Giải thích các bước giải:
Bài 4:
a.Ta có:
$\dfrac{a}{b}=\dfrac bc=\dfrac cd=\dfrac{a+b+c}{b+c+d}$
$\to (\dfrac{a+b+c}{b+c+d})^3=\dfrac ab.\dfrac bc.\dfrac cd=\dfrac ad$
$\to \dfrac{(a+b+c)^3}{(b+c+d)^3}=\dfrac ad$
b.Ta có:
$\dfrac{a}{b}=\dfrac bc=\dfrac cd$
$\to \dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac ab.\dfrac bc.\dfrac cd$
$\to \dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac ad$
Bài 8.
c.Ta có:
$\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}$
$\to \dfrac{2x+3y-1}{6x}=\dfrac{2x+3y-1}{12}$
$\to 2x+3y-1=0$
$\to\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=0$
$\to 2x+1=3y-2=0$
$\to x=-\dfrac12, y=\dfrac23$
Hoặc $6x=12\to x=2$
$\to \dfrac{2\cdot 2+1}{5}=\dfrac{3y-2}{7}$
$\to \dfrac{3y-2}{7}=1$
$\to 3y-2=7$
$\to 3y=9$
$\to y=3$
Bài 9:
Ta có: $\dfrac x2=\dfrac y5=\dfrac z7=k$
$\to x=2k, y=5k, z=7k$
a.Ta có:
$P=\dfrac{2k-5k+3\cdot 7k}{2k+2\cdot 5k-7k}$
$\to P=\dfrac{18k}{5k}$
$\to P=\dfrac{18}{5}$
b.Ta có:
$Q=\dfrac{(2k)^2+3(5k)^2-5(7k)^2}{7(2k)^2+2(7k)^2}$
$\to Q=\dfrac{-166k^2}{126k^2}$
$\to Q=-\dfrac{83}{63}$
