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Ta có: \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c};c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)
Theo TCDTSBN ta có:
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(1\right)\)
Lại có: \(\frac{a^3}{b^3}=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\left(2\right)\)
Từ (1) và (2) => đpcm
Đặt \(\frac{a}{b}=\frac{c}{d}=k\) ,ta có:
\(a=bk,c=dk\)
\(\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}=\frac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\frac{b^2}{d^2}\)(1)
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2) suy ra:
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)(đpcm)
1, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{3a}{3c}=\frac{b}{d}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{c}=\frac{3a+b}{3c+d}\Rightarrow\frac{a}{3a+b}=\frac{c}{3c+d}\)
2, a, Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{a}{c}\cdot\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ab}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ab}{cd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)
b, Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a-b}{c-d}\cdot\frac{a-b}{c-d}\Rightarrow\frac{ab}{cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c},c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(1\right)\)
\(\frac{a^3}{b^3}=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\left(2\right)\)
=> đpcm
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\left(1\right)\)
\(c^2=bd\Rightarrow\frac{b}{c}=\frac{c}{d}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{abc}{bcd}=\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(đpcm\right)\)
b, Tỉ số = nhau + tất vào là xông
Ta có : \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}\)(1)
Áp dụng tính chất của dãy tỉ số bằng nhau cho ( 1 ) ta có :
\(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{c}{d}=\frac{ac}{bd}\)
=> Điều phải chứng minh
a
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)
b
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)
c
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{5a^2}{5b^2}=\frac{3c^2}{3d^2}=\frac{5a^2+3c^2}{3d^2+5b^2}\)
Đặt\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=k\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}=k\Rightarrow\left(\frac{a-b}{c-d}\right)^{2013}=k^{2013}\)(1)
Mặt khác:\(\frac{a}{c}=\frac{b}{d}=k\Rightarrow\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=k^{2013}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}=k^{2013}\)(2)
Từ (1);(2) ta có: \(\left(\frac{a-b}{c-d}\right)^{2013}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\left(=k^{2013}\right)\)
có \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)=>\(\frac{a^{2013}}{c^{2013}}=\frac{\left(a-b\right)^{2013}}{\left(c-d\right)^{2013}}\)
ngược lại cũng có \(\frac{a^{2013}}{c^{2013}}=\frac{b^{2013}}{d^{2013}}=\frac{a^{2013}+b^{2013}}{c^{2013}+d^{2013}}\)
=> đpcm :V