a) (x5+4x3-6x2):4x2
b)x(2x2-3)-x2(5x+1)+x2
c)(x-2)2-(x-1)(x+1)-x(1-x)
d)\(\frac{1}{2}\)x2(6x-3)-x(x2+ \(\frac{1}{2}\))+ \(\frac{1}{2}\)(x+4)
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Những câu hỏi liên quan
10 tháng 12 2020
a) \(\left(x^5+4x^3-6x^2\right):4x^2\)
\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)
\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)
b)
Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)
c) (-2x5 : 2x2) + (3x2 : 2x2) + (-4x^3 : 2x^2)
= \(-x^3+\dfrac{3}{2}-2x\)
d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)
\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)
\(=x-4\)
(dùng hẳng đẳng thức thứ 7)
Bài 2 :
a) 3x(x - 2) - 5x(1 - x) - 8(x2 - 3)
= 3x2 - 6x - 5x + 5x2 - 8x2 + 24
= (3x2 + 5x2 - 8x2) + (-6x - 5x) + 24
= -11x + 24
b) (x - y)(x2 + xy + y2) + 2y3
= x3 - y3 + 2y3
= x3 + y3
c) (x - y)2 + (x + y)2 - 2(x - y)(x + y)
= (x - y)2 - 2(x - y)(x + y) + (x + y)2
= [(x - y) + x + y)2 = [x - y + x + y] = (2x)2 = 4x2
18 tháng 10 2021
Bài 1 :
a]= \(\frac{1}{4}\)x3 + x - \(\frac{3}{2}\).
b] => [x3 + x2 -12 ] = [ x2 +3 ][x-2] + [-6]
c]= -x3 -2x +\(\frac{3}{2}\).
d] = [ x3 - 64 ] = [ x2 + 4x + 16][ x- 4].
23 tháng 2 2022
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
9 tháng 9 2021
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
9 tháng 9 2021
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
14 tháng 3 2020
Bài 2
a. (x-2y)2 =2x-4y
b. (2x^2 +3)2 =4x^2+6
c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)
d. (2x-1)3 = 6x-3
Xin lỗi mik chỉ lm ổn bài 2 thôi!
DT
18 tháng 2 2022
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
18 tháng 2 2022
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
TT
12 tháng 12 2020
Bạn chú ý đăng lẻ câu hỏi! 1/
a/ \(=x^3-2x^5\)
b/\(=5x^2+5-x^3-x\)
c/ \(=x^3+3x^2-4x-2x^2-6x+8=x^3=x^2-10x+8\)
d/ \(=x^2-x^3+4x-2x+2x^2-8=3x^2-x^3+2x-8\)
e/ \(=x^4-x^2+2x^3-2x\)
f/ \(=\left(6x^2+x-2\right)\left(3-x\right)=17x^2+5x-6-6x^3\)
7 tháng 11 2021
a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
ND
Nguyễn Đức Trí
VIP
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
ND
Nguyễn Đức Trí
VIP
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) \(-6x^2\left(x+5\right)^2-\left(x-3\right)^2+\left(x^2-2\right)\left(2x^2+1\right)-4x^2\left(3x-4\right)^2\)
\(-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)
\(-6x^4-60x^3+150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)
\(-40x^4+36x^3+82x^2+6x-11\)
1 tháng 4 2018
a. x(2x2 – 3) – x2(5x + 1) + x2
= 2x3 – 3x – 5x3 – x2 + x2 = -3x – 3x3
b. 3x(x – 2) – 5x(1 – x) – 8(x2 – 3)
= 3x2 – 6x – 5x + 5x2 – 8x2 + 24
= - 11x + 24
c. 1/2 x2(6x – 3) – x( x2 + 1/2 (x + 4)
= 3x3 - 3/2 x2 – x3 - 1/2 x + 1/2 x + 2
= 2x3 - 3/2 x2 + 2
1 tháng 4 2018
a, x(2x2-3)-x2(5x+1)x2
=2x3-3x-5x3- x2+x2=-3x-3x3
học tốt nhé!!
a) (x^5 + 4x^3 - 6x^2) : 4x^2
= (x^5 : 4x^2) + (4x^3 : 4x^2) - (6x^2 : 4x^2)
= 1/4x^3 + x + 3/2
b) x(2x^2 - 3) - x^2(5x + 1) + x^2
= 2x^3 - 3x - 5x^3 - x^2 + x^2
= -3x^3 - 3x
c) (x - 2)^2 - (x - 1)(x + 1) - x(1 - x)
= x^2 - 4x + 4 - x^2 + 1 - x + x^2
<=> x^2 - 5x + 5
d) 1/2x^2(6x - 3) - x(x^2 + 1/2) + 1/2(x + 4)
= \(\frac{x^2}{2}\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{x+4}{2}\)
= \(\frac{x^2\left(6x-3\right)}{2}-x\left(x^2+\frac{1}{2}\right)+\frac{x+4}{2}\)
= \(-x\left(x^2+\frac{1}{2}\right)+\frac{x^2\left(6x-3\right)+x+4}{2}\)
= \(\frac{4x^3-3x^2+4}{2}\)