a, Có: \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{c}=\frac{b}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{a}{c}=\frac{b}{d}=\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\)

Có: \(\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\Leftrightarrow\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

b, Co: \(\frac{a}{c}=\frac{b}{d}\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\Rightarrow\frac{ab}{cd}\)

Lại có:\(\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\left(1\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\left(2\right)\)

Tu (1)&(2),có: \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

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

Xét \(VT=\frac{\left(bk\right)^2-b^2}{\left(dk\right)^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(1\right)\)

Xét \(VP=\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2k}{d^2k}=\frac{b^2}{d^2}\left(2\right)\)

Từ (1) và (2) =>Đpcm

LINK dưới đây bạn nhé

http://olm.vn/hoi-dap/question/143125.html

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

                                       =>\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)(3)

                                      =>\(\frac{a+b}{c+d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(4)

=>Từ (1),(2),(3),(4)=>\(\frac{a}{b}=\frac{a^2-b^2}{c^2-d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)(đpcm)

chứng minh này chị ngu lắm em

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)

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

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

\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(2)

Từ (1) và (2), ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

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

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

\(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)(2)

Từ (1) và (2), ta có: \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{ab}{cd}\)

a) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)

mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)

\(\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

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

mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)

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

1/h=1/2(1/a+1/b)=1/2a+1/2b=(a+b)/2ab

=>(a+b/)2ab-1/h=0

quy dong len ta co

(a+b)h/2abh-2ab/2abh=0=> (ah+bh-2ab)/2abh=0 =>ah+bh-2ab=0

                                                                       =>ah+bh-ab-ab=0

                                                                         =>a(h-b)-b(a-h)=0  

                                                                           =>a(h-b)=b(a-h)

                                                                              =>a/b=(a-h)(h-b)

Đặt x = a - b ; y = b - c ; z = c - a thì x + y + z = a - b + b - c + c - a = 0

Ta có : \(\sqrt{\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{y^2}}\)

\(=(\frac{1}{x}+\frac{1}{y}+\frac{1}{y})^2-2(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx})\)

\(=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2-2\frac{x+y+z}{xyz}\)

\(=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2=(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a})^2(đpcm)\)

Chúc bạn học tốt

Ta có: \(F=\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\)

\(\Leftrightarrow F=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{da+db}\)  

\(\Leftrightarrow F\ge\frac{\left(a+b+c+d\right)^2}{2ac+2bd+\left(a+c\right)\left(b+d\right)}=P\)

\(\Leftrightarrow P=\frac{a^2+b^2+c^2+2ab+2bc+2cd+2ad+2ac+2bd}{ab+ac+bc+bd+cd+ac+ad+bd}\)

\(\Leftrightarrow P=\frac{\left(a^2+c^2\right)+\left(b^2+d^2\right)+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+bc+cd+ad}\)

(Vì \(a^2+c^2\ge2ac\Leftrightarrow\left(a-c\right)^2\ge0\)luôn đúng; \(b^2+d^2\ge2bd\Leftrightarrow\left(b-d\right)^2\ge0\)luôn đúng)

\(\Leftrightarrow P\ge\frac{2ac+2bd+2ab+2bc+2cd+2ad+2ac+2bd}{2ac+2bd+ab+cd+ad+ac+bd}\)

\(\Leftrightarrow P\ge\frac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\)

\(\Leftrightarrow F\ge P\ge2\)

\(\LeftrightarrowĐPCM\)