a. ĐK: a, b, c khác 0.

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

\(\Leftrightarrow\left[\frac{a^2+b^2-c^2}{2ab}-1\right]+\left[\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2-\left(a^2-b^2\right)}{b}+\frac{c^2+\left(a^2-b^2\right)}{a}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{1}{2c}\left[\frac{c^2\left(a+b\right)-\left(a^2-b^2\right)\left(a-b\right)}{ab}\right]=0\)

\(\Leftrightarrow\frac{\left(a-b\right)^2-c^2}{2ab}+\frac{\left(a+b\right)\left(c^2-\left(a-b\right)^2\right)}{2abc}=0\)

\(\Leftrightarrow\left[\left(a-b\right)^2-c^2\right]\left(1-\frac{a+b}{c}\right)=0\)

\(\Leftrightarrow\left(a-b-c\right)\left(a-b+c\right)\left(c-a-b\right)=0\)

\(\Leftrightarrow a=b+c\)hoặc \(b=a+c\)hoặc \(c=a+b\).

b) Không mất tính tổng quả. G/s: a = b + c

Khi đó ta có:

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

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

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

=> Điều phải chứng minh.

(a² + b²) / (c² + d²) = ab/cd 
<=> (a² + b²)cd = ab(c² + d²) 
<=> a²cd + b²cd = abc² + abd² 
<=> a²cd - abc² - abd² + b²cd = 0 
<=> ac(ad - bc) - bd(ad - bc) = 0 
<=> (ac - bd)(ad - bc) = 0 
<=> ac - bd = 0 hoặc ad - bc = 0 
<=> ac = bd hoặc ad = bc 
<=> a/b = d/c hoặc a/b = c/d (đpcm)

hello ib ko

C2: Đặt \(\frac{a}{b}.\frac{c}{d}=k=>a=bk,c=dk\)

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

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

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

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

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

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

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

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

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

a, \(\dfrac{a^2+2ab+b^2}{4}\ge ab\)

\(\Leftrightarrow\)a^2+2ab+b^2>=4ab

\(\Leftrightarrow\)a^2-2ab+b^2>=0

\(\Leftrightarrow\)(a-b)^2>=0 (luôn đúng)

b,\(a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2\ge0\) 

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) luôn đúng

Đặt \(\frac{b^2+c^2-a^2}{2bc}=A,\frac{c^2+a^2-b^2}{2ac}=B;\frac{a^2+b^2-c^2}{2ab}=C.\)

Theo giả thiết : \(A+B+C=1\)

Suy ra \(S=\left(A-1\right)+\left(B-1\right)+\left(C+1\right)=0\)

\(A-1=\frac{\left(b-c-a\right)\left(b-c+a\right)}{2bc};\)

\(B-1=\frac{\left(a-c-b\right)\left(a-c+b\right)}{2ac};\)

\(C+1=\frac{\left(a+b+c\right)\left(a+b-c\right)}{2ab}\)

\(S=\frac{a+b-c}{2abc}\left[c\left(a+b+c\right)+b\left(a-c-b\right)+a\left(b-c-a\right)\right]\)

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

Có 3 khả năng xảy ra :

TH1 : \(a+b-c=0\Rightarrow A-1=B-1=C+1=0\left(đpcm\right)\)

TH2 :

\(b+c-a=0\).Ta xét : \(A+1=B-1=C-1=0\left(đpcm\right)\)

TH3:

\(c+a-b=0\). Ta xét : \(S=\left(A-1\right)+\left(B+1\right)+\left(C-1\right)=0\)

và \(\Rightarrow A-1=B+1=C-1=0\left(đpcm\right)\)

Ta có :

\(a^2+b^2+1\ge ab+a+b\)

\(\Leftrightarrow a^2+b^2+1-ab-a-b\ge0\)

\(\Leftrightarrow2a^2+2b^2-2ab-2a-2b+2\ge0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) ( đúng)

giúp mình với

Đặt $\frac{a}{b}=\frac{c}{d}=k$

Khi đó ta có :

$a=bk$ và $c=dk$

Suy ra :

$\frac{{}^{}-{}^{}}{{}^{}-{}^{}}=\frac{{}^{}-{}^{}}{{}^{}-{}^{}}$

$=\frac{{}^{}{}^{}-{}^{}}{{}^{}{}^{}-{}^{}}$

$=\frac{{}^{}.({}^{}-1)}{{}^{}.({}^{}-1)}$

$=\frac{{}^{}}{{}^{}}\left(1\right)$

Ta lại có :

Đặt $\frac{a}{b}=\frac{c}{d}=k$

Khi đó ta có :

$a=bk$ và $c=dk$

Suy ra :

$\frac{{}^{}-{}^{}}{{}^{}-{}^{}}=\frac{{}^{}-{}^{}}{{}^{}-{}^{}}$

$=\frac{{}^{}{}^{}-{}^{}}{{}^{}{}^{}-{}^{}}$

$=\frac{{}^{}.({}^{}-1)}{{}^{}.({}^{}-1)}$

$=\frac{{}^{}}{{}^{}}\left(1\right)$

Ta lại có :

a) VT = ( a + b + a − b ) ( a + b − a + b ) 4 = 2 a . 2 b 4 = 4 = VP => đpcm.

b) VP = x 2   +   2 xy   +   y 2   +   x 2   –   2 xy   +   y 2   =   2 ( x 2   +   y 2 ) = VT => đpcm.