3xy(x-y)+6x²(y-x)=?
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Giúp mình với mình cần gấp kết quả ạ 1, (x²y + 6x) . (x² - 3xy) a, x⁴y - 3x³y² + 6x³ - 18x²y b, x²y - x³y² + 6x³ - 18x²y C,x⁴y - 3x³y² + 6x³ + 18x²y d, x⁴y + 3x³y² - 6x³ - 18x²y 2, Tìm x biết x . (2x - 4) - 2x² + 9x - 7 = 3 a, x = 1 b, x = 2 C, x = 3 d, x = 4 3, tính giá trị của biểu thức sau tại x = 3 ; y=2 7x . (x² - 2y) + 3xy - 7x³ a, 24 b, -4 c, 6 d, -24 Cảm ơn đã giúp đỡ mình ✨
cho x+y =1 . tinh gia tri cua bieu thuc A=x^3+y^3+3xy
chox-y=1. tinh gia tri cua bieu thuc B=x^3-y^3-3xy
cho x+y=1 . tinh gia tri cua bieu thuc C=x^3+y^3+3xy(x^2+y^2)+6x^2*y^2(x+y)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
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17 tháng 2 2021 lúc 10:55
a) \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)
b)\(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)
a) ĐKXĐ: \(x\ne0\)
\(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)
\(=\dfrac{2\left(4x+1\right)+2x-3}{6x}\)
\(=\dfrac{10x-1}{6x}\)
b) ĐKXĐ: \(x,y\ne0\)
\(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)
\(=\dfrac{\left(x-y\right).\left(x+y\right)}{6x^2y^2}.\dfrac{3xy}{x+y}\)
\(=\dfrac{x-y}{2xy}\)
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a) Ta có: \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)
\(=\dfrac{2\left(4x+1\right)}{6x}+\dfrac{2x-3}{6x}\)
\(=\dfrac{8x+2+2x-3}{6x}\)
\(=\dfrac{10x-1}{6x}\)
b) Ta có: \(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{6x^2y^2}\cdot\dfrac{3xy}{x+y}\)
\(=\dfrac{x-y}{2xy}\)
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cho x+y=1 tính
x^3+y^3+3xy(x^2+y^2)+6x^2y^2(x+y)
\(x^3+y^3+3xy\left(x^2+y^2\right)+6x^2y^2\left(x+y\right)\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\left[\left(x^2+2xy+y^2\right)-2xy\right]+6x^2y^2\)
\(=1-3xy+3xy\left(x+y\right)^2-6x^2y^2+6x^2y^2\)
\(=1-3xy+3xy\)
\(=1\)
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Bài 3: phân tích thành nhân tử:1/ 9x^3-xy^22/x^2-3xy-6x+18y3/x^2-3xy-6x+18y 3/6x(x-y)-9y^2+9xy4/ 6xy-x^2+36-9y^25/ x^4-6x^2+56/ 9x62-6x-y^2+2yBài 4:Tìm x, biết:1/ (x-1)(x^2+x+1)-x^3-6x112/ 16x^2-(3x-4)^203/ x^3-x^2+3-3x04/ x-1/x+2x+2/x+15/1/x+2/x+106/ 9-x^2/x : (x-3)1Bài5: 1/ 12x^3y^2/18xy^52/10xy-5x^2/2x^2-8y^23/ x^2-xy-x+y/x^2+xy-x-y4/ (x+1)(x^2-2x+1)/(6x^2-6)(x^3-1)5/ 2x^2-7x+3/1-4x^2
Đọc tiếp
Bài 3: phân tích thành nhân tử:
1/ 9x^3-xy^2
2/x^2-3xy-6x+18y
3/x^2-3xy-6x+18y 3/6x(x-y)-9y^2+9xy
4/ 6xy-x^2+36-9y^2
5/ x^4-6x^2+5
6/ 9x62-6x-y^2+2y
Bài 4:Tìm x, biết:
1/ (x-1)(x^2+x+1)-x^3-6x=11
2/ 16x^2-(3x-4)^2=0
3/ x^3-x^2+3-3x=0
4/ x-1/x+2=x+2/x+1
5/1/x+2/x+1=0
6/ 9-x^2/x : (x-3)=1
Bài5: 1/ 12x^3y^2/18xy^5
2/10xy-5x^2/2x^2-8y^2
3/ x^2-xy-x+y/x^2+xy-x-y
4/ (x+1)(x^2-2x+1)/(6x^2-6)(x^3-1)
5/ 2x^2-7x+3/1-4x^2
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
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3xy - 3xz-y 2 +yz
x 4 –x 3 +x 2 –x
xy+xz + y 2 +yz
x 2 -6x+9
x 3 +6x 2 +12x+8
4x 2 - (x - y) 2
5x(y + 1) - 2(y + 1)
x 2 - 4x + 4
x 4 -2x 2
3x 2 - 12xy
\(a,=3x\left(y-z\right)-y\left(y-z\right)=\left(3x-y\right)\left(y-z\right)\\ b,=x^3\left(x-1\right)+x\left(x-1\right)=x\left(x^2+1\right)\left(x-1\right)\\ c,=x\left(y+z\right)+y\left(y+z\right)=\left(x+y\right)\left(y+z\right)\\ d,=\left(x-3\right)^2\\ e,=\left(x+2\right)^3\\ f,=\left(2x-x+y\right)\left(2x+x-y\right)=\left(x+y\right)\left(3x-y\right)\\ g,=\left(y+1\right)\left(5x-2\right)\\ h,=\left(x+2\right)^2\\ i,=x^2\left(x^2-2\right)\\ k,=3x\left(x-4y\right)\)
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Tìm x;y để
3xy+y+6x=7
tim x,y
3xy -6x +y+3 =0
tim x,y:
3xy-6x+y+3=0
3xy - 6x + y + 3 = 0
=> 3x.(y - 2) + y - 2 + 5 = 0
=> (y - 2).(3x + 1) = -5
Ta có bảng sau:
| 3x+1 | -5 | -1 | 1 | 5 |
| x | -2 | -2/3 | 0 | 4/3 |
| y-2 | 1 | 5 | -5 | -1 |
| y | 3 | 7 | -3 | 1 |
Vậy...
*Thiếu điều kiện x,y thuộc ...? , bổ sung rồi loại bớt.
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