( 6 x 3   –   5 x 2 + 4x – 1) : ( 2 x 2 – x + 1) = 3x – 1  

Đáp án cần chọn là: A

\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)

\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)

A(x)=(7x²-5x²-2x²)-(6x³-10x³)+2x-12

A(x)=-4x³+2x-12

Sắp xếp:-4x³+2x-12

\(A\left(x\right)=\left(7x^2-5x^2-2x^2\right)-\left(6x^3-10x^3\right)+2x-12\)

\(A\left(x\right)=-4x^3+2x-12\)

Sắp xếp:\(-4x^3+2x-12\)

Bạn xem đã viết đúng đề chưa vậy?

a: Sửa đề: \(\left(6x^3+x^2\right):2x-3x+2=0\)

ĐKXĐ: x<>0

Ta có: \(\left(6x^3+x^2\right):2x-3x+2=0\)

=>\(\frac{6x^3}{2x}+\frac{x^2}{2x}-3x+2=0\)

=>\(3x+\frac12x-3x+2=0\)

=>\(\frac12x+2=0\)

=>\(\frac12x=-2\)

=>x=-4(nhận)

b: Sửa đề: \(\left(5x^4-3x^2\right):x^2-x\left(5x+6\right)=0\)

ĐKxĐ: x<>0

Ta có: \(\left(5x^4-3x^2\right):x^2-x\left(5x+6\right)=0\)

=>\(5x^2-3-5x^2-6x=0\)

=>-6x-3=0

=>6x+3=0

=>3(2x+1)=0

=>2x+1=0

=>2x=-1

=>\(x=-\frac12\) (nhận)

c: \(\dfrac{x^4-x-14}{x-2}\)

\(=\dfrac{x^4-2x^3+2x^3-4x^2+4x^2-8x+7x-14}{x-2}\)

\(=x^3+2x^2+4x+7\)

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

\(=10x^4-15x^3-4x^3+6x^2+2x^2-3x\)

\(=10x^4-19x^3+8x^2-3x\)

b) \(\left(18x^4y^3-6x^2y^3+12x^3y^4z\right)\)

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

\(\Rightarrow\left(18x^4y^3-6x^2y^3+12x^3y^4z\right):6x^2y^3\)

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

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

a)(5x²-2x+1).(2x²-3x)

=10x⁴-4x³+2x²-15x³+6x²-3x

=10x⁴-19x³+8x²-3x

b)(18x⁴y³-6x²y³+12x³y⁴z):6x²y³

=(18x⁴y³:6x²y³)-(6x²y³:6x²y³)+(12x³y⁴z:6x²y³)

=3x²y-xy+2xyz

c) x ∈ Z và  x 2 - 5 x 2 - 24 < 0

Ta có:  x 2 - 5 > 0 ; x 2 - 24 < 0 ⇒ x 2 > 5 ; x 2 < 24 Nên  x 2 ∈ 9 ; 16

x 2 = 9 ⇒ x = ± 3 ; x = 16 ⇒ x = ± 4

Vậy  x ∈ - 3 ; 3 ; - 4 ; 4