Bài 1: Cho 4 số a,b,c,d thỏa mãn \(b^2=ac;c^2=bd\\ \) . Chứng minh \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)
Bài 2 : Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh
a) \(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
b) \(\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
Bài 3 : CMR : Nếu a(y+z)=b(z+x)=c(x+y) trong đó a,b,c là các số thực khác nhau thì \(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)
Bài 4 : Cho \(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\). Chứng minh \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
Bài 5 : CMR : Nếu \(\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4b+c}\) thì \(\dfrac{a}{x+2y+z}=\dfrac{b}{2x+y-z}=\dfrac{c}{4x-4y+z}\)
Ta có:
\(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\left(1\right)\)
\(c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\left(2\right)\)
Từ (1) và (2), suy ra: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\)
Vậy \(\dfrac{a}{d}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\)(đpcm)
~ Học tốt!~
