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Áp dụng tính chất.......
a/b=b/c=c/d=a+b+c/b+c+d suy ra (a/b)^3=(b/c)^3=(c/d)^3=(a+b+c)^3/(b+c+d)^3(1)
a/b= b/c=c/dsuy ra a^3/b^3=b^3/c^3=c^3/d^3(2)
Áp dụng tính chất .....
a^3/b^3=b^3/c^3=c^3/d^3=a^3+b^3+c^3/b^3+c^3+d^3 (3)
Từ 1,2 và 3 suy ra :a^3+b^3+c^3/b^3+c^3+d^3=(a+b+c)^3/(b+c+d)^3
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b-c}{b+c-d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b-c}{b+c-d}\right)^3\)
Mà \(\left(\frac{a}{b}\right)^3=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\)
=>\(\left(\frac{a+b-c}{b+c-d}\right)^3=\frac{a}{d}\Rightarrow\frac{\left(a+b-c\right)^3}{\left(b+c-d\right)^3}=\frac{a}{d}\Rightarrow\frac{\left(a+b-c\right)^3}{a}=\frac{\left(b+c-d\right)^3}{d}\) (đpcm)
Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
\(\Rightarrow k=\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\Rightarrow k^3=\left(\frac{a+b+c}{b+c+d}\right)^3\left(1\right)\)
Mà \(k^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\left(2\right)\)
Từ (1),(2)\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\left(đpcm\right)\)
Ta có:\(\frac{a}{b}\)=\(\frac{b}{c}\)=\(\frac{c}{d}\)
\(\Rightarrow\)\(\frac{a}{b}\)3=\(\frac{b}{c}\)3=\(\frac{c}{d}\)3=\(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)=\(\left(\frac{a+b+c}{b+c+d^{ }}\right)\)3
\(\Rightarrow\)\(\left(\frac{a+b+c}{b+c+d^{ }}\right)\)3=\(\frac{a}{b}\).\(\frac{b}{c}\).\(\frac{c}{d}\)=\(\frac{a}{d}\)
\(\Rightarrow\)đpcm
Có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)(Tính chất dãy tỉ số bằng nhau)
=> \(\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{c^2}{d^2}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}=\frac{abc}{bcd}=\frac{a}{d}\)
=> \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
=> Đpcm
Ta có: \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\frac{abc}{bcd}=\frac{a}{d}\)( theo TC dãy TSBN)
Mà:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\)(theo TC dãy TSBN)
=>\(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\left(=\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3\right)\)(đpcm)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)
Do đó :
\(\frac{a}{b}=\frac{a+b+c}{b+c+d}\)\(\Rightarrow\)\(\left(\frac{a}{b}\right)^3=\left(\frac{a+b+c}{b+c+d}\right)^3\) \(\left(1\right)\)
Lại có :
\(\left(\frac{a}{b}\right)^3=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{abc}{bcd}=\frac{a}{d}\) \(\left(2\right)\)
Từ (1) và (2) suy ra \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\) \((\) cùng bằng \(\left(\frac{a}{b}\right)^3\) \()\)
Vậy \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\) ( điều phải chứng minh )
Chúc bạn học tốt ~
\(\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{a+b+c}{b+c+d}. \frac{a}{b} . \frac{b}{c} . \frac{c}{d} = \frac{a+b+c}{b+c+d} + \frac{a+b+c}{b+c+d} + \frac{a+b+c}{b+c+d}. \Leftrightarrow \frac{a}{d} = (\frac{a+b+c}{b+c+d} )^3 \RightarrowĐPCM\)
cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\) CMR : \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)
theo dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)
\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}=\frac{a}{d}\)
theo t/c của dãy t/s ta có:
a/b=b/c=c/d=a+b+c/b+c+d=(a+b+c/b+c+d)^3=a/b
=>Đpcm