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2 tháng 7 2021
1) (2x + 1)(3x – 2) = (5x – 8)(2x + 1)
⇔ (2x + 1)(3x – 2) – (5x – 8)(2x + 1) = 0
⇔ (2x + 1).[(3x – 2) – (5x – 8)] = 0
⇔ (2x + 1).(3x – 2 – 5x + 8) = 0
⇔ (2x + 1)(6 – 2x) = 0
⇔\(\left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=3\end{matrix}\right.\)
Vậy.....
2) 4x2 -1 = (2x + 1)(3x - 5)
⇔ (2x-1)(2x+1)-(2x+1)(3x-5)=0
⇔ (2x+1)(2x-1-3x+5)=0
⇔ (2x+1)(4-x)=0
⇔ \(\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=4\end{matrix}\right.\)
Vậy...
3)
(x + 1)2 = 4(x2 – 2x + 1)
⇔ (x + 1)2 - 4(x2 – 2x + 1) = 0
⇔ x2 + 2x +1- 4x2 + 8x – 4 = 0
⇔ - 3x2 + 10x – 3 = 0
⇔ (- 3x2 + 9x) + (x – 3) = 0
⇔ -3x (x – 3)+ ( x- 3) = 0
⇔ ( x- 3) ( - 3x + 1) = 0
⇔\(\left[{}\begin{matrix}x-3=0\\-3x+1=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy......
5 tháng 7 2023
A) -2x(3x+2)(3x-2)+5(x+2)2 - (x-1)(2x+1)(2x+1)
= -2x(9x2-4)+5(x2+4x+4) - (x-1)(4x2-1)
= -18x3+8x+5x2+20x+20-(4x3-x-4x2+1)
= -18x3+5x2+28x+20-4x3+x+4x2+1
= -22x3+9x2+29x+21
B) (7x-8)(7x+8)-10(2x+3)2+5x(3x-2)2-4x(x-5)2
= 49x2 - 64 -10(4x2+ 12x + 3) + 5x(9x2 - 12x +4) - 4x(x2 - 10x +25)
= 49x2 - 64 -40x2 - 120x - 30 + 45x3 - 60x2 - 20x - 4x3 + 40x2 -100x
= 41x3 -11x2 -240x -94
6 tháng 7 2023
C) \(\left(x^2-3\right)\left(x^2+3\right)-5x^2\left(x+1\right)^2-\left(x^2-3x\right)\left(x^2-2x\right)+4x\left(x+2\right)^2\)
\(\left(x^4-9\right)-5x^2\left(x^2+2x+1\right)-\left(x^4-2x^3-3x^3+6x^2\right)+4x\left(x^2+4x+4\right)\)
\(x^4-9-5x^4-10x^3-5x^2-x^4+5x^3-6x^2+4x^3+16x^2+16x\)
\(-5x^4-x^3+5x^2+20x-9\)
D) \(-6x^2\left(x+5\right)^2-\left(x-3\right)^2+\left(x^2-2\right)\left(2x^2+1\right)-4x^2\left(3x-4\right)^2\)
\(-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)
\(-6x^4-60x^3+150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)
\(-40x^4+36x^3+82x^2+6x-11\)
4 tháng 1 2023
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2=x.5x^2+x.\left(-3x\right)+x.2+\left(-1\right).5x^2+\left(-1\right)\left(-3x\right)+\left(-1\right).2=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)+(−1).2
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
4 tháng 1 2023
(x−1)(5x2−3x+2)=x(5x2−3x+2)−1(5x2−3x+2)
=x.5x2+x.(−3x)+x.2+(−1).5x2+(−1)(−3x)
=5x3−3x2+2x−5x2+3x−2=5x^3-3x^2+2x-5x^2+3x-2=5x3−3x2+2x−5x2+3x−2
=5x3−8x2+5x−2=5x^3-8x^2+5x-2=5x3−8x2+5x−2.
CM
23 tháng 12 2018
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
MS
phân tích các đa thức thành nhân tử
a) ( x2 + 4)2- 16x2
b) ( x+3)2- 8x3
c) (4x2-3x -18)2- ( 4x2 + 3x)2
1
LN
10 tháng 10 2021
\(a,\left(x^2+4\right)^2-16x^2=\left(x^2+4\right)-\left(4x\right)^2=\left(x^2+4-4x\right).\left(x^2+4+4x\right)=\left(x-2\right)^2.\left(x+2\right)^2\)
\(b,\left(x+3\right)^3-8x^3=\left(x+3\right)^3-\left(2x\right)^3=\left(x+3-2x\right).\left[x^2+\left(x+3\right).2x+\left(2x\right)^2\right]=\left(3-x\right).\left(x^2+2x^2+6x+4x^2\right)\)
\(c,\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=\left(4x^2-3x-18-4x^2-3x\right).\left(4x^2-3x-18+4x^2+3x\right)=\left(-6x-18\right).\left(8x^2-18\right)\)
HC
1-3x/2x + 3x-2/2x-1 + 2-3x/4x2-2x
2
TM
20 tháng 12 2021
\(2x;2x-1;4x^2-2x=2x\left(2x-1\right)\)
\(MTC=2x\left(2x-1\right)\)
\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{2-3x}{4x^2-2x}\)
\(=\dfrac{\left(1-3x\right).2x\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{\left(3x-2\right).2x}{\left(2x-1\right).2x}+\dfrac{2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{2x\left(1-3x\right)\left(2x-1\right)+2x\left(2x-2\right)+2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{-8x^2+4x+4x^2-4x+2-3x}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2-3x+2}{2x\left(2x-1\right)}\)
#AEZn8
20 tháng 12 2021
\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{2-3x}{4x^2-2x}=\dfrac{\left(1-3x\right)\left(2x-1\right)}{2x\left(2x-1\right)}+\dfrac{2x\left(3x-2\right)}{2x\left(2x-1\right)}+\dfrac{2-3x}{2x\left(2x-1\right)}=\dfrac{-6x^2+5x-1}{2x\left(2x-1\right)}+\dfrac{6x^2-4x}{2x\left(2x-1\right)}+\dfrac{2-3x}{2x\left(2x-1\right)}=\dfrac{\left(-6x^2+6x^2\right)+\left(5x-4x-3x\right)+\left(-1+2\right)}{2x\left(2x-1\right)}=\dfrac{-2x}{2x\left(2x-1\right)}=\dfrac{-1}{2x-1}\)
em c.ơn trước
25 tháng 3 2020
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
=>(5x^2+3x-2-4x^2+3x+2)(5x^2+3x+2+4x^2-3x-2)=0
=>(x^2+6x)(9x^2)=0
=>x=0; x=-6
\(\left(5x^2+3x-2\right)^2=\left(4x^2-3x-2\right)^2\)
\(\Leftrightarrow5x^2+3x-2=4x^2-3x-2\)
\(\Leftrightarrow5x^2+3x-2-4x^2+3x+2=0\)
\(\Leftrightarrow x^2+6x=0\)
\(\Leftrightarrow x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)