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\(a,\left(2x+1\right)^2-\left(x-1\right)^2\\ =\left(2x+1-x+1\right)\left(2x+1+x-1\right)\\ =\left(x+2\right)3x\)
\(b,9\left(x+5\right)^2-\left(x-7\right)^2\\ =\left[3\left(x+5\right)\right]^2-\left(x-7\right)^2\\ =\left(3x+15-x+7\right)\left(3x+15+x-7\right)\\ =\left(2x+22\right)\left(4x+8\right)\)
\(c,x^2y+xy^2-x-y\\ =xy\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(xy-1\right)\)
Các câu sau tương tự
\(1\hept{\begin{cases}6x^2-8x+3x-4\\2x\left(3x-4\right)+\left(3x-4\right)\\\left(3x-4\right)\left(2x+1\right)\end{cases}}\)
\(2\hept{\begin{cases}7x^2-7xy-5x+5y+6xy\\7x\left(x-y\right)-5\left(x-y\right)+\frac{6xy\left(x-y\right)}{\left(x-y\right)}\\\left(x-y\right)\left(7x-5+\frac{6xy}{\left(x-y\right)}\right)\end{cases}}\)
\(3\hept{\begin{cases}5x\left(x-y\right)-15\left(x-y\right)\\\left(x-y\right)\left(5x-15\right)\end{cases}}\)
\(4,,2x^2+x=x\left(2x+1\right)\)
\(5\hept{\begin{cases}x^3-4x-3x^2+12\\x\left(x^2-4\right)-3\left(x^2-4\right)\\\left(x+2\right)\left(x-2\right)\left(x-3\right)\end{cases}}\)
\(6\hept{\begin{cases}2x+2y+x^2-y^2\\2\left(x+y\right)+\left(x+y\right)\left(x-y\right)\\\left(x+y\right)\left(2+x-y\right)\end{cases}}\)
\(7\hept{\begin{cases}\left(x^2y-2xy\right)-\left(xy-2y\right)+\left(xy-y\right)\\xy\left(x-2\right)-y\left(x-2\right)+y\left(x-1\right)\\y\left(X-2\right)\left(x-1\right)+y\left(x-1\right)\end{cases}}\Leftrightarrow y\left(x-1\right)\left(x-2+1\right)\)
\(8\hept{\begin{cases}x\left(2-y\right)+z\left(2-y\right)\\\left(2-y\right)\left(x+1\right)\end{cases}}\)
a) Đặt \(x^2+3x+1=y\) khi đó ta có:
\(y\left(y-4\right)-5\)
\(=y^2-4y-5\)
\(=y\left(y-5\right)+\left(y-5\right)\)
\(=\left(y+1\right)\left(y-5\right)\)
Thay \(y=x^2+3x+1\):
\(\left(x^2+3x+1+1\right)\left(x^2+3x+1-5\right)\)
\(=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)
\(=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x-1\right)+4\left(x-1\right)\right]\)
\(=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x+4\right)\)
b) Biến đổi 3 số sau có chứa x2 + 2x rồi đặt ẩn.
c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+7=y'\)
Khi đó ta đc:
\(y'\left(y'+8\right)+15\)
\(=\left(y'\right)^2+8y'+15\)
\(=y'\left(y'+3\right)+5\left(y'+3\right)\)
\(=\left(y'+5\right)\left(y'+3\right)\)
....
d) \(x^2-2xy+y^2-7x+7y+12\)
Biến đổi chứa x - y rồi đặt ẩn.
Đỗ thị như quỳnh: làm tương tự thôi mà, nếu bạn ko hiểu chỗ nào thì nói đi :)
1, x2+3xy+2y2= x2+xy+2xy+2y2=x(x+y)+2y(x+y)=(x+2y)(x+y)
2, x(x+2)(x+3)(x+5)+9=x(x+5)(x+2)(x+3)+9=(x2+5x)(x2+5x+6)+9
Đặt x2+5x=t, ta có
t(t+6)+9=t2+6t+9=(t+3)2=(x2+5x+3)2=(x2+8)2
3, x2+2xy+y2+2x+2y-15=(x+y)2+2(x+y)-15=(x+y)2+2(x+y)+1-16=(x+y+1)2-42
= (x+y+1-4)(x+y+1+4)=(x+y-3)(x+y+5)
4, 4x4y4+1=4x4y4+4x2y2+1-4x2y2=(2x2y2+1)2-(2xy)2=(2x2y2+1-2xy)(2x2y2+1+2xy)
a,(x-y)^2-2(x+y)+1 b, x^2-y^2+4x+4 c, 4x^2-y^2+8(y-2)
=(x-y-1)^2 =(x^2+4x+4)-y^2 =4x^2-y^2+8y-16
=(x+2)^2-y^2 =4x^2-(y^2-8y+16)
=(x+2-y)(x+2+y) =4x^2-(y-4)^2
a) (x+y)2-2(x+y)+1=(x+y-1)2
b) x2-y2+4x+4 = (x2+4x+4)-y2=(x+2)2-y2=(x+y+2)(x-y+2)
c)4x2-y2+8(y-2) = 4x2-(y2-8y+16) = (2x)2-(y-4)2=(2x+y-4)(2x-y+4)
d)x3-2x2+2x-4 = x2(x-2)+2(x-2) = (x-2)(x2+2)
e)xy-4+2x-2y=x(y+2) - 2(y+2) = (x-2)(y+2)
Bài 1 :
\(e,x^2+2xy+y^2-2x-2y+1\)
\(=\left(x+y-1\right)^2\)
Bài 2:
\(b,2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow\left(2x^3+2x\right)+\left(3x^2+3\right)=0\)
\(\Leftrightarrow2x\left(x^2+1\right)+3\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\left(x^2+1>0\right)\)
\(\Leftrightarrow x=-\dfrac{3}{2}\)
A . 5(x-y)-y(x-y)
=(x6-y)(5-y)
B . x^2 - xy - 8x+8y
=(x^2-xy)-(8x-8y))
=x(x-y) - 8(x-y)
C. x^2-10x+25 - y^2
=(x^2 - 10x + 25 ) - y^2
=(x-5)^2 - y^2
=(x-5+y)(x-5-y)
D . x^3 - 3x^2-4x+12
=(x^3 - 3x^2 ) - (4x - 12)
=x^2 (x-3)-4(x-3)
=(x^2-4)(x-3)
=(x+2)(x-2)(x-3)
D . 2x^2-2y^2- 6x-6y
=(2^x - 2y^2) - (6x+ 6y)
=2(x^2 - y^2) - 6(x+y)
=2(x+y)(x-y) - 6(x+y)
=2(x+y)(x-y-3)
E . x^3 - 3x^2 + 3x - 1
=(x-1)^3
D.x^2+3x+2
=x^2+2x+x+2
=(x^2+2x)+(x+2)
=x(x+2)+(x+2)
=(x+2)(x+1)
\(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(\Leftrightarrow\left(x+2\right)3x\)
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