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=(a+c+d+c+d+a+d+a+b+a+b+c)/(a+b+c+d)=3.(a+b+c+d)/a+b+c+d=3
=> k=3
Có nhiều cách:
mình làm 1 câu thôi
a) + \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
+\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\Rightarrow ad+bd=bc+bd\Rightarrow\left(a+b\right)d=\left(c+d\right)b\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\)
+Đặt \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=k\Rightarrow a=ck;b=dk\Rightarrow\frac{a+b}{b}=\frac{ck+dk}{dk}=\frac{c+d}{d}\)
cong tat ca lai ta co
a/b+c+d=b/c+d+a=c/a+b+d=d/a+b+c
=>a+b+c/3(a+b+c)
tom lai ket qua la 4
=>\(\frac{a}{d+b+c}\)+1=\(\frac{b}{c+a+d}\)+1=\(\frac{c}{a+b+d}\)+1=\(\frac{d}{a+b+c}\)+1
=>\(\frac{a+b+c+d}{b+c+d}\)=\(\frac{b+a+c+d}{a+c+d}\)=\(\frac{c+a+b+d}{a+b+d}\)=\(\frac{a+b+c+d}{a+b+c}\)
=>b+c+d=a+c+d=a+b+d=a+b+c
=>a=b=c=d
=>\(\frac{a+b}{c+d}\)=\(\frac{b+c}{a+d}\)=\(\frac{c+d}{a+b}\)=\(\frac{d+a}{b+c}\)=1
=>\(\frac{a+b}{c+d}\)=\(\frac{b+c}{a+d}\)=\(\frac{c+d}{a+b}\)=\(\frac{d+a}{b+c}\)=1+1+1+1=4
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có:
b+c+d/a=c+d+a/b=d+a+b/c=a+b+c/d=b+c+d+c+d+a+d+a+b+a+b+c/a+b+c=3(a+b+c+d)/a+b+c=k
Vì a+b+c+d khác 0 nên suy ra k=3
(Mong các bạn tham khảo )
\(\Rightarrow\frac{a}{b+c+d}+1=\frac{b}{a+c+d}+1=\frac{c}{a+b+d}+1=\frac{d}{a+b+c}+1\)
\(\Leftrightarrow\frac{a+b+c+d}{b+c+d}=\frac{a+b+c+d}{a+c+d}=\frac{a+b+c+d}{a+b+d}=\frac{a+b+c+d}{a+b+c}\)
=> b+c+d = a+c+d =a+b+d =a+b+c
=> a=b=c=d
Vậy T =1+1+1+1 =4
\(\frac{a}{b+c+d}=\frac{b}{c+d+a}=\frac{c}{a+b+d}=\frac{d}{a+b+c}=\frac{a+b+c+d}{b+c+d+c+d+a+a+b+d+a+b+c}=\frac{\left(a+b+c+d\right)}{3\left(a+b+c+d\right)}=3\rightarrow a=b=c=d\rightarrow\frac{a+c}{b+d}+\frac{a+b}{c+d}+\frac{a+c}{b+d}+\frac{b+c}{a+d}=1+1+1+1=4\)
Ta có k=b+c+d/a=c+d+a/b=d+a+b/c=a+b+c/d=(b+c+d+c+d+a+d+a+b+a+b+c/)(a+b+c+d)
(3(a+b+c+d))/(a+b+c+d)=3
=>k=3
k sẽ bằng 3