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a) ( -5x2 +3xy + 7) + ( -6x2y + 4xy2 - 5)=4*x*y^2-6*x^2*y+3*a*x*y-5*a*x^2+7*a-5
b) ( 2,4x3 - 10x2y) + (7x2y - 2,4x3 + 3xy2)=3*x*y^2-3*x^2*y
c) ( 15x2y - 7xy2 - 6y2) + (2x2 - 12x2y + 7xy2)=-6*y^2+3*x^2*y+2*x^2
d) ( 4x2 + x2y - 5y3) + (5/3 x3 - 6xy2 - x2y) + (x3/3 + 10y3) + ( 6y3-15xy2 - 4x2y - 10x3)=11*y^3-21*x*y^2-4*x^2*y-8*x^3+4*x^2
Bài 1:
a)\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
b)\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
c)Đề sai hoàn toàn
d) \(2x^2+4xy+2y^2-8z^2=2\left(x^2+2xy+y^2-4z^2\right)=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]=2\left(x+y-2z\right)\left(x+y+2z\right)\)e) \(3x-3a+yx-ya=3\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(3+y\right)\)
f)\(\left(x^2+y^2\right)^2-4x^2y^2=\left(x-y\right)^2\left(x+y\right)^2\)
g)\(2x^2-5x+2=2x^2-x-4x+2=x\left(2x-1\right)-2\left(2x-1\right)=\left(2x-1\right)\left(x-2\right)\)
i)\(14x\left(x-y\right)-21y\left(y-x\right)+28z\left(x-y\right)=14x\left(x-y\right)+21y\left(x-y\right)+28z\left(x-y\right)=7\left(x-y\right)\left(2x+3y+4z\right)\)
a) (x3 + 8y3) : (2y + x)
= (x + 2y)(x2 - 2xy + 4y2) : (2y + x)
= x2 - 2xy + 4y2
b) (x3 + 3x2y + 3xy2 + y3) : (2x + 2y)
= (x + y)3 : 2(x + y)
= \(\dfrac{\left(x+y\right)^2}{2}\)
c) (6x5y2 - 9x4y3 + 15x3y4) : 3x3y2
= 3x3y2(2x2 - 3xy + 5y2) : 3x3y2
= 2x2 - 3xy + 5y2
a) (12x6y4 + 9x5y3 - 15x2y3) : 3x2y3
= (12x6y4 : 3x2y3) + (9x5y3 : 3x2y3) - (15x2y3 : 3x2y3)
= 4x4y + 3x3 - 5
b) (x2 - 2)(1 - x) + (x + 3)(x2 - 3x + 9)
= x2 - x3 - 2 + 2x + x3 - 3x2 + 9x + 3x2 - 9x + 27
= x2 + 25 + 2x
* 45x(3 - x) = 15x(x - 3)3
\(\Leftrightarrow\) 45x(3 - x) - 15x(x - 3)3 = 0
\(\Leftrightarrow\) 45x(3 - x) + 15x(3 - x)3 = 0
\(\Leftrightarrow\) 15x(3 - x)[3 + (3 - x)2] = 0
\(\Leftrightarrow\left[{}\begin{matrix}15x=0\\3-x=0\\3+\left(3-x\right)^2=0\end{matrix}\right.\)
Vì 3 + (3 - x)2 > 0 với mọi x
\(\Rightarrow\) 15x = 0 hoặc 3 - x = 0
\(\Leftrightarrow\) x = 0 và x = 3
Vậy S = {0; 3}
* 7x2 + 14x + 7 = 3x2 + 3x
\(\Leftrightarrow\) 7(x2 + 2x + 1) = 3x(x + 1)
\(\Leftrightarrow\) 7(x + 1)2 = 3x(x + 1)
\(\Leftrightarrow\) 7(x + 1)2 - 3x(x + 1) = 0
\(\Leftrightarrow\) (x + 1)[7(x + 1) - 3x] = 0
\(\Leftrightarrow\) (x + 1)(7x + 7 - 3x) = 0
\(\Leftrightarrow\) (x + 1)(4x + 7) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\4x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\frac{-7}{4}\end{matrix}\right.\)
Vậy S = {-1; \(\frac{-7}{4}\)}
* 3x2 - 12x + 12 = x4 - 8x
\(\Leftrightarrow\) 3(x2 - 4x + 4) = x(x3 - 8)
\(\Leftrightarrow\) 3(x - 2)2 = x(x - 2)(x2 + 2x + 4)
\(\Leftrightarrow\) 3(x - 2)2 - x(x - 2)(x2 + 2x + 4) = 0
\(\Leftrightarrow\) (x - 2)[3(x - 2) - x(x2 + 2x + 4)] = 0
\(\Leftrightarrow\) (x - 2)(3x - 6 - x3 - 2x2 - 4x) = 0
\(\Leftrightarrow\) (x - 2)(-x3 - 2x2 - x - 6) = 0
\(\Leftrightarrow\) -1(x - 2)(x3 + 2x2 + x + 6) = 0
\(\Leftrightarrow\) (x - 2)[x(x2 + 2x + 1) + 6] = 0
\(\Leftrightarrow\) (x - 2)[x(x + 1)2 + 6] = 0
Ta có: x(x + 1)2 + 6 = 0
\(\Leftrightarrow\) x(x + 1)2 = -6
Nếu x = -2 thì (x + 1)2 = 3 hay (x + 1)2 + 3 = 0
mà (x + 1)2 + 3 > 0 với mọi x nên x không thỏa mãn giá trị trên
Nếu x = 2 thì (x + 1)2 = -3 (loại vì KTM)
Nếu x = 1 thì (x + 1)2 = -6 (loại vì KTM)
Nếu x = -1 thì (x + 1)2 = 6
Thay x = -1 vào pt (x + 1)2 = 6 ta được:
(-1 + 1)2 = 6
\(\Leftrightarrow\) 0 = 6 (KTM)
Từ đó suy ra phương trình x(x + 1)2 + 6 = 0 vô nghiệm
\(\Rightarrow\) x - 2 = 0
\(\Leftrightarrow\) x = 2
Vậy S = {2}
* y2 - x2 = x3 - 3x2y + 3xy2 - y3
\(\Leftrightarrow\) (y - x)(y + x) = (x - y)3
\(\Leftrightarrow\) (y - x)(y + x) - (x - y)3 = 0
\(\Leftrightarrow\) (y - x)(y + x) + (y - x)3 = 0
\(\Leftrightarrow\) (y - x)[y + x + (y - x)2] = 0
Vì y + x + (y - x)2 > 0 với mọi x
\(\Rightarrow\) y - x = 0
\(\Leftrightarrow\) x = y
Vậy S = {y}
Chúc bn học tốt!!
1)\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}=\frac{2y}{5\left(x+y\right)^2}\)
2) \(\frac{15x\left(x+y\right)^2}{20x^2\left(x+5\right)}=\frac{3\left(x^2+2xy+y^2\right)}{4x\left(x+5\right)}=\frac{3\left(x+y\right)^2}{4x^2+20x}\)
3) \(\frac{15x\left(x-y\right)}{3\left(y-x\right)}=\frac{5x\left(x-y\right)}{-3\left(x-y\right)}=-\frac{5x}{3}\)
4)\(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}=\frac{\left(y-x\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x-y\right)\left(x+y\right)}{\left(x-y\right)^3}=\frac{-\left(x+y\right)}{\left(x-y\right)^2}\)