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3 tháng 6 2023
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: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
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) \(-6x^2\left(x+5\right)^2-\left(x-3\right)^2+\left(x^2-2\right)\left(2x^2+1\right)-4x^2\left(3x-4\right)^2\)
\(-6x^2\left(x^2+10x+25\right)-\left(x^2-6x+9\right)+2x^4-3x^2-2-4x^2\left(9x^2-24x+16\right)\)
\(-6x^4-60x^3+150x^2-x^2+6x-9+2x^4-3x^2-2-36x^4+96x^3-64x^2\)
\(-40x^4+36x^3+82x^2+6x-11\)
x2 ( 2x3 – 4x + 3)
24 tháng 12 2021
a: \(=15x^5-25x^4+15x^3\)
b: \(=2x^3+10x^2-8x-x^2-5x+4\)
\(=2x^3+9x^2-13x+4\)
NM
1
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
a) \(\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right):\dfrac{3}{5}ax^3\)
\(=\dfrac{\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5}{\dfrac{3}{5}ax^3}\)
\(=\dfrac{\dfrac{3}{5}ax^3\left(a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}x^2\right)}{\dfrac{3}{5}ax^3}\)
\(=a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}x^2\)
b) \(\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)\cdot y^2\)
\(=\dfrac{3x^2y\left(3y^2-5x^2y^3\right)}{3x^2y}-2y^2+3x^2y^3\)
\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)
\(=y^2-2x^2y^3\)
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
c) \(\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)
\(=\dfrac{6x^2-xy}{x}+\dfrac{2x^3y+3xy^2}{xy}-x\left(2x-1\right)\)
\(=\dfrac{x\left(6x-y\right)}{x}+\dfrac{xy\left(2x^2+3y\right)}{xy}-2x^2+x\)
\(=6x-y+2x^2+3y-2x^2+x\)
\(=7x+2y\)
d) \(\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^2\right):\dfrac{3}{2}x^2y^3\)
\(=\dfrac{x^2-xy}{x}+\dfrac{6x^2y^5-9x^3y^4+15x^4y^2}{\dfrac{3}{2}x^2y^3}\)
\(=\dfrac{x\left(x-y\right)}{x}+\dfrac{\dfrac{3}{2}x^2y^2\left(4y^3-6xy^2+10x^2\right)}{\dfrac{3}{2}x^2y^3}\)
\(=x-y+\dfrac{4y^3-6xy^2+10x^2}{y}\)
6 tháng 3 2021
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
15 tháng 10 2023
2:
a: \(9x^2-1=\left(3x\right)^2-1=\left(3x-1\right)\left(3x+1\right)\)
b: \(2\left(x-1\right)+x^2-x\)
\(=2\left(x-1\right)+x\left(x-1\right)\)
\(=\left(x-1\right)\left(x+2\right)\)
c: \(3x^2+14x-5\)
\(=3x^2+15x-x-5\)
\(=3x\left(x+5\right)-\left(x+5\right)=\left(x+5\right)\left(3x-1\right)\)
3:
a: \(2x\left(x-1\right)-2x^2=4\)
=>\(2x^2-2x-2x^2=4\)
=>-2x=4
=>x=-2
b: \(x\left(x-3\right)-\left(x+2\right)\left(x-1\right)=5\)
=>\(x^2-3x-\left(x^2+x-2\right)=5\)
=>\(x^2-3x-x^2-x+2=5\)
=>-4x=3
=>x=-3/4
c: \(4x^2-25+\left(2x+5\right)^2=0\)
=>\(\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\)
=>\(\left(2x+5\right)\left(2x-5+2x+5\right)=0\)
=>4x(2x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)
CM
29 tháng 8 2018
a) M = 4 x 5 y – 6 x 3 y 2 + 10 x 3 y 2 z .
b) N = x 4 y 3 – 2 x 2 y 4 + x 2 y 3