(2x2+3x-1)2-5(2x2+3x+3)+24=0
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PT
2
15 tháng 1 2017
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 \(2x^2+3x-1=1\)
\(\Rightarrow\left(x+2\right)\left(2x-1\right)=0\)\(\Rightarrow\left[\begin{matrix}x=-2\\x=\frac{1}{2}\end{matrix}\right.\)
15 tháng 1 2017
Bài 2:
\(\left(x^2-4\right)\left(x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Rightarrow\left(x^2-4\right)\left(x+3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left[x+3-\left(x-1\right)\right]=0\)
\(\Rightarrow4\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+2\right)=0\)\(\Rightarrow\left[\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Y
12 tháng 2 2023
Đặt \(t=2x^2-3x-1\)
\(\Rightarrow t^2-3\left(t-4\right)-16=0\)
\(\Rightarrow t^2-3t+12-16=0\)
\(\Rightarrow t^2-3t-4=0\)
\(\Rightarrow\left\{{}\begin{matrix}t_1=-1\\t_2=4\end{matrix}\right.\)
\(TH_1:t=-1\)
\(\Leftrightarrow2x^2-3x-1=-1\)
\(\Leftrightarrow2x^2-3x=0\)
\(\Leftrightarrow x\left(2x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(TH_2:t=4\)
\(\Leftrightarrow2x^2-3x-1=4\)
\(\Leftrightarrow2x^2-3x-5=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=\dfrac{5}{2}\end{matrix}\right.\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
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
9 tháng 12 2021
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
27 tháng 10 2021
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
Đặt \(2x^2+3x+1=y\).Ta có:
\(\left(y-2\right)^2-5\left(y+2\right)+24=0\)
\(\Leftrightarrow y^2-4y+4-5y-10+24=0\)
\(\Leftrightarrow y^2-9y+18=0\)
\(\Leftrightarrow\left(y-3\right)\left(y-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y=3\\y=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2+3x+1=3\\2x^2+3x+1=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-1\right)\left(x+2\right)=0\\\left(2x+5\right)\left(x-1\right)=0\end{cases}}\)
Vậy PT có 4 nghiệm là:\(\frac{1}{2}\)\(,\)\(-2,-\frac{5}{2},1\)