Ta có: \(\frac{\left(a^2+b^2\right)}{\left(c^2+d^2\right)}=\frac{ab}{cd}.\)

\(\Rightarrow\left(a^2+b^2\right).cd=ab.\left(c^2+d^2\right)\)

\(\Rightarrow a^2cd+b^2cd=abc^2+abd^2\)

\(\Rightarrow a^2cd+b^2cd-abc^2-abd^2=0\)

\(\Rightarrow\left(a^2cd-abc^2\right)-\left(abd^2+b^2cd\right)=0\)

\(\Leftrightarrow ac.\left(ad-bc\right)-bd.\left(ad-bc\right)=0\)

\(\Leftrightarrow\left(ad-bc\right).\left(ac-bd\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}ad-bc=0\\ac-bd=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}ad=bc\\ac=bd\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{matrix}\right.\left(đpcm\right).\)

Chúc bạn học tốt!

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

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

\(\Rightarrow a=b=c\)

Ta có a^2 + b^2 + (a - b)^2= c^2 + d^2 + (c - d)^2.
=> a^4+b^4+(a-b)^4+2[a^2b^2+a^2(a-b)^2+b^2(a-b)2]=

=c^4+d^4+(c-d)^4+2[c^2d^2+c^2(c-d)^2+d^2(c-d)^2

<=>a^4+b^4+(a-b)^4+2[a^2b^2+(a^2+b^2)(a-b)^2]

=c^4+d^4+(c-d)^4+2[c^2d^2+(c^2+d^2)(c-d)^2

Lại có a^2 + b^2 + (a - b)^2 = c^2 + d^2 + (c - d)^2.

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

=>a^2+b^2-ab =c^2+d^2-cd

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

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

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

Từ đó bạn sẽ có đpcm

2)

Xét hiệu:

\(A^2+B^2+C^2+D^2+4-2A-2B-2C-2D\)

\(=\left(A^2-2A+1\right)+\left(B^2-2B+1\right)+\left(C^2-2C+1\right)+\left(D^2-2D+1\right)\)

\(=\left(A-1\right)^2+\left(B-1\right)^2+\left(C-1\right)^2+\left(D-1\right)^2\ge0\)

=> BĐT luôn đúng

Vậy \(A^2+B^2+C^2+D^2+4\ge2\left(A+B+C+D\right)\)

1)

Áp dụng BĐT Cauchy cho 2 số không âm, ta có:

\(\dfrac{AB}{C}+\dfrac{BC}{A}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{BC}{A}}=2B\) (1)

\(\dfrac{BC}{A}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{BC}{A}.\dfrac{AC}{B}}=2C\) (2)

\(\dfrac{AB}{C}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{AC}{B}}=2A\) (3)

Từ (1)(2)(3) cộng vế theo vế:

\(2\left(\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\right)\ge2\left(A+B+C\right)\)

\(\Rightarrow\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\ge A+B+C\)

Câu 1 

Ta có : \(\frac{a}{b}=\frac{c}{d}=>\left(\frac{a}{b}+1\right)=\left(\frac{c}{d}+1\right)\left(=\right)\frac{a+b}{b}=\frac{c+d}{d}\)

=> ĐPCM

Câu 2

Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{b}{a}=\frac{d}{c}=>\left(\frac{b}{a}+1\right)=\left(\frac{d}{c}+1\right)\left(=\right)\frac{b+a}{a}=\frac{d+c}{c}=>\frac{a}{b+a}=\frac{c}{d+c}\)

=> ĐPCM

Câu 3

Câu 3

Ta có \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(=) (a+b).(c-d)=(a-b).(c+d)(=)ac-ad+bc-bd=ac+ad-bc-bd(=)-ad+bc=ad-bc(=) bc+bc=ad+ad(=)2bc=2ad(=)bc=ad=> \(\frac{a}{b}=\frac{c}{d}\)

=> ĐPCM

Câu 4 

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(=>\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Ta có \(\frac{ac}{bd}=\frac{bk.dk}{bd}=k^2\left(1\right)\)

Lại có \(\frac{a^2+c^2}{b^2+d^2}=\frac{b^2k^2+c^2k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)

Từ (1) và (2) => ĐPCM

Mày là thằng anh tuấn lớp 7c trường THCS yên lập đúng ko