Theo đề bài ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\) ( 1 )

Theo tính chất dãy tỉ số bằng nhau ta có :

\(k=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

\(k^2=\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)  ( 2 )

Mà từ ( 1 ) = > \(k^2=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\) ( 3 )

Từ ( 2 ) , ( 3 ) 

 = > \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) ( đpcm )

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk,c=dk\)

Ta có: \(\dfrac{ab}{cd}=\dfrac{bkb}{dkd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}=\dfrac{b}{d}\left(1\right)\)

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{bk-b}{dk-d}=\dfrac{b\left(k-1\right)}{d\left(k-1\right)}=\dfrac{b}{d}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\Rightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\\ \dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)

Cách giải:

1+1=3

6-6=0

9-9=0

Vậy => 6-6=9-9

(3-3)+(3-3) = 3x3 - 3x3

(1+1)=3

1+1=3

giúp mình với

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (*)

a) Từ (*)suy ra:

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)\(=\dfrac{b^2}{d^2}\) (1)

\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\dfrac{b^2}{d^2}\)(2)

Từ (1) và (2) suy ra: \(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) (đpcm)

b) Tương tự câu a nhé bạn!

Câu b giải chi tiết như sau nhé:

b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Từ đó, ta suy ra:

\(\dfrac{\left(a-b\right)^4}{\left(c-d\right)^4}=\dfrac{\left(bk-b\right)^4}{\left(dk-d\right)^4}=\dfrac{\left[b\left(k-1\right)\right]^4}{\left[d\left(k-1\right)\right]^4}=\dfrac{b^4.\left(k-1\right)^4}{d^4.\left(k-1\right)^4}=\dfrac{b^4}{d^4}\)(1)

\(\dfrac{a^4+b^4}{c^4+d^4}=\dfrac{\left(bk\right)^4+b^4}{\left(dk\right)^4+d^4}=\dfrac{b^4.k^4+b^4}{d^4.k^4+d^4}=\dfrac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\dfrac{b^4}{d^4}\)

(2)

Từ (1) và (2) suy ra: \(\dfrac{\left(a-b\right)^4}{\left(c-d\right)^4}=\dfrac{a^4+b^4}{c^4+d^4}\)

Ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{3a^2}{3b^2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a^2+c^2}{3b^2+d^2}\left(1\right)\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\)

Lại có : \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\left(2\right)\)

Từ (1); (2) và vì b + d khác 0 \(\RightarrowĐpcm\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

Ta có VT:

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}\)

\(=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\) (1)

VT: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\) (2)

Từ (1) và (2) 

\(\Rightarrow\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\left(đpcm\right)\)

Có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow ab=cd\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)\(\Leftrightarrow\left(\dfrac{a}{c}\right)^2=\left(\dfrac{b}{d}\right)^2=\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)

Vậy...

Bài 1:

Áp dụng t.c của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ =\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a^3}{b^3}=\dfrac{a.b.c}{b.c.d}=\dfrac{a}{d}\left(dpcm\right)\)

Bài 1:

Ta có: \(\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}=\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\)

Áp dụng bđt Cauchy Schwarz có:

\(\dfrac{a^2}{a\sqrt{a^2+8bc}}+\dfrac{b^2}{b\sqrt{b^2+8ac}}+\dfrac{c^2}{c\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+8bc}+b\sqrt{b^2+8bc}+c\sqrt{c^2+8bc}}\)

Lại sử dụng bđt Cauchy schwarz ta có:

\(a\sqrt{a^2+8bc}+b\sqrt{b^2+8ac}+c\sqrt{c^2+8ab}=\sqrt{a}\cdot\sqrt{a^3+8abc}+\sqrt{b}\cdot\sqrt{b^3+8abc}+\sqrt{c}\cdot\sqrt{c^3+8abc}\ge\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}\)

\(\Rightarrow\dfrac{a}{\sqrt{a^2+8bc}}+\dfrac{b}{\sqrt{b^2+8ac}}+\dfrac{c}{\sqrt{c^2+8ab}}\ge\dfrac{\left(a+b+c\right)^2}{\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+24abc\right)}}=\sqrt{\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}}\)

=> Ta cần chứng minh: \(\left(a+b+c\right)^3\ge a^3+b^3+c^3+24abc\)

hay \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)

Áp dụng bđt Cosi ta có:

\(a+b\ge2\sqrt{ab};b+c\ge2\sqrt{bc};c+a\ge2\sqrt{ca}\)

Nhân các vế của 3 bđt trên ta đc:

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge2\sqrt{ab}\cdot2\sqrt{bc}\cdot2\sqrt{ca}=8\sqrt{a^2b^2c^2}=8abc\)

=> Đpcm

a) \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

\(\Rightarrow ad=cb\)

=> \(ad+bd=bc+bd\)

\(\Rightarrow d\left(a+b\right)=b\left(c+d\right)\)

\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\dfrac{a+b}{b}=\dfrac{bk+b}{b}=\dfrac{b\left(k+1\right)}{b}=k+1\)

\(\Rightarrow\dfrac{c+d}{d}=\dfrac{dk+d}{d}=\dfrac{d\left(k+1\right)}{d}=k+1\)

\(\Rightarrow\dfrac{a+b}{b}=\dfrac{c+d}{d}\rightarrowđpcm\)

\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\dfrac{b^2}{d^2}\)

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\)

\(\Rightarrow\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\rightarrowđpcm\)