a,3xy+x+15y+5
b,9-x^2-2xy-y^2
c,x^3-5x^2+x-5
d,x^2+y^2-2xy-1
tìm x
(5x-1)=(1-5x)^2
9-(x-2)^2=0
a,3xy+x+15y+5
b,9-x^2-2xy-y^2
c,x^3-5x^2+x-5
d,x^2+y^2-2xy-1
tìm x
(5x-1)=(1-5x)^2
9-(x-2)^2=0
a. \(3xy+x+15y+15=x\left(3y+1\right)+15\left(y+1\right)=\left(x+15\right)\left(3y+1\right)\)
b.\(9-x^2-2xy-y^2=9-\left(x+y\right)^2=\left(3+x+y\right)\left(3-x-y\right)\)
c.\(x^3-5x^2+x-5=x^2\left(x-5\right)+\left(x-5\right)=\left(x^2+1\right)\left(x-5\right)\)
d.\(x^2-2xy+y^2-1=\left(x-y\right)^2-1=\left(x-y+1\right)\left(x-y-1\right)\)
\(\left(5x-1\right)=\left(1-5x\right)^2\)
\(\left(5x-1\right)=\left(5x-1\right)^2\)
\(\left(5x-1\right)\left(1-5x+1\right)=0\)
\(\left(5x-1\right)\left(2-5x\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=\frac{2}{5}\end{array}\right.\)
\(9-\left(x-2\right)^2=0\)
\(\left(3+x-2\right)\left(3-x+2\right)=0\)
\(\left(1+x\right)\left(5-x\right)=0\)
\(\left[\begin{array}{nghiempt}x=-1\\x=5\end{array}\right.\)
Dạng 1: Phân tích đa thức sau thành nhân tử 1) x ^ 2 - 9 2) 5x - 5y + ax - ay 3) x ^ 2 + 6x + 9 4) 10x * (x - y) - 7y * (y - x) 5) 5x - 15y 6) x ^ 2 - 2xy + y ^ 2 - z ^ 2
\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
chứng minh rằng nếu các cặp x,y thoả mãn các đẳng thức :
x2-3xy+2y2+x-y=0 (1) và x2-2xy+y2-5x+7y=0 (2) thì cũng thoả mãn đẳng thức xy-12x+15y=0
Đặt \(xy-12x+15y\)là (*)
Từ phương trình (1) ta có \(x^2-3xy+2y^2+x-y=0\Leftrightarrow\left(x-y\right)\left(x-2y\right)+\left(x-y\right)=0\)
\(\Leftrightarrow\left(x-y\right)\left(x-2y+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=y\\x=2y-1\end{cases}}\)
Với \(x=y\)thay vào (2) ta có \(x^2-2x^2+x^2-5x+7x=0\Leftrightarrow x=0\Rightarrow x=y=0\)
Thay \(x=y=0\)vào (*) ta thấy 0.0-12.0+15.0=0(tm)
Với \(x=2y-1\Rightarrow\left(2y-1\right)^2-2\left(2y-1\right)y+y^2-5\left(2y-1\right)+7y=0\)
\(\Leftrightarrow4y^2-4y+1-4y^2+2y+y^2-10y+5+7y=0\)
\(\Leftrightarrow y^2-5y+6=0\Leftrightarrow\left(y-2\right)\left(y-3\right)=0\Leftrightarrow\orbr{\begin{cases}y=2\\y=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}}\)
Với \(x=3;y=2\)thay vào (*) ta thấy \(3.2-12.3+15.0=0\left(tm\right)\)
Với \(x=5;y=3\)thay vào (*) ta thấy \(5.3-12.5+15.3=0\left(tm\right)\)
Vậy .....
Rút gọn rồi tính giá trị của biểu thức :
A = 5x(x-y) - y(5x-y) với x =-1 ; y =3
B = 4y(x2 - 3xy + 3y2) - 2xy(2x-6y-3) với x = 5; y = -1
C = 5x2(x-y2) + 3x(xy2-y) -5x3 với x=-2 ; y=-5
D = 6x2(y2-xy+2x2y) -3xy(2xy-x2+4x3) với x=11; y=-1
A = 5x(x - y) - y(5x - y)
A = 5x2 - 5xy - 5xy + y2
A = 5x2 - 10xy + y2 (1)
Thay x = -1; y = 3 vào (1), ta có:
5.(-1)2 - 10.(-1).3 + 32 = 44
B = 4y(x2 - 3xy + 3y2) - 2xy(2x - 6y - 3)
B = 4x2y - 12x2 + 12y3 - 4x2y + 12xy2 + 6xy
B = 12y3 + 6xy (1)
Thay x = 5; y = -1 vào (1), ta có:
12.(-1)3 + 6.5.(-1) = -42
C = 5x2(x - y2) + 3x(xy2 - y) - 5x3
C = 5x3 - 5x2y2 + 3x2y2 - 3xy - 5x3
C = -2x2y2 - 3xy (1)
Thay x = -2; y = -5 vào (1), ta có:
-2.(-2)2.(-5)2 - 3.(-2).(-5) = -230
D = 6x2(y2 - xy + 2x2y) - 3xy(2xy - x2 + 4x3)
D = 6x2y2 - 6x3y + 12x4y - 6x2y2 + 3x3y - 12x4y
D = -3x3y (1)
Thay x = 11; y = -1 vào (1), ta có:
-3.113.(-1) = 3993
1/tìm các cặp số nguyên (x;y) thỏa mãn:\(5x^2+2xy+y^2-4x-40=0\)0
2/tìm các số nguyên x;y thỏa mãn:\(3xy+x+15y-44=0\)
3/gtp nghiệm nguyên :\(2x^2+3xy-2y^2=7\)
\(3xy+x+15y-44=0\)
\(3y\left(x+5\right)+\left(x+5\right)-49=0\)
\(\left(x+5\right)\left(3y+1\right)=49\)
Vì x;y là số nguyên \(\Rightarrow\hept{\begin{cases}x+5\in Z\\3y+1\in Z\end{cases}}\)
Có \(\left(x+5\right)\left(3y+1\right)=49\)
\(\Rightarrow\left(x+5\right)\left(3y+1\right)\in\text{Ư}\left(49\right)=\left\{\pm1;\pm7;\pm49\right\}\)
b tự lập bảng nhé~
phân tích đa thức thành nhân tử:a)x^5-x^3+x^2-1/5
b)5x^3-45x
c)16x^4y^2+2xy^5
d)a^3-8+6a^2-12a
e)x^4+x^3+x+1
a: Ta có: \(x^5-x^3+x^2-1\)
\(=x^3\left(x^2-1\right)+\left(x^2-1\right)\)
\(=\left(x-1\right)\cdot\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
b: Ta có: \(5x^3-45x\)
\(=5x\left(x^2-9\right)\)
\(=5x\left(x-3\right)\left(x+3\right)\)
c: Ta có: \(16x^4y^2+2xy^5\)
\(=2xy^2\left(8x^3+y^3\right)\)
\(=2xy^2\cdot\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
d: Ta có: \(a^3-8+6a^2-12a\)
\(=\left(a-2\right)\left(a^2+2a+4\right)+6a\left(a-2\right)\)
\(=\left(a-2\right)\left(a^2+8a+4\right)\)
e: Ta có: \(x^4+x^3+x+1\)
\(=x^3\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)^2\cdot\left(x^2-x+1\right)\)
a) 3x 2 (2x 3 – x + 5)
b) (4xy + 3y – 5x)x 2 y
c) (3x 2 y – 6xy + 9x)(- 3
4
xy)
d) - 3
1
xz(- 9xy + 15yz) + 3x 2 (2yz 2 – yz)
e) (x 3 + 5x 2 – 2x + 1)(x – 7)
f) (2x 2 – 3xy + y 2 )(x + y)
g) (x – 2)(x 2 – 5x + 1) – x(x 2 + 11)
h) [(x 2 – 2xy + 2y 2 )(x + 2y) - (x 2 + 4y 2 )(x – y)] 2xy
Mọi người giúp em với ạ
Đề bài là gì sao không ghi rõ??
phân tích thành nhân tử
`3x^2 -3xy-5x+5y`
`2x^3 y-2xy^3 -4xy^2 -2xy`
`x^2 -1+2x-y^2`
`x^2 +4x-2xy-4y+4y^2`
`x^3 -2x^2 +x`
`2x^2 +4x+2-2y^2`
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
Thuc hien phep tinh
a, [ x2y2- 1/3xy+3y13xy+3y ] (x-3y)
b, (x^2+xy+y^2) (x-y)
c, (1/5x-1) (x^2-5x+2)
d, (x^2-2xy+y^2) (x-y)
d. ( x2 - 2xy + y2 ) ( x-y )
= ( x- y )2 ( x- y )
= ( x - y )3
b, (x^2+xy+y^2) (x-y)
= ( x+ y )2 ( x- y )
= ( x2 - y 2 ) ( x +y )