x/2=y/3 và 2x+2xy=16
Những câu hỏi liên quan
a)x^3-2x62+x
b)2x^2+4x+2-2y^2
c)2xy-x^2-4^2+16
d)x^3+2x^2y+xy^2-yx
e)2x-2y-x^2+2xy-y^2
x mũ 2 + x mũ 2 . y - y - 1
x mũ 2 + y mũ 2 - 2xy - 25
( 2x - 1 ) ( x mũ 2 + 2x - 1) - ( 1 - 2x ) (x - 3)
a mũ 2 + x mũ 2 -16 + 2ax
1) x2 + x2y - y - 1
= x2( 1 + y ) - ( 1 + y )
= ( 1 + y )( x2 - 1 )
= ( 1 + y )( x - 1 )( x + 1 )
2) x2 + y2 - 2xy - 25
= ( x2 - 2xy + y2 ) - 25
= ( x - y )2 - 52
= ( x - y - 5 )( x - y + 5 )
3) ( 2x - 1 )( x2 + 2x - 1 ) - ( 1 - 2x )( x - 3 )
= ( 2x - 1 )( x2 + 2x - 1 ) + ( 2x - 1 )( x - 3 )
= ( 2x - 1 )( x2 + 2x - 1 + x - 3 )
= ( 2x - 1 )( x2 + 3x - 4 )
= ( 2x - 1 )( x2 - x + 4x - 4 )
= ( 2x - 1 )[ x( x - 1 ) + 4( x - 1 ) ]
= ( 2x - 1 )( x - 1 )( x + 4 )
4) a2 + x2 - 16 + 2ax
= ( a2 + 2ax + x2 ) - 16
= ( a + x )2 - 42
= ( a + x - 4 )( a + x + 4 )
PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ
a.2x^2-4x-8y^2+2
b.16+2xy-x^2-y^2
c.x^2-4+3.(x-2)^2
d.x^4+2x^2-15
c: \(x^2-4+3\left(x-2\right)^2\)
\(=\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(3x-6\right)\)
\(=\left(x-2\right)\left(x+2+3x-6\right)\)
\(=\left(4x-4\right)\left(x-2\right)\)
\(=4\left(x-1\right)\left(x-2\right)\)
Đúng 1
Bình luận (0)
Tìm x,y
a) x2(x+3)+y2(y+5)-(x+y) (x2-xy+y2)=0
b) (2x-y) (4x2+2xy+y2) +(2x+y) (4x2-2xy+y2)-16(x2-y)=32
a ) \(x^2\left(x+3\right)+y^2\left(y+5\right)-\left(x+y\right)\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow x^3+3x^2+y^3+5y^2-\left(x^3+y^3\right)=0\)
\(\Leftrightarrow3x^2+5y^2=0\)
Do \(\left\{{}\begin{matrix}3x^2\ge0\forall x\\5y^2\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow3x^2+5y^2\ge0\forall x;y\)
Dấu " = " xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}3x^2=0\\5y^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)
Vậy \(x=0;y=0\)
b )\(\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(-16\left(x^3-y\right)=32\)
\(\Leftrightarrow\left[\left(2x\right)^3-y^3\right]+\left[\left(2x\right)^3+y^3\right]-16x^3+16y=32\)
\(\Leftrightarrow8x^3-y^3+8x^3+y^3-16x^3+16y=32\)
\(\Leftrightarrow16y=32\)
\(\Leftrightarrow y=2\)
Vậy \(y=2\)
Đúng 0
Bình luận (0)
a) x^2+4x+4-y^2
b) x^2-16-4xy+4y^2
c) x^3+2x^2y +xy^2
d) 5x+5y-x^2-2xy-y^2
e) x^5-x^4+x^3-x^2
a) \(x^2+4x+4-y^2\)
\(=\left(x^2+2.x.2+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
Đúng 0
Bình luận (0)
\(a,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\\ b=\left(x-2y\right)^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\\ c,=x\left(x^2+2xy+y^2\right)=x\left(x+y\right)^2\\ d,=5\left(x+y\right)-\left(x+y\right)^2=\left(5-x-y\right)\left(x+y\right)\\ e,=x^4\left(x-1\right)+x^2\left(x-1\right)\\ =x^2\left(x^2+1\right)\left(x-1\right)\)
Đúng 2
Bình luận (0)
a: \(x^2+4x+4-y^2=\left(x+2-y\right)\left(x+2+y\right)\)
b: \(x^2-4xy+4y^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\)
c: \(x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)=x\left(x+y\right)^2\)
Đúng 0
Bình luận (0)
Cho \(x+y=1\). Tính :
a) \(A=x^4-xy^3+yx^3-y^4+y^3-x^3-2\)
b) \(B=3x+3y+2x^2y+2xy^2-2xy+5x^3y^2+5x^2y^3-5x^2y^2+3\)
c) \(C=3xy\left(x+y\right)+2x^3y+2x^2y^2-2x^2y+\sqrt{16}-3xy\)
tính giá trị của biểu thức : (-> nhân tử)
a) x^2+x-2x-2 tại x=-1
b)3x^2-2x+9x-6 tại x=7
c)2x^2-3xy-xy^2 tại x=2, y=3
d)x^2+y^2+2x-2y-2xy tại x=-5, y=1
e) 2(x+y)^3 + 16 tại x=1,y=3
Xem chi tiết
a: \(x^2+x-2x-2\)
\(=x\left(x+1\right)-2\left(x+1\right)\)
\(=\left(x+1\right)\left(x-2\right)=\left(-1+1\right)\left(-1-2\right)=0\)
b: \(3x^2-2x+9x-6\)
\(=x\left(3x-2\right)+3\left(3x-2\right)\)
\(=\left(3x-2\right)\left(x+3\right)=\left(3\cdot7-2\right)\left(7+3\right)\)
\(=19\cdot10=190\)
c: \(2x^2-3xy-xy^2\)
\(=x\left(2x-3y-y^2\right)\)
\(=2\left(2\cdot2-3\cdot3-9\right)\)
\(=2\cdot\left(4-18\right)=-28\)
Đúng 0
Bình luận (0)
Cho x+y=1x+y=1. Tính :
a) \(A=x^4-xy^3+yx^3-y^4+y^3-x^3-2\)
b) \(B=3x+3y+2x^2y+2xy^2-2xy+5x^3y^2+5x^2y^3-5x^2y^2+3\)
c) \(C=3xy\left(x+y\right)+2x^3y+2x^2y^2-2x^2y+\sqrt{16}-3xy\)
Giải phương trình nghiệm nguyên: a)\(2x^4+3x^2=x^3+x^2y+x+y+16\) b)\(2x^3=x^2+2xy+13x+y+86\)