Ta có: \(b^2=ac\)

=>\(\frac{a}{b}=\frac{b}{c}\) (1)

\(c^2=bd\)

=>\(\frac{b}{c}=\frac{c}{d}\) (2)

Từ (1),(2) suy ra \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)

=>c=dk; \(b=ck=dk\cdot k=dk^2\) và \(a=bk=dk^2\cdot k=dk^3\)

\(\frac{3a^3-4b^3+5c^3}{3b^3-4c^3+5d^3}=\frac{3\cdot\left(dk^3\right)^3-4\cdot\left(dk^2\right)^3+5\cdot\left(dk\right)^3}{3\cdot\left(dk^2\right)^3-4\cdot\left(dk\right)^3+5\cdot d^3}\)

\(=k^3\cdot\frac{3\cdot d^3\cdot k^6-4\cdot d^3\cdot k^3+5d^3}{3d^3\cdot k^6-4\cdot d^3\cdot k^3+5\cdot d^3}=k^3\)

\(\frac{a}{d}=\frac{dk^3}{d}=k^3\)

Do đó: \(\frac{3a^3-4b^3+5c^3}{3b^3-4c^3+5d^3}=\frac{a}{d}\)

vì -1 hơn 1 hai số cho nên;

a) a/b và c/d ^2 =ab/cd hơn kém nhau 2

b) dựa theo tính chất kết hợp (a+b/c+d ) ^3 = a ^3 ...

(a+b+c+d)(a+d-b-c)=(a-b+c-d)(a+b-c-d)

=>(a+d)^2-(b+c)^2=(a-d)^2-(b-c)^2

=>(a+d)^2-(a-d)^2=(b+c)^2-(b-c)^2

=>(a+d-a+d)(a+d+a-d)=(b+c+b-c)(b+c-b+c)

=>4ad=4bc

=>ad=bc

=>a/c=b/d

a: Sửa đề: \(\frac{a}{b}=\frac{a+c}{b+d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a}{b}=\frac{bk}{b}=k\)

\(\frac{a+c}{b+d}=\frac{bk+dk}{b+d}=\frac{k\left(b+d\right)}{b+d}=k\)

Do đó: \(\frac{a}{b}=\frac{a+c}{b+d}\)

b: \(\frac{a-b}{a}=\frac{bk-b}{bk}=\frac{b\left(k-1\right)}{bk}=\frac{k-1}{k}\)

\(\frac{c-d}{c}=\frac{dk-d}{dk}=\frac{d\left(k-1\right)}{dk}=\frac{k-1}{k}\)

Do đó: \(\frac{a-b}{a}=\frac{c-d}{c}\)

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)

P/s : 

Đề thiếu rồi bạn ơi : 

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