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a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=kb\\c=kd\end{cases}\)
=> \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\) (1)
\(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{k.b^2}{k.d^2}=\frac{b^2}{d^2}\) (1)
Từ (1) và (2) => \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
b) Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)
Ta có: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)
Mà: \(k^3=\frac{a}{d}\) => \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)
a)Ta có:\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\left(đpcm\right)\)
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
a)Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)\(\Rightarrow\left(\frac{a-b}{c-d}\right)^2=\frac{ab}{cd}\Leftrightarrow\left(\frac{bk-b}{dk-d}\right)^2=\frac{bkb}{dkd}\)
Xét VT \(\left(\frac{bk-b}{dk-d}\right)^2=\left(\frac{b\left(k-1\right)}{d\left(k-1\right)}\right)^2=\left(\frac{b}{d}\right)^2=\frac{b^2}{d^2}\left(1\right)\)
Xét VP \(\frac{bkb}{dkd}=\frac{b^2}{d^2}\left(2\right)\)
Từ (1) và (2) =>Đpcm
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
Ta có:
\(a=bk\)
\(c=dk\)
a) Ta có:
\(\left(\frac{a-b}{c-d}\right)^2=\left(\frac{bk-b}{dk-d}\right)^2=\left[\frac{b\left(k-1\right)}{d\left(k-1\right)}\right]^2=\left(\frac{b}{d}\right)^2\) (1)
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2.k}{d^2.k}=\frac{b^2}{d^2}=\left(\frac{b}{d}\right)^2\) (2)
Từ (1) và (2) suy ra \(\left(\frac{a-b}{c-d}\right)^2=\frac{ab}{cd}\left(đpcm\right)\)
b) Ta có:
\(\left(\frac{a+b}{c+d}\right)^3=\left(\frac{bk+b}{dk+d}\right)^3=\left[\frac{b\left(k+1\right)}{d\left(k+1\right)}\right]^3=\left(\frac{d}{b}\right)^3\) (1)
\(\frac{a^3-b^3}{c^3-d^3}=\frac{\left(bk\right)^3-b^3}{\left(dk\right)^3-d^3}=\frac{b^3.k^3-b^3}{d^3.k^3-d^3}=\frac{b^3.\left(k^3-1\right)}{d^3.\left(k^3-1\right)}=\frac{b^3}{d^3}=\left(\frac{b}{d}\right)^3\) (2)
Từ (1) và (2) suy ra\(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\) (đpcm)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\Rightarrow\frac{a^3}{c^3}=\frac{b^3}{d^3}\)
áp dụng t.c dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\Rightarrow\left(\frac{a^2}{c^2}\right)^3=\frac{\left(a^2+b^2\right)^3}{\left(a^2+d^2\right)^3}=\frac{a^6}{c^6}\left(1\right)\)
\(\frac{a^3}{c^3}=\frac{b^3}{d^3}=\frac{a^3+b^3}{c^3+d^3}\Rightarrow\left(\frac{a^3}{c^3}\right)^2=\frac{\left(a^3+b^3\right)^2}{\left(a^3+d^3\right)^2}=\frac{a^6}{c^6}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{\left(a^2+b^2\right)^3}{\left(c^2+d^2\right)^3}=\frac{\left(a^3+b^3\right)^2}{\left(c^3+d^3\right)^2}\left(đpcm\right)\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\left(k\in Z\right)\)
\(\Rightarrow a=bk,c=dk\)
Có :
\(\left(\frac{a+b}{c+d}\right)^3=\frac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\frac{\left[b\left(k+1\right)\right]^3}{\left[d\left(k+1\right)\right]^3}=\frac{b^3\left(k+1\right)^3}{d^3\left(k+1\right)^3}=\frac{b^3}{d^3}\)
\(\frac{a^3+b^3}{c^3+d^3}=\frac{\left(bk\right)^3+b^3}{\left(dk\right)^3+d^3}=\frac{b^3k^3+b^3}{d^3k^3+d^3}=\frac{b^3\left(k^3+1\right)}{d^3\left(k^3+1\right)}=\frac{b^3}{d^3}\)
\(\Rightarrow\left(\frac{a+b}{c+d}\right)^3=\frac{a^3+b^3}{c^3+d^3}\left(=\frac{b^3}{d^3}\right)\)
Vậy ...
a/b = c/d =K ( K thuộc N* )
a = bK
c = dK
thay vào 2 cái cần so sanh đó là ok
k cho mik nha