\(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\)

Có:

• \(\frac{ab+ad}{b\left(b+d\right)}< \frac{ab+bc}{b\left(b+d\right)}\)

\(\Rightarrow\frac{a\left(b+d\right)}{b\left(b+d\right)}< \frac{b\left(a+c\right)}{b\left(b+d\right)}\)

\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}\left(1\right)\)

• \(\frac{ad+cd}{d\left(b+d\right)}< \frac{bc+cd}{d\left(b+d\right)}\)

\(\Rightarrow\frac{d\left(a+c\right)}{d\left(b+d\right)}< \frac{c\left(b+d\right)}{d\left(b+d\right)}\)

\(\Rightarrow\frac{a+c}{b+d}< \frac{c}{d}\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)

a/ \(\left(a^2-b^2\right)\left(c^2-d^2\right)=a^2c^2-a^2d^2-b^2c^2+b^2d^2\)

\(=\left(a^2c^2+2abcd+b^2d^2\right)-\left(a^2d^2+2abcd+b^2c^2\right)\)

\(=\left(ac+bd\right)^2-\left(ad+bc\right)^2\)

b/ \(x^2+y^2+z^2=xy+yz+zx\)

\(\Leftrightarrow2x^2+2y^2+2z^2=2xy+2yz+2zx\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\)

\(\Leftrightarrow x=y=z\)

#)Giải :

Ta có : \(a^4+b^4+c^4+d^4=4abcd\)

\(\Leftrightarrow a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4+2a^2b^2-4abcd+2c^2d^2=0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(c^2-d^2\right)+2\left(ab-cd\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}a^2=b^2\\c^2=d^2\\ab=cd\end{cases}}\)

Do a, b, c, d > 0

\(\Leftrightarrow a=b=c=d\left(đpcm\right)\)

bạn có thể áp dụng cái cuối

Ta có:

\(a^3+b^3+c^3=3abc\)

\(\Rightarrow\left(a^3+b^3\right)+c^3-3abc=0\)

\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3+3abc=0\)

\(\Rightarrow[\left(a+b\right)^3+c^3]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)[\left(a+b\right)^2-\left(a+b\right)c+c^2]-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc-3ab\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ac=0\left(2\right)\end{cases}}\)

Từ (1) => a = b = c (vì a ; b ; c là các số dương)

Giải (2) ta có:

\(2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Rightarrow2a^2+2b^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)

Vì \(\left(a-b\right)^2\ge\forall a,b\)

\(\left(a-c\right)^2\ge\forall a,c\)

\(\left(b-c\right)^2\ge\forall b,c\)

\(\Rightarrow\)Ta có: \(a-b=a-c=b-c\Rightarrow a=b=c\)

a) phải là a.d<b.c

 chứ ko phải a,d<b,c đâu