Rút gọn :
A,\(\dfrac{2ax^2-4ax+2a}{5b-5b^2}\)
B,\(\dfrac{(x+y)^2-z^2}{x+y+z}\)
Những câu hỏi liên quan
Rút gọn:
\(a,\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(b,\dfrac{\left(x+y^2\right)-z^2}{x+y+z}\)
\(a,\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{2a\left(x-1^2\right)}{5b\left(x-1\right)\left(1+x\right)}\)
\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)
\(b,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
\(=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)
Đúng 3
Bình luận (0)
Rút gọn:
\(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)
\(\dfrac{2a\cdot x^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{-2a\left(x-1\right)^2}{5b\left(x-1\right)\left(x+1\right)}=\dfrac{-2a\left(x-1\right)}{5b\left(x+1\right)}\)
\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
\(=\dfrac{4x\cdot x-4x\cdot y}{5x^2\cdot x-5x^2\cdot y}\)
\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}\)
=x+y-z
\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3+y^3\right)\left(x^3-y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
Đúng 2
Bình luận (0)
Rút gọn
1). \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
2). \(\dfrac{x^2+4x+3}{2x+6}\)
3). \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
4). \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
1) \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{-5b\left(x^2-1\right)}\)
\(=\dfrac{2a\left(x-1\right)^2}{-5b\left(x-1\right)\left(x+1\right)}=\dfrac{2a\left(x-1\right)}{-5b\left(x+1\right)}\)
2) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+x+3x+3}{2\left(x+3\right)}=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{2\left(x+3\right)}\)\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
3)\(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
4) \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)Học tốt nha you<3
p/s: tớ ko bk đã rút gọn hết chưa:(
Đúng 0
Bình luận (0)
Rút gọn
1). \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
2). \(\dfrac{x^2+4x+3}{2x+6}\)
3). \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
4). \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
5). \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)
1, \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)
\(=\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\)
\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)
2, \(\dfrac{x^2+4x+3}{2x+6}\)
\(=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}\)
\(=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
3, \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
4, \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)
5, \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)
\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)
Đúng 0
Bình luận (0)
Rút gọn
A= \(\frac{2ax^2-4ax+2a}{5b-bx^2}\)
B=\(\frac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\)
B=\(\frac{5\left(x-y\right)-3\left(x-y\right)}{10\left(x-y\right)}\)
B=\(\frac{\left(x-y\right)\left(5-3\right)}{10\left(x-y\right)}\)
B= \(\frac{\left(x-y\right)2}{10\left(x-y\right)}\)
B= 5
vậy B=5
Đúng 0
Bình luận (0)
Bài 6. Rút gọn các phân thức sau
c) \(\frac{2ax^2-4ax+2a}{5b-5b^2}\)
f) \(\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)
a) dfrac{x^2-16}{4x-x^2}
b) dfrac{x^2+4x+3}{2x+6}
c) dfrac{5left(x-yright)-3left(y-xright)}{10left(x-yright)}
d) dfrac{2ax^2-4ax+2a}{5b-5bx^2}
e) dfrac{left(x+yright)^2-z^2}{x+y+z}
f) dfrac{4x^2-4xy}{5x^3-5x^2y}
Nguyễn Huy Tú, Hoàng Ngọc Anh, Toshiro Kiyoshi,...
Đọc tiếp
a) \(\dfrac{x^2-16}{4x-x^2}\)
b) \(\dfrac{x^2+4x+3}{2x+6}\)
c) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\)
d) \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)
e) \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)
f) \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)
Nguyễn Huy Tú, Hoàng Ngọc Anh, Toshiro Kiyoshi,...
a) \(\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left[-\left(x-4\right)\right]}\)
\(=\dfrac{\left(-1\right)\left(x+4\right)}{x}=\dfrac{-x-4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
d) \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}=\dfrac{2a\left(x-1\right)^2}{5b\left[-\left(x-1\right)\right]\left(1+x\right)}\)
\(=\dfrac{2a\left[-\left(x-1\right)\right]}{5b\left(1+x\right)}=-\dfrac{2ax-2a}{50\left(1+x\right)}=-\dfrac{-\left(2a-2ax\right)}{5b\left(1+x\right)}-\dfrac{2a-2ax}{5b-5bx}\)
e) \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=z+y-z\)
f) \(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4x}{5x^2}=\dfrac{4}{5x}\)
Đúng 0
Bình luận (1)
a, \(\dfrac{x^2-16}{4x-x^2}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(4-x\right)}=\dfrac{-x-4}{x}=-1-\dfrac{4}{x}\)
b, \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+x+3x+3}{2\left(x+3\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x+1\right)}{2\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{x+1}{2}\)
c,\(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{4}{5}\)
d, \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}=\dfrac{a\left(2x^2-4x+2\right)}{-5b\left(x^2-1\right)}\)
\(=\dfrac{2a\left(x-1\right)^2}{-5b\left(x-1\right)\left(x+1\right)}=\dfrac{2a\left(x-1\right)}{-5b\left(x+1\right)}\)
e, \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
f, \(\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
Chúc bạn học tốt!!!
Đúng 0
Bình luận (2)
rút gọn phân thức
2ax mũ 2 - 4ax + 2a / 5b - 5bx mũ 2
cứu dzớiiiiii
\(\frac{2ax^2-4ax+2a}{5b-5bx^2}\)
\(=-\frac{2a\left(x-1\right)^2}{5b\left(x+1\right)\left(x-1\right)}\)
\(=-\frac{2a\left(x-1\right)}{5b\left(x+1\right)}\)
1, Rút gọn các phân thức sau :
a, dfrac{x^2-xy}{3xy-3y^2} (x # y, y # 0)
b, dfrac{2ax^2-4ax+2a}{5b-5bx^2} (b # 0, x # pm1)
c, dfrac{4x^2-4xy}{5x^3-5x^2y} ( x 3 ), x # y)
d, dfrac{left(x+yright)^2-z^2}{x+y+z} (x+y+z # 0)
e, dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6} ( x # 0, x # pm y)
2, Rút gọn, rồi tính giá trị các phân thức sau :
a, A dfrac{2x^2+2xleft(x-2right)^2}{left(x^3-4xright)left(x+1right)} với x dfrac{1}{2}
b, Bdfrac{x^3-x^2y+xy^2}{x^3+y^3} với x -5; y 10
3, Rút gọn các phân thức sa...
Đọc tiếp
1, Rút gọn các phân thức sau :
a, \(\dfrac{x^2-xy}{3xy-3y^2}\) (x # y, y # 0)
b, \(\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\) (b # 0, x # \(\pm1\))
c, \(\dfrac{4x^2-4xy}{5x^3-5x^2y}\) ( x 3 ), x # y)
d, \(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\) (x+y+z # 0)
e, \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\) ( x # 0, x # \(\pm y\))
2, Rút gọn, rồi tính giá trị các phân thức sau :
a, A= \(\dfrac{2x^2+2x\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\) với x = \(\dfrac{1}{2}\)
b, B=\(\dfrac{x^3-x^2y+xy^2}{x^3+y^3}\) với x = -5; y = 10
3, Rút gọn các phân thức sau :
a, \(\dfrac{\left(a+b\right)^2-c^2}{a+b+c}\)
b, \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)
c, \(\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
Câu 1:
\(\text{a) }\dfrac{x^2-xy}{3xy-3y^2}=\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}=\dfrac{x}{3y}\)
\(\text{b) }\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\\ =\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\\ =\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)^2}{5b\left(x-1\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\\ =-\dfrac{2ax-2a}{5bx+5b}\)
\(\text{c) }\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)
\(\text{d) }\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)
\(\text{e) }\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x+y\right)^3}\\ =\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\\ =\dfrac{x^3+y^3}{x^4-xy^3}\)
Đúng 0
Bình luận (0)
Câu 3:
\(\text{ a) }\dfrac{\left(a+b\right)^2-c^2}{a+b+c}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)
\(\text{b) }\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\\ =\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}\\ =\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\\ =\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}\\ =\dfrac{a+b-c}{a-b+c}\)
\(\text{c) }\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\\ =\dfrac{2x^3-x^2-6x^2+3x-15x+45}{3x^3-10x^2-9x^2+3x+30x-9}\\ =\dfrac{\left(2x^3-x^2-15x\right)-\left(6x^2-3x-45\right)}{\left(3x^3-10x^2+3x\right)-\left(9x^2-30x+9\right)}\\ =\dfrac{x\left(2x^2-x-15\right)-3\left(2x^2-x-15\right)}{x\left(3x^2-10x+3\right)-3\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-6x+5x-15\right)}{\left(x-3\right)\left(3x^2-9x-x+3\right)}\\ =\dfrac{\left(x-3\right)\left[\left(2x^2-6x\right)+\left(5x-15\right)\right]}{\left(x-3\right)\left[\left(3x^2-9x\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left[x\left(x-3\right)+5\left(x-3\right)\right]}{\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x-3\right)\left(3x-1\right)}\\ =\dfrac{x+5}{3x-1}\)
Đúng 0
Bình luận (0)
Câu 2:
\(A=\dfrac{2x^2+2x\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\\ A=\dfrac{2x\left(x+x^2-4x+4\right)}{x\left(x^2-4\right)\left(x+1\right)}\\ A=\dfrac{2x\left(x^2-3x+4\right)}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\\ \RightarrowĐKXĐ:x\left(x-2\right)\left(x+2\right)\left(x+1\right)\ne0\\ \Leftrightarrow\left[{}\begin{matrix}x-2\ne0\\x+2\ne0\\x+1\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne2\\x\ne-2\\x\ne-1\\x\ne0\end{matrix}\right.\\ \Rightarrow x=\dfrac{1}{2}\text{ }thõa\text{ }mãn\text{ }với\text{ }ĐKXĐ\text{ }của\text{ }A\\ Thay\text{ }x=\dfrac{1}{2}\text{ }vào\text{ }biểu\text{ }thức,\text{ }ta\text{ }\text{ được: }\\ A=\dfrac{2\cdot\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2-3\cdot\dfrac{1}{2}+4\right]}{\dfrac{1}{2}\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)\left(\dfrac{1}{2}+1\right)}\\ A=\dfrac{\dfrac{23}{4}}{-\dfrac{45}{16}}=-\dfrac{1035}{64}\\ \text{Vậy }A=-\dfrac{1035}{64}\text{ }tại\text{ }x=\dfrac{1}{2}\)
\(\text{b) }B=\dfrac{x^3-x^2y+xy^2}{x^3+y^3}\\ B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\\ B=\dfrac{x}{x+y}\\ \RightarrowĐKCD\text{ }của\text{ }B:x+y\ne0\\ \Leftrightarrow x\ne-y\\ \Rightarrow x=-5;y=10\text{ }thõa\text{ }mãn\text{ }với\text{ }ĐKCĐ\text{ }của\text{ }B\\ Thay\text{ }x=-5;y=10\text{ }vào\text{ }biểu\text{ }thức,\text{ }ta\text{ được }:\\ B=\dfrac{-5}{-5+10}=-1\\ \text{ Vậy }B=-1\text{ }tại\text{ }x=-5;y=10\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời