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( 6 x 3 – 5 x 2 + 4x – 1) : ( 2 x 2 – x + 1) = 3x – 1
Đáp án cần chọn là: A
a) (15x2-1+9x4-6x3+2x) :( 5 + 3x2-2x)
b) ( -19x+ 10+ 3x4- 5x2+11x3) : ( 3x+ x2-2)
c) (x4-14-x) : (x-2)
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) \(\left(5x^2-2x+1\right)\left(2x^2-3x\right)\)
\(=10x^4-15x^3-4x^3+6x^2+2x^2-3x\)
\(=10x^4-19x^3+8x^2-3x\)
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
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
a: \(\dfrac{2x^3-5x^2-x+1}{2x+1}\)
\(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}\)
\(=x^2-3x+1\)
b: \(\dfrac{x^3-2x+4}{x+2}\)
\(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}\)
\(=x^2-2x+2\)
\(a,x+5x^2=0\\ \Rightarrow a,x\left(1+5x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\\ b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=0\\ \Rightarrow x^2+6x+9+16-x^2=0\\ \Rightarrow6x+25=0\\ \Rightarrow6x=-25\\ \Rightarrow x=-\dfrac{25}{6}\)
\(c,5x\left(x-1\right)=x-1\\ \Rightarrow c,5x\left(x-1\right)-\left(x-1\right)\\ \Rightarrow\left(x-1\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ d,x^2-2x-3=0\\ \Rightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Rightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)
Ta có:
- 4 x 2 ( 6 x 3 + 5 x 2 – 3 x + 1 ) = ( - 4 x 2 ) . 6 x 3 + ( - 4 x 2 ) . 5 x 2 + ( - 4 x 2 ) . ( - 3 x ) + ( - 4 x 2 ) . 1 = - 24 x 5 – 20 x 4 + 12 x 3 – 4 x 2
Đáp án cần chọn là: C
:)) làm rồi :poop: