐ x + 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 + x 2 ‐ 1﴿ 6, x 5‐x 4‐1

= x 5 ‐ x 3 ‐ x 2 ‐ x 4 + x 2 + x + x 3 ‐ x ‐ 1

= x 2 ﴾x 3 ‐ x ‐ 1﴿ ‐ x﴾x 3 ‐ x ‐ 1﴿ + ﴾x 3 ‐ x ‐ 1﴿

= ﴾x 2 ‐ x + 1﴿﴾x 3 ‐ x ‐ 1﴿ 

Câu 1: Đa thức không phân tích được thành nhân tử

Câu 2:

\(x^3-8x^2+x+42\)

\(=x^3+2x^2-10x^2-20x+21x+42\)

\(=x^2(x+2)-10x(x+2)+21(x+2)\)

\(=(x^2-10x+21)(x+2)\)

\(=(x^2-3x-7x+21)(x+2)\)

\(=[x(x-3)-7(x-3)](x+2)=(x-3)(x-7)(x+2)\)

Câu 3:

Đa thức không phân tích được thành nhân tử dù sửa dấu = thành cộng hoặc trừ.

a) x^4+3x^3-7x^2-27x-18 = x^4+3x^3-3x^2-4x^2-9x-12x-6x-18

                                         = (x^4+3x^3)-(3x^2+9x)-(4x^2+12x)-(6x+18)

                                         = x^3*(x+3)-3x*(x+3)-4x*(x+3)-6(x+3)

                                         = ( x^3-3x-4x-6)*(x+3)

Những câu kia chưa tính ra

a) \(x^4+3x^3-7x^2-27x-18\)

\(=\left(x^4+3x^3+2x^2\right)-\left(9x^2-27x-18\right)\)

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

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

b) \(x^4+5x^3-7x^2-41x-30\)

\(=\left(x^4+2x^3-15x^2\right)+\left(3x^3+6x^2-45x\right)+\left(2x^2+4x-30\right)\)

\(=x^2\left(x^2+2x-15\right)+3x\left(x^2+2x-15\right)+2\left(x^2+2x-15\right)\)

\(=\left(x^2+2x-15\right)\left(x^2+3x+2\right)=\left(x^2+5x-3x-15\right)\left(x^2+x+2x+2\right)\)

\(=\left(x+5\right)\left(x-3\right)\left(x+1\right)\left(x+2\right)\)

c) \(x^6-14x^4+49x^2-36\)

\(=\left(x^6-9x^4\right)+\left(-5x^4+45x^2\right)+\left(4x^2-36\right)\)

\(=x^4\left(x^2-9\right)-5x^2\left(x^2-9\right)+4\left(x^2-9\right)\)

\(=\left(x^2-9\right)\left(x^4-5x^2+4\right)=\left(x^2-9\right)\left(x^4-4x^2-x^2+4\right)\)

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

Dài 166

b) 2x²+3x-27=2x²-6x+9x-27=2x(x-3)+9(x-3)=(x-3)(2x+9)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2