f(x,y)=(2x−5y+4)(5x+2y−1)
Rút gọn các phân thức sau:
a) \(\dfrac{5x}{10}\)
b)\(\dfrac{4xy}{2y}\) (y≠0)
c)\(\dfrac{5x-5y}{3x-3y}\) (x≠y)
d) \(\dfrac{x^2-y^2}{x+y}\)(chưa có điều kiện xác định)
e) \(\dfrac{x^3-x^2+x-1}{x^2-1}\)(chưa có điều kiện xác định)
f) \(\dfrac{x^2+4x+4}{2x+4}\)(chưa có điều kiện xác định)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)
d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)
e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)
f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)
giải các hệ phương trình sau
a.{ x + 3y = -2
{ 5x - 4y = 11
b.{ 3xy = 5
{ 5x + 2y = 23
c.{ 3x +5y = 1
{ 2x - y = -8
d.{ x - 2y + 6 = 0
{ 5x - 3y - 5 = 0
e.{ 2(x + y) + 3(x - y) = 4
{ (x + y) + 2(x - y) = 5
\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)
\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
1, Rút gọn các phân thức sau :
a, \(\dfrac{5x}{10}\)
b, \(\dfrac{4xy}{2y}\) ( y # 0)
c, \(\dfrac{21x^2y^3}{6xy}\) ( xy # 0)
d, \(\dfrac{2x+2y}{4}\)
e, \(\dfrac{5x-5y}{3x-3y}\) ( x # y)
f, \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}\) ( x # y)
2, Rút gọn các phân thức sau :
a, \(\dfrac{x^2-16}{4x-x^2}\) ( x # 0, x # 4)
b, \(\dfrac{x^2+4x+3}{2x+6}\) ( x # -3)
c, \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}\) ( y + ( x+y) # 0)
d, \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\) ( x # y)
e, \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}\) (x # -y)
1)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)
d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)
e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)
f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)
\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)
2)
a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )
d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
1/2.(6x-2y).(3x+y)
(2/3z-2/5x).(1/3z+1/5x).1/2
(5y-3x).1/4.(12x+20y)
(3/4y-1/2x).(x+3/2y).2
(a+b+c).(a+b-c)
(x-y+z).(x+y-z)
mng giúp mình vs ạ
\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
a: \(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\left(3x-y\right)\cdot\left(3x+y\right)=9x^2-y^2\)
b: \(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\cdot\dfrac{1}{2}\)
\(=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\)
\(=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
c: \(\left(5y-3x\right)\cdot\dfrac{1}{4}\cdot\left(12x+20y\right)\)
\(=\left(5y-3x\right)\left(5y+3x\right)\)
\(=25y^2-9x^2\)
d: \(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(\dfrac{3}{2}y+x\right)\cdot2\)
\(=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)\)
\(=\dfrac{9}{4}y^2-x^2\)
e: \(\left(a+b+c\right)\left(a+b-c\right)\)
\(=\left(a+b\right)^2-c^2\)
\(=a^2+2ab+b^2-c^2\)
BT10: Thực hiện phép tính
\(a,\dfrac{4}{5}y^2x^5-x^3.x^2y^2\)
\(b,-xy^3-\dfrac{2}{7}y^2.xy\)
\(c,\dfrac{5}{6}xy^2z-\dfrac{1}{4}xyz.y\)
\(d,15x^4+7x^4-20x^2.x^2\)
\(e,\dfrac{1}{2}x^5y-\dfrac{3}{4}x^5y+xy.x^4\)
\(f,13x^2y^5-2x^2y^5+x^6\)
a: =-1/5x^5y^2
b: =-9/7xy^3
c: =7/12xy^2z
d: =2x^4
e: =3/4x^5y
f: =11x^2y^5+x^6
Phân tích mỗi đa thức sau thành nhân tử
a)x^3-2x^2y+xy^2+xy
b)x^3+4x^2y+4xy^2-9x
c)x^3-y^3+x-y
d)4x^2-4xy+2x-y+y^2
e)9x^2-3x+2y-4y^2
f)3x^2-6xy+3y^2-5x+5y
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
chung minh rang x,y thuoc Z va (5x-2y)x(2x-5y) chia het cho 7 thi tich (5x-2y)x(2x-5y) co nhieu hon 2 uoc chinh phuong
chung minh rang x,y thuoc Z va (5x-2y)x(2x-5y) chia het cho 7 thi tich (5x-2y)x(2x-5y) co nhieu hon 2 uoc chinh phuong
dùng tính chất cơ bản của phân thức ,hãy điền 1 đa thức thích hợp vào chỗ trống trong các đẳng thức sau:
a)x-x^2/5x^2-5 =x/....
b)x^2+8/2x-1=3x^3+24x/..........
c)....../x-y=3x^2-3xy/3(y-x)^2
d)-x^2+2xy-y^2/x+y=....../y^2+x^2
e)x^3+x^2/x^2-1=..../x-1
f)5x+5y/.....=5x^2-5y^2/2y-2x