Những câu hỏi liên quan
Thơ Nụ =))
Xem chi tiết

\(\dfrac{2a\cdot x^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{-2a\left(x-1\right)^2}{5b\left(x-1\right)\left(x+1\right)}=\dfrac{-2a\left(x-1\right)}{5b\left(x+1\right)}\)

\(\dfrac{4x^2-4xy}{5x^3-5x^2y}\)

\(=\dfrac{4x\cdot x-4x\cdot y}{5x^2\cdot x-5x^2\cdot y}\)

\(=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)

\(\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

\(=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}\)

=x+y-z

\(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3+y^3\right)\left(x^3-y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)

Ru Bakaa
Xem chi tiết
Nguyen My Van
11 tháng 5 2022 lúc 10:22

\(a,\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\)

\(=\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\)

\(=\dfrac{2a\left(x-1^2\right)}{5b\left(x-1\right)\left(1+x\right)}\)

\(=\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\)

\(b,\dfrac{\left(x+y\right)^2-z^2}{x+y+z}\)

\(=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{x+y+z}=x+y-z\)

Thu Phương
Xem chi tiết
Akai Haruma
24 tháng 11 2018 lúc 9:30

a)

\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)

b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)

c)

\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)

d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)

Akai Haruma
24 tháng 11 2018 lúc 9:34

e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)

f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)

g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)

\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)

Akai Haruma
24 tháng 11 2018 lúc 9:37

h)

\(\frac{4x^2-4xy}{5x^3-5x^2y}=\frac{4x(x-y)}{5x^2(x-y)}=\frac{4}{5x}\)

i) \(\frac{(x+y)^2-z^2}{x+y+z}=\frac{(x+y-z)(x+y+z)}{x+y+z}=x+y-z\)

k) \(\frac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\frac{(x^3)^2+2.x^3.y^3+(y^3)^2}{x(x^6-y^6)}\)

\(=\frac{(x^3+y^3)^2}{x(x^3-y^3)(x^3+y^3)}=\frac{x^3+y^3}{x(x^3-y^3)}\)

Kathy Nguyễn
Xem chi tiết
Nguyễn Huy Tú
13 tháng 8 2017 lúc 12:57

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, \(\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\)

\(=\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x^3+y^3\right)}=\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\)

Kathy Nguyễn
Xem chi tiết
T.Thùy Ninh
13 tháng 8 2017 lúc 15:30

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:(

Phương Trần Lê
Xem chi tiết
Nguyễn Lê Phước Thịnh
26 tháng 12 2021 lúc 23:32

a: \(=\dfrac{x-z}{2}\)

b: \(=\dfrac{3x}{4y^3}\)

Ánh Vũ Ngọc
Xem chi tiết
Trần Quốc Lộc
10 tháng 11 2017 lúc 8:54

Câu 1:

\(\text{a) }\dfrac{x^2-xy}{3xy-3y^2}=\dfrac{x\left(x-y\right)}{3y\left(x-y\right)}=\dfrac{x}{3y}\)

\(\text{b) }\dfrac{2ax^2-4ax+2a}{5b-5bx^2}\\ =\dfrac{2a\left(x^2-2x+1\right)}{5b\left(1-x^2\right)}\\ =\dfrac{2a\left(x-1\right)^2}{5b\left(1-x\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)^2}{5b\left(x-1\right)\left(1+x\right)}\\ =-\dfrac{2a\left(x-1\right)}{5b\left(x+1\right)}\\ =-\dfrac{2ax-2a}{5bx+5b}\)

\(\text{c) }\dfrac{4x^2-4xy}{5x^3-5x^2y}=\dfrac{4x\left(x-y\right)}{5x^2\left(x-y\right)}=\dfrac{4}{5x}\)

\(\text{d) }\dfrac{\left(x+y\right)^2-z^2}{x+y+z}=\dfrac{\left(x+y+z\right)\left(x+y-z\right)}{x+y+z}=x+y-z\)

\(\text{e) }\dfrac{x^6+2x^3y^3+y^6}{x^7-xy^6}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^6-y^6\right)}\\ =\dfrac{\left(x^3+y^3\right)^2}{x\left(x^3-y^3\right)\left(x+y\right)^3}\\ =\dfrac{x^3+y^3}{x\left(x^3-y^3\right)}\\ =\dfrac{x^3+y^3}{x^4-xy^3}\)

Trần Quốc Lộc
10 tháng 11 2017 lúc 9:13

Câu 3:

\(\text{ a) }\dfrac{\left(a+b\right)^2-c^2}{a+b+c}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{a+b+c}=a+b-c\)

\(\text{b) }\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\\ =\dfrac{\left(a^2+2ab+b^2\right)-c^2}{\left(a^2+2ac+c^2\right)-b^2}\\ =\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\\ =\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}\\ =\dfrac{a+b-c}{a-b+c}\)

\(\text{c) }\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\\ =\dfrac{2x^3-x^2-6x^2+3x-15x+45}{3x^3-10x^2-9x^2+3x+30x-9}\\ =\dfrac{\left(2x^3-x^2-15x\right)-\left(6x^2-3x-45\right)}{\left(3x^3-10x^2+3x\right)-\left(9x^2-30x+9\right)}\\ =\dfrac{x\left(2x^2-x-15\right)-3\left(2x^2-x-15\right)}{x\left(3x^2-10x+3\right)-3\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\\ =\dfrac{\left(x-3\right)\left(2x^2-6x+5x-15\right)}{\left(x-3\right)\left(3x^2-9x-x+3\right)}\\ =\dfrac{\left(x-3\right)\left[\left(2x^2-6x\right)+\left(5x-15\right)\right]}{\left(x-3\right)\left[\left(3x^2-9x\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left[x\left(x-3\right)+5\left(x-3\right)\right]}{\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]}\\ =\dfrac{\left(x-3\right)\left(x-3\right)\left(x+5\right)}{\left(x-3\right)\left(x-3\right)\left(3x-1\right)}\\ =\dfrac{x+5}{3x-1}\)

Trần Quốc Lộc
10 tháng 11 2017 lúc 9:30

Câu 2:

\(A=\dfrac{2x^2+2x\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\\ A=\dfrac{2x\left(x+x^2-4x+4\right)}{x\left(x^2-4\right)\left(x+1\right)}\\ A=\dfrac{2x\left(x^2-3x+4\right)}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}\\ \RightarrowĐKXĐ:x\left(x-2\right)\left(x+2\right)\left(x+1\right)\ne0\\ \Leftrightarrow\left[{}\begin{matrix}x-2\ne0\\x+2\ne0\\x+1\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne2\\x\ne-2\\x\ne-1\\x\ne0\end{matrix}\right.\\ \Rightarrow x=\dfrac{1}{2}\text{ }thõa\text{ }mãn\text{ }với\text{ }ĐKXĐ\text{ }của\text{ }A\\ Thay\text{ }x=\dfrac{1}{2}\text{ }vào\text{ }biểu\text{ }thức,\text{ }ta\text{ }\text{ được: }\\ A=\dfrac{2\cdot\dfrac{1}{2}\left[\left(\dfrac{1}{2}\right)^2-3\cdot\dfrac{1}{2}+4\right]}{\dfrac{1}{2}\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)\left(\dfrac{1}{2}+1\right)}\\ A=\dfrac{\dfrac{23}{4}}{-\dfrac{45}{16}}=-\dfrac{1035}{64}\\ \text{Vậy }A=-\dfrac{1035}{64}\text{ }tại\text{ }x=\dfrac{1}{2}\)

\(\text{b) }B=\dfrac{x^3-x^2y+xy^2}{x^3+y^3}\\ B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\\ B=\dfrac{x}{x+y}\\ \RightarrowĐKCD\text{ }của\text{ }B:x+y\ne0\\ \Leftrightarrow x\ne-y\\ \Rightarrow x=-5;y=10\text{ }thõa\text{ }mãn\text{ }với\text{ }ĐKCĐ\text{ }của\text{ }B\\ Thay\text{ }x=-5;y=10\text{ }vào\text{ }biểu\text{ }thức,\text{ }ta\text{ được }:\\ B=\dfrac{-5}{-5+10}=-1\\ \text{ Vậy }B=-1\text{ }tại\text{ }x=-5;y=10\)

Hồ Hữu Duyy
Xem chi tiết
Nguyễn Hoàng Minh
13 tháng 12 2021 lúc 22:13

\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)

Trần Quang Minh
Xem chi tiết
Trần Quang Minh
8 tháng 12 2017 lúc 10:42

Giúp mình nhé mọi người ! leuleu

Trần Nguyễn Bảo Quyên
8 tháng 12 2017 lúc 11:10

\(1.\)

\(a.\)

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)

\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)

\(=x-1\)

\(b.\)

\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)

\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)

\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)

\(=\dfrac{2y}{\left(x-y\right)}\)

Tương tự các câu còn lại