Question

1,tìm x,y:

a)0,2 : 1\(\dfrac{1}{6}\)= \(\dfrac{2}{3}\): (6x+7)

b)\(\dfrac{a}{a+2b}\)= \(\dfrac{c}{c+2d}\). Tính \(\dfrac{a^2.d^2-4b^2.c^2}{abcd}\)

c)\(\dfrac{a}{b}\)= \(\dfrac{a^2+c^2}{ab+cd}\).CMR:\(\dfrac{a}{b}\)= \(\dfrac{c}{d}\)(b,d khác 0)

Answer 1

a: \(\dfrac{2}{3}:\left(6x+7\right)=0.2:1\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{2}{3}:\left(6x+7\right)=\dfrac{1}{5}:\dfrac{7}{6}=\dfrac{6}{35}\)

\(\Leftrightarrow6x+7=\dfrac{35}{9}\)

=>6x=-28/9

hay x=-28/54=-14/27

b: \(\dfrac{a}{a+2b}=\dfrac{c}{c+2d}\)

\(\Leftrightarrow a\left(c+2d\right)=c\left(a+2b\right)\)

\(\Leftrightarrow ac+2ad=ac+2bc\)

=>2ad=2bc

=>ad=bc

=>a/b=c/d

Đặt a/b=c/d=k

=>a=bk; c=dk

\(A=\dfrac{a^2\cdot d^2-4b^2\cdot c^2}{abcd}=\dfrac{b^2k^2\cdot d^2-4\cdot b^2\cdot d^2k^2}{bk\cdot b\cdot dk\cdot d}\)

\(=\dfrac{-3b^2k^2d^2}{b^2k^2d^2}=-3\)