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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:\(\left(\dfrac{a+b}{c+d}\right)^3=\left(\dfrac{bk+b}{dk+d}\right)^3=\left(\dfrac{b.\left(k+1\right)}{d.\left(k+1\right)}\right)^3=\dfrac{b^3}{d^3}\)(1)
Lại có :\(\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{b^3k^3+b^3}{d^3k^3+d^3}=\dfrac{b^3.\left(k^3+1\right)}{d^3.\left(k^3+1\right)}=\dfrac{b^3}{d^3}\)(2)
Từ (1) và (2) => ĐPCM
Từ a/b=c/d
=>a/c=b/d=a+b/c+d
<=>a^3/c^3=b^3/d^3=(a+b)^3(c+d)^3
=a^3+b^3/c^3+d^3
Vậy
(a+b)^3(c+d)^3=a^3+b^3/c^3+d^3 (đpcm)
Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng:
\(\frac{a^3}{c^3}=\frac{\left(2a-b\right)^3+b^3}{\left(2c-d\right)^3+d^3}\)
BÀI LÀM:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có: \(\frac{\left(2a-b\right)^3+b^3}{\left(2c-d\right)^3+d^3}=\frac{\left(2bk-b\right)^3+b^3}{\left(2dk-d\right)^3+d^3}=\frac{b^3.\left(2k-1\right)^3+b^3}{d^3.\left(2k-1\right)^3+d^3}=\frac{b^3.\left[\left(2k-1\right)^3+1\right]}{d^3.\left[\left(2k-1\right)^3+1\right]}=\frac{b^3}{d^3}\left(1\right)\)
Vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{c^3}{d^3}\left(2\right)\)
Từ (1) và (2) => \(\frac{a^3}{c^3}=\frac{\left(2a-b\right)^3+b^3}{\left(2c-d\right)^3+d^3}\left(đpcm\right)\)
Cách 1:
Ta xét tích a(c-d) và c(a-b)
Ta có: a(c-d)=ac-ad (1)
c(a-b)=ac-bc(2)
Ta lại có \(\dfrac{a}{c}=\dfrac{c}{d}\)=>ad=bc (3)
Từ (1), (2), (3) ta có a(c-d)=c(a-d). Do đó \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Cách 2:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)=k thì a=bk, c=dk.
Xét \(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\left(1\right)\)
Xét \(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\left(2\right)\)
Từ (1) và (2)=> \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Cách 3: Ta có
\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}\)
Aps dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a-b}{c-d}\)
=>\(\dfrac{a}{c}=\dfrac{a-b}{c-d}=>\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
hay \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
\(\left\{{}\begin{matrix}b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\\c^2=bd\Rightarrow\dfrac{b}{c}=\dfrac{c}{d}\end{matrix}\right.\)\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}\left(1\right)\)
Và \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a^3}{b^3}=\dfrac{b^3}{c^3}=\dfrac{c^3}{d^3}=\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\left(\dfrac{a+b+c}{b+c+d}\right)^3\left(đpcm\right)\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{c}{d}\right)^3=\frac{a^3}{b^3}=\frac{c^3}{d^3}=\frac{a^3+c^3}{b^3+d^3}\left(1\right)\)
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\Rightarrow\frac{a^3}{b^3}=\frac{c^3}{d^3}=\frac{\left(a+c\right)^3}{\left(b+d\right)^3}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a^3+c^3}{b^3+d^3}=\frac{\left(a+c\right)^3}{\left(b+d\right)^3}\left(dpcm\right)\)