ta có: \(\frac{2}{x+1}=\frac{3}{y+2}=\frac{4}{z+3}\)
\(=\frac{4}{2.\left(x+1\right)}=\frac{9}{3.\left(y+2\right)}=\frac{12}{4\left(z+3\right)}=\frac{4}{2x+2}=\frac{9}{6+3y}=\frac{16}{12+4z}\)
ADTCDTSBN
có: \(\frac{4}{2x+2}=\frac{9}{6+3y}=\frac{16}{12+4z}=\frac{4+9+16}{\left(2x+3y+4z\right)+\left(2+6+13\right)}=\frac{29}{20+21}=\frac{29}{41}\)
\(\Rightarrow\frac{2}{x+1}=\frac{29}{41}\Rightarrow x+1=2:\frac{29}{41}=\frac{82}{29}\Rightarrow x=\frac{53}{29}\)
\(\frac{3}{y+2}=\frac{29}{41}\Rightarrow y+2=3:\frac{29}{41}=3\Rightarrow y=1\)
\(\frac{4}{z+3}=\frac{29}{41}\Rightarrow z+3=4:\frac{29}{41}=\frac{164}{29}\Rightarrow z=\frac{164}{29}-3=\frac{77}{29}\)
KL:.................................
