Violympic toán 8

Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài
lê nhật duẫn

bài 2 : thực hiện phép tính

a. \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)

b. \(\frac{12x}{5y^3}.\frac{15y^4}{8x^3}\)

c.\(\frac{4y^2}{11x^4}.\left(-\frac{3x^2}{8y}\right)\)

d.\(\frac{x^{2-4}}{3x+12}.\frac{x+4}{2x-4}\)

e.\(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}\)

f.\(\frac{x^2-36}{2x+10}.\frac{3}{6-x}\)

g.\(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6}\)

h.\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)

i.\(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}\)

j.\(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}\)

k.\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)

Nguyễn Ngọc Lộc
28 tháng 3 2020 lúc 19:25

ĐKXĐ bạn tự tìm nha : )

k, Ta có : \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}\)

\(=\frac{3x\left(1-2x\right)\left(1+2x\right)}{2x\left(x+4\right)\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)

j, Ta có : \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}:\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}=\frac{x+y}{y-x}.\frac{3\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\)

\(=\frac{3\left(x-y\right)\left(x+y\right)}{x\left(y-x\right)}=\frac{3\left(x-y\right)\left(x+y\right)}{-x\left(x-y\right)}=\frac{-3\left(x+y\right)}{x}\)

i, Ta có : \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right)}{-\left(a-b\right)}:\frac{a+b}{2\left(a^2-b^2\right)}=\frac{a\left(a+b\right)}{-\left(a-b\right)}.\frac{2\left(a-b\right)\left(a+b\right)}{a+b}\)

\(=\frac{2a\left(a+b\right)\left(a-b\right)}{-\left(a-b\right)}=-2a\left(a+b\right)\)

h, = k,

f, Ta có : \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{-3}{x-6}=\frac{-3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(x-6\right)}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)

Khách vãng lai đã xóa
💋Amanda💋
28 tháng 3 2020 lúc 19:27
https://i.imgur.com/1LeIfCN.jpg
Khách vãng lai đã xóa
Bùi Lan Anh
28 tháng 3 2020 lúc 19:29

a. \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right).2\left(2-x\right)}{4\left(x-2\right)\left(x+2\right)}=\frac{-5}{2}\)

b. \(\frac{12x}{5y^3}.\frac{15y^4}{8x^3}=\frac{12x.15y^4}{5y^3.8x^3}=\frac{3.3y}{2x^2}=\frac{9y}{2x^2}\)

c. \(\frac{4y^2}{11x^4}.\left(\frac{-3x^2}{8y}\right)=\frac{4y^2.\left(-3x^2\right)}{11x^4.8y}=\frac{-3y}{22x^2}\)

d. \(\frac{x^2-4}{3x+12}.\frac{x+4}{2x-4}=\frac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}=\frac{x+2}{6}\)

f. \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x+6\right)\left(x-6\right).3}{\left(2x+10\right)\left(6-x\right)}=\frac{-3x-18}{2x+10}\)

g. \(\frac{x^2-9y^2}{x^2y^2}.\frac{3xy}{2x-6}=\frac{\left(x^2-9y^2\right).3xy}{x^2y^2.\left(2x-6\right)}=\frac{3x^2-27y^2}{2x^2y-6xy}\)

h. \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right).3x}{x\left(x+4\right).2\left(1-2x\right)}=\frac{3+6x}{2x+8}\)

i. \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right).2\left(a-b\right)\left(a+b\right)}{\left(b-a\right)\left(a+b\right)}=-2a^2-2ab\)

j. \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{\left(x+y\right).3\left(x-y\right)\left(x+y\right)}{\left(y-x\right).x\left(x+y\right)}=\frac{-3x-3y}{x}\)

Khách vãng lai đã xóa
Hoàng Ngọc Anh
28 tháng 3 2020 lúc 19:48

a)

\(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)

\(\Leftrightarrow\frac{\left(5x+10\right)\cdot-\left(2x-4\right)}{\left(4x-8\right)\cdot\left(x+2\right)}=\frac{5\cdot\left(x+2\right)\cdot-2\cdot\left(x-2\right)}{4\cdot\left(x-2\right)\cdot\left(x+2\right)}=\frac{-10}{4}=\frac{-5}{2}\)

b)

\(\frac{12x}{5y^3}\cdot\frac{15y^4}{8x^3}=\frac{xy^3\cdot\left(12\cdot15y\right)}{xy^3\cdot\left(5\cdot8x^2\right)}=\frac{180y}{40x^2}=\frac{9y}{2x^2}\)

c)

\(\frac{4y^2}{11x^4}\cdot\left(-\frac{3x^2}{8y}\right)=\frac{x^2y\cdot\left(4y\cdot3\right)}{x^2y\cdot\left(11x^2\cdot8\right)}=\frac{12y}{88x^2}=\frac{3y}{22x^2}\)

d)

\(\frac{x^2-4}{3x+12}\cdot\frac{x+4}{2x-4}=\frac{\left(x-2\right)\cdot\left(x+2\right)\cdot\left(x+4\right)}{3\cdot\left(x+4\right)\cdot2\cdot\left(x-2\right)}=\frac{x+2}{6}\)

e)

\(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}=\frac{5\cdot\left(x+2\right)\cdot-2\cdot\left(x-2\right)}{4\cdot\left(x-2\right)\cdot\left(x+2\right)}=\frac{-10}{4}=\frac{-5}{2}\)

f)

\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\cdot\left(x+6\right)\cdot-3}{\left(2x+10\right)\cdot\left(x-6\right)}=\frac{-3x-18}{2x+10}\)

g)

\(\frac{x^2-9y^2}{x^2y^2}\cdot\frac{3xy}{2x-6}=\frac{xy\cdot3\cdot\left(x^2-9y^2\right)}{xy\cdot xy\cdot\left(2x-6\right)}=\frac{3x^2-27y^2}{2x^2y-6xy}\)

h)

\(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{1-4x^2}{x^2+4x}\cdot\frac{3x}{2-4x}=\frac{\left(1-2x\right)\cdot\left(1+2x\right)\cdot3x}{2x\cdot\left(x+4\right)\cdot\left(1-2x\right)}=\frac{3+6x}{2x+8}\)

i)

\(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\cdot\left(a+b\right)\cdot2\cdot\left(a+b\right)\cdot\left(a-b\right)}{-\left(a-b\right)\cdot\left(a+b\right)}=\frac{2a\cdot\left(a+b\right)^2\cdot\left(a-b\right)}{-\left(a-b\right)\cdot\left(a+b\right)}\)

\(=\frac{2a^2+2ab}{-1}=-2a^2-2ab\)

j)

\(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}\cdot\frac{3x^2-3y^2}{x^2+xy}=\frac{\left(x+y\right)\cdot3\cdot\left(x+y\right)\cdot\left(x-y\right)}{-\left(x-y\right)\cdot x\cdot\left(x+y\right)}=\frac{3\cdot\left(x+y\right)^2\cdot\left(x-y\right)}{-x\cdot\left(x-y\right)\cdot\left(x+y\right)}=\frac{3x+3y}{-x}\)

k) câu h

Khách vãng lai đã xóa

Các câu hỏi tương tự
lê nhật duẫn
Xem chi tiết
lê nhật duẫn
Xem chi tiết
Hoàng Diệu Anh
Xem chi tiết
mi tra
Xem chi tiết
Thanh Thanh
Xem chi tiết
Cao Thị Minh Vui
Xem chi tiết
Đỗ Lệ Huyền
Xem chi tiết
nguyễn hoài thu
Xem chi tiết
Trần Tuấn Kiệt
Xem chi tiết