Violympic toán 8

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

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💋Amanda💋
28 tháng 3 2020 lúc 19:27
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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. \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{\left(x+y\right).3\left(x-y\right)\left(x+y\right)}{\left(y-x\right).x\left(x+y\right)}=\frac{-3x-3y}{x}\)

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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

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