3xy(x+4)-2x(x+4)
Bai 1 TINH
a) x ^2 . x -2x^3
b) 6 x^2y . 3xy - 2y^2.x +y
C) 4x^2 +5x -1 .2x^3 -3x
d)-8x^3y + 2y^4 . 3xy^3 - 2x^4 +7y^4
GIUP MINH NHÀ
(x^2y^2 - x^2y + 4xy + 2x - 4) + (-x^2y^2 - 6x^2y - xy + 2x+4) - (2x^2y^2 - 3xy +x - 4)
=
Tìm x,yϵZ:
y-3xy+2x=4
\(y-3xy+2x=4\)
\(\Leftrightarrow3y-9xy+6x=12\)
\(\Leftrightarrow3y\left(1-3x\right)-2\left(1-3x\right)=10\)
\(\Leftrightarrow\left(1-3x\right)\left(3y-2\right)=10\)
1-3x | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
3y-2 | -1 | -2 | -5 | -10 | 10 | 5 | 2 | 1 |
x | 11/3 | 2 | 1 | 2/3 | 0 | -1/3 | -4/3 | -3 |
y | 1/3 | 0 | -1 | -8/3 | 4 | 7/3 | 4/3 | 1 |
Vậy \(\left(x;y\right)=\left(2;0\right);\left(1;-1\right);\left(0;4\right);\left(-3;1\right)\)
phân tích đa thức sau thành nhân tử1,3x 2 x 22, 2x 2 3xy 2y 23, 2x 2 3xy 2y 24, x 2 4xy 2x 3y 2 65, x 8 x 1Tìm x,y biết1, x 2 2x 5 y 2 4y 02,4x 2 y 4 20x 2y 26 0
đa thức lớp 5 hả bạm
mình ghi sao đề, các bạn ko cần làm đâu
3. Thực hiện phép tính:
a. 5/(2x ^ 2 * y) + 2/(3xy) - y/(x ^ 3)
b. (2x - 7)/(10x - 4) - (3x + 5)/(4 - 10x)
c. x ^ 2 + 1 - (x ^ 4 - 3x ^ 2)/(x ^ 2 - 1)
a: \(=\dfrac{5}{2x^2y}+\dfrac{2}{3xy}-\dfrac{y}{x^3}\)
\(=\dfrac{5\cdot3\cdot x}{6x^3y}+\dfrac{2\cdot2\cdot x^2}{6x^3y}-\dfrac{6y^2}{6x^3y}\)
\(=\dfrac{15x+4x^2-6y^2}{6x^3y}\)
b: \(=\dfrac{2x-7+3x+5}{10x-4}=\dfrac{5x-2}{10x-4}=\dfrac{1}{2}\)
c: \(=\dfrac{x^4-1-x^4+3x^2}{x^2-1}=\dfrac{3x^2-1}{x^2-1}\)
Tìm x thỏa mãn điều kiện
(2x+1)^3-(2x+1).(4x^2-2x+1)-3.(2x-1)^2=15
y.(y+3)^2-(y+2).(y^2-2y+4)-6.(y+5).(y-5)=97
(x-3)^3-(x-3).(x^2+3x+9)+9.(x+1)^2=18
x.(x-4).(x+4)-(x-5).(x^2+5x+25)=13
2.Rút gọn biểu thức rồi tính giá trị
3.(x-1).(x^2+x+1)+(x-1)^3-4x.(x+1).(x-1) tại x=-1
(3xy-2).(9x^2y^2+6xy+4)-3xy.(3xy+1)^2 tại x=-2010,y=-1/2010
Bài 2:
a: \(3\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)\)
\(=3\left(x^3-1\right)+x^3-3x^2+3x-1-4x\left(x^2-1\right)\)
\(=3x^3-3+x^3-3x^2+3x-1-4x^3+4x\)
\(=-3x^2+7x-4\)
\(=-3\cdot\left(-1\right)^2+7\cdot\left(-1\right)-4\)
=-3-4-7=-14
b: \(=27x^3y^3-8-3xy\left(9x^2y^2+6xy+1\right)\)
\(=27x^3y^3-8-27x^3y^3-18x^2y^2-3xy\)
\(=-18x^2y^2-3xy-8\)
\(=-18\cdot\left[\left(-2010\right)\cdot\left(-\dfrac{1}{2010}\right)\right]^2-3\cdot\left(-2010\right)\cdot\dfrac{-1}{2010}-8\)
\(=-18-3-8=-29\)
a,(2x+3)^3 b,(x-3y)^3 c.(x+4)(x^2-4x+15) d,(1/3x+1y)(1/9x^2-2/3xy+4y) e,(x-3y)(x^2+3xy+9y^2)
a: \(\left(2x+3\right)^3=8x^3+36x^2+54x+27\)
b: \(\left(x-3y\right)^3=x^3-9x^2y+27xy^2-27y^3\)
phân tích đa thức sau thành nhân tử
1,3x^2+x-2
2, 2x^2-3xy-2y^2
3, 2x^2-3xy-2y^2
4, x^2+4xy+2x+3y^2+6
5, x^8+x+1
Tìm x,y biết
1, x^2+2x+5+y^2-4y=0
2,4x^2+y^4-20x-2y=26=0
mik ko bít
I don't now
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Ai giải đúng chỗ mình mình sẽ đánh giá 5 sao và đúng mình cần gấp lắm a)(x+2)(x^2-24+4)(x^3+8) b)(2x-1/2)(4x^2+x+1/4) c)(x^2+y)(x^2-y)+y^2+x^4 d)(x+3)(x^2-3x+9)-x^3 e)(3x+y)(9x^2-3xy+y^2)-26x^3 g)(x+3y)(x^2-3xy+9y^2)+(3x-y)(9x^2+3xy+y^2)
a) \(\left(x+2\right)\left(x^2-24+4\right)\left(x^3+8\right)\)
\(=\left(x+2\right)\left(x^2-20\right)\left(x^3+8\right)\)
\(=\left(x^3-20x+2x^2-40\right)\left(x^3+8\right)\)
\(=x^6+8x^3-20x^4+160x+2x^5+16x^2-40x^3-120\)
\(=x^6+2x^5-20x^4-32x^3+16x^2+160x-120\)
b) \(\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
\(=8x^3+2x^2+\dfrac{1}{2}x-2x^2-\dfrac{1}{2}x-\dfrac{1}{8}\)
\(=8x^3-\dfrac{1}{8}\)
c) \(\left(x^2+y\right)\left(x^2-y\right)+y^2+x^4\)
\(=\left(x^2\right)^2-y^2+y^2+x^4\)
\(=x^4-y^2+y^2+x^4\)
\(=2x^4\)
d) \(\left(x+3\right)\left(x^2-3x+9\right)-x^3\)
\(=\left(x+3\right)\left(x^2-3\cdot x+3^2\right)-x^3\)
\(=x^3+3^3-x^3\)
\(=27\)
e) \(\left(3x+y\right)\left(9x^2-3xy+y^2\right)-26x^3\)
\(=\left(3x+y\right)\left[\left(3x\right)^2-3x\cdot y+y^2\right]-26x^3\)
\(=\left(3x\right)^3+y^3-26x^3\)
\(=27x^3+y^3-26x^3\)
\(=x^3+y^3\)
g) \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=\left(x+3y\right)\left[x^2-x\cdot3y+\left(3y\right)^2\right]+\left(3x-y\right)\left[\left(3x\right)^2+3x\cdot y+y^2\right]\)
\(=\left[x^3+\left(3y\right)^3\right]+\left[\left(3x\right)^3-y^3\right]\)
\(=x^3+27y^3+27x^3-y^3\)
\(=28x^3+26y^3\)
a) Sửa đề:
(x + 2)(x² - 2x + 4)(x³ + 8)
= (x³ + 8)(x³ + 8)
= (x³ + 8)²
b) (2x - 1/2)(4x² + x + 1/4)
= (2x)³ - (1/2)³
= 8x³ - 1/8
c) (x² + y)(x² - y) + y² + x⁴
= (x²)² - y² + y² + x⁴
= 2x⁴
d) (x + 3)(x² - 3x + 9) - x³
= x³ + 3³ - x³
= 27
e) (3x + y)(9x² - 3xy + y²) - 26x³
= (3x)³ + y³ - 26x³
= 27x³ + y³ - 26x³
= x³ + y³
g) (x + 3y)(x² - 3xy + 9y²) + (3x - y)(9x² + 3xy + y²)
= x³ + (3y)³ + (3x)³ - y³
= x³ + 27y³ + 27x³ - y³
= 28x³ + 26y³