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`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

Bài 1:

Đặt \(t=2x^2+3x-1\) ta có:

\(t^2-5\left(t+4\right)+24=0\)

\(\Rightarrow t^2-5t-20+24=0\)

\(\Rightarrow t^2-5t+4=0\)

\(\Rightarrow\left(t-4\right)\left(t-1\right)=0\)\(\Rightarrow\left[\begin{matrix}t=4\\t=1\end{matrix}\right.\)

*)Xét \(2x^2+3x-1=4\)

\(\Rightarrow\left(x-1\right)\left(2x+5\right)=0\)\(\Rightarrow\left[\begin{matrix}x=1\\x=-\frac{5}{2}\end{matrix}\right.\)

*)Xét \(2x^2+3x-1=1\)

\(\Rightarrow\left(x+2\right)\left(2x-1\right)=0\)\(\Rightarrow\left[\begin{matrix}x=-2\\x=\frac{1}{2}\end{matrix}\right.\)

Bài 2:

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

\(\Rightarrow\left(x^2-4\right)\left(x+3\right)-\left(x^2-4\right)\left(x-1\right)=0\)

\(\Rightarrow\left(x^2-4\right)\left[x+3-\left(x-1\right)\right]=0\)

\(\Rightarrow4\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)\(\Rightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Đặt \(t=2x^2-3x-1\)

\(\Rightarrow t^2-3\left(t-4\right)-16=0\)

\(\Rightarrow t^2-3t+12-16=0\)

\(\Rightarrow t^2-3t-4=0\)

\(\Rightarrow\left\{{}\begin{matrix}t_1=-1\\t_2=4\end{matrix}\right.\)

\(TH_1:t=-1\)

\(\Leftrightarrow2x^2-3x-1=-1\)

\(\Leftrightarrow2x^2-3x=0\)

\(\Leftrightarrow x\left(2x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(TH_2:t=4\)

\(\Leftrightarrow2x^2-3x-1=4\)

\(\Leftrightarrow2x^2-3x-5=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=\dfrac{5}{2}\end{matrix}\right.\)

a) 2x(x+3) – 3x²(x+2) + x(3x² + 4x – 6)

= (2x . x + 2x . 3) – (3x² . x + 3x² . 2) + (x . 3x² + x . 4x – x . 6)

= 2x² + 6x – (3x³ + 6x²) + (3x³ + 4x² - 6x)

= 2x² + 6x – 3x³ – 6x² + 3x³ + 4x² - 6x

= (– 3x³ + 3x³ ) + (2x²  - 6x² + 4x² ) + (6x – 6x)

= 0 + 0 + 0

= 0

b) 3x(2x² – x) – 2x²(3x+1) + 5(x² – 1)

= [3x . 2x² + 3x . (-x)] – (2x² . 3x + 2x² . 1) + [5x² + 5 . (-1)]

= 6x³ – 3x² – (6x³ +2x²) + 5x² – 5

= 6x³ – 3x² – 6x³ - 2x² + 5x² – 5

= (6x³ – 6x³ ) + (-3x² – 2x² + 5x²) – 5

= 0 + 0 – 5

= - 5

Chọn D

a: \(=2x^3-14x^2-6x\)

c: \(=-10x^5-15x^4+25x^3\)

a) 2x. (x2 – 7x -3)

= 2x³- 14x²- 6x

b) ( -2x3 + y2 -7xy). 4xy2 

= -8x⁴y²+ 4xy⁴- 28x²y³

c)(-5x3).(2x2+3x-5)

= -10x⁵-15x⁴+25x³

d) (2x2 - xy+ y2).(-3x3)

=-6x⁵+ 3x⁴y -3x³y²

e)(x2 -2x+3). (x-4) 

=x³-2x²+3x -4x²+8x-12

=x³-6x²+11x-12

f) ( 2x3 -3x -1). (5x+2)

=10x⁴-15x²-5x +4x³-6x-2

=10x⁴+4x³-15x²-11x-2

\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)

Yêu cầu đề là gì vậy bạn?