Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải
1) 3xy2 : 5x = \(\frac{3}{5}\)y2
2) 15x4yz3 : 4xyz = \(\frac{15}{4}\)x3z2
3) (4x2y2 - 12xy3 - 7x) : 3x = \(\frac{4}{3}\)xy2 - 4y3 - \(\frac{7}{3}\)
4) (14x4y2 - 12xy3 - x) : 4x = \(\frac{7}{2}\)x3y2 - 3y3 - \(\frac{1}{4}\)
5) (6x2 + 13x - 5) : (2x + 5) = (3x - 1)(2x + 5) : (2x + 5) = 3x - 1
6) (2x4 + x3 - 5x2 - 3x - 3) : (x2 - 3)
= 2x4 + x2 - 6x2 + x3 - 3 - 3x : x2 - 3
= x2(2x2 + x + 1) - 3(2x2 + x + 1) : x2 - 3
= (2x2 + x + 1)(x2 - 3) : x2 - 3
= 2x2 + x + 1
a) Ta có: \(-3x^2\left(2x^2-\frac{1}{3}x+2\right)\)
\(=-6x^4+x^3-6x^2\)
b) Ta có: \(2xy^2\left(x-3y+xy\right)\)
\(=2x^2y^2-6xy^3+2x^2y^3\)
c) Ta có: \(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-10x^2-4x^2+8x\)
\(=5x^3-14x^2+8x\)
d) Ta có: \(-\left(2-x\right)\left(2x+3\right)\)
\(=\left(x-2\right)\left(2x+3\right)\)
\(=2x^2+3x-4x-6\)
\(=2x^2-x-6\)
e) Ta có: \(\left(3x^3-2x^2+x\right):\left(-2x\right)\)
\(=\frac{-3}{2}x^2+x-\frac{1}{2}\)
f) Ta có: \(\left(15x^2y^2-21x^3y+2x^2y\right):\left(3x^2y\right)\)
\(=5y-7x+\frac{2}{3}\)
g) 
a) Ta có: \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)\)
\(=x^2+y^2\)
b) Ta có: \(\left(49x^2-81y^2\right):\left(7x+9y\right)\)
\(=\frac{\left(7x+9y\right)\left(7x-9y\right)}{7x+9y}\)
\(=7x-9y\)
c) Ta có: \(\left(x^3+3x^2y+3xy^2+y^3\right):\left(x+y\right)\)
\(=\left(x+y\right)^3:\left(x+y\right)\)
\(=\left(x+y\right)^2=x^2+2xy+y^2\)
d) Ta có: \(\left(x^3-3x^2y+3xy^2-y^3\right):\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3:\left(x-y\right)^2\)
\(=\left(x-y\right)\)
e)Sửa đề: \(\left(8x^3+1\right):\left(2x+1\right)\)
Ta có: \(\left(8x^3+1\right):\left(2x+1\right)\)
\(=\frac{\left(2x+1\right)\left(4x^2-2x+1\right)}{2x+1}\)
\(=4x^2-2x+1\)
f) Ta có: \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)
\(=\frac{\left(2x-1\right)\left(4x^2+2x+1\right)}{4x^2+2x+1}\)
\(=2x-1\)
a, (x4 + 2x2y2 + y4) : (x2 + y2)
= (x2 + y2)2 : (x2 + y2)
= x2 + y2
b, (49x2 - 81y2) : (7x + 9y)
= (7x - 9y)(7x + 9y) : (7x + 9y)
= 7x - 9y
c, (x3 + 3x2y + 3xy2 + y3) : (x + y)
= (x + y)3 : (x + y)
= (x + y)2
d, (x3 - 3x2y + 3xy2 - y3) : (x2 - 2xy + y2)
= (x - y)3 : (x - y)2
= x - y
Phần e thiếu thì phải
f, (8x3 - 1) : (4x2 + 2x + 1)
= (2x - 1)(4x2 + 2x + 1) : (4x2 + 2x + 1)
= 2x - 1
Chúc bn học tốt!
1 ) x3 - 2x2 + x
= x( x2 - 2x + 1 )
= x ( x-1)2
2) 4x3 - 25x
= x ( 4x2 - 25)
= x( 2x-5) ( 2x +5)
11) \(x^2-y^2-4x+4\)
\(=\left(x^2-4x+4\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-y-2\right)\left(x+y-2\right)\)
13) \(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
1,4x2.(5x3+2x-1)
=4x2.5x3+4x2.2x-4x2.1
20x5+8x3-4x2
2,4x3y2:x2
=4xy2
3,(15x2y3-10x3y3+6xy):5xy
15x2y3:5xy-10x3y3:5xy+6xy:5xy
3xy2-2x2y2+\(\dfrac{6}{5}\)
1: \(=20x^5+8x^3-4x^2\)
2: \(=4xy^2\)
3: \(=3xy^2-2x^2y^2+\dfrac{6}{5}\)
4: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
5: \(=\dfrac{7}{2x}+\dfrac{11}{3y^2}=\dfrac{21y^2+22x}{6xy^2}\)
6: \(=\dfrac{4x^2-7x+3}{\left(4x-7\right)\left(x+2\right)}\)
7: \(=\dfrac{3x+3y-2x^3+2x^2y}{\left(x-y\right)\left(x+y\right)}\)
8: \(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)
9: \(=\left(x-\dfrac{1}{4}\right)\left(4x-1\right)=4\left(x-\dfrac{1}{4}\right)^2=4\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=4x^2-2x+\dfrac{1}{4}\)
10: \(=\dfrac{3x^2+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
11: \(=\dfrac{x+1}{2}-\dfrac{3}{x-1}\)
\(=\dfrac{x^2-7}{2\left(x-1\right)}\)
12: \(=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)
15:=x^3-y^3+2
1: \(=20x^5+8x^3-4x^2\)
2: \(=4xy^2\)
3: \(=3xy^2-2x^2y^2+\dfrac{6}{5}\)
4: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
5: \(=\dfrac{7}{2x}+\dfrac{11}{3y^2}=\dfrac{21y^2+22x}{6xy^2}\)
6: \(=\dfrac{4x^2-7x+3}{\left(4x-7\right)\left(x+2\right)}\)
7: \(=\dfrac{3x+3y-2x^3+2x^2y}{\left(x-y\right)\left(x+y\right)}\)
8: \(=\dfrac{1}{2}x^2y^2\left(4x^2-y^2\right)=2x^4y^2-\dfrac{1}{2}x^2y^4\)
9: \(=\left(x-\dfrac{1}{4}\right)\left(4x-1\right)=4\left(x-\dfrac{1}{4}\right)^2=4\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)
\(=4x^2-2x+\dfrac{1}{4}\)
10: \(=\dfrac{3x^2+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
11: \(=\dfrac{x+1}{2}-\dfrac{3}{x-1}\)
\(=\dfrac{x^2-7}{2\left(x-1\right)}\)
12: \(=\dfrac{x^2-xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{x}{x+y}\)
15:=x^3-y^3+2
* 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!!