a) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

\(\Leftrightarrow a^2x^2+b^2x^2+a^2y^2+b^2y^2=a^2x^2+b^2y^2+2abxy\)

\(\Leftrightarrow b^2x^2-2abxy+a^2y^2=0\)

\(\Leftrightarrow\left(bx\right)^2-2\cdot bx\cdot ay+\left(ay\right)^2=0\)

\(\Leftrightarrow\left(bx-ay\right)^2=0\Rightarrow bx=ay\Rightarrow\left(\frac{a}{x}=\frac{b}{y}\right)\)

b) \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)

\(\Leftrightarrow a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2\)

\(=a^2x^2+b^2y^2+c^2z^2+2abxy+2bcyz+2acxz\)

\(\Leftrightarrow b^2x^2-2bxay+a^2y^2+b^2z^2-2bzcy+c^2y^2+a^2z^2-2azcx+c^2x^2=0\)

\(\Leftrightarrow\left(bx-ay\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)

\(\hept{\begin{cases}bx=ay\\bz=cy\\az=cx\end{cases}\Rightarrow\hept{\begin{cases}\frac{a}{x}=\frac{b}{y}\\\frac{b}{y}=\frac{c}{z}\\\frac{a}{x}=\frac{c}{z}\end{cases}}\Rightarrow\left(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\right)}\)

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

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

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

\(\Leftrightarrow\left(a-b\right)^2=0\Leftrightarrow a=b\)

a,  Tương đương   :   \(a^2x^2+a^2y^2+b^2x^2+b^2y^2\)   =   \(a^2x^2+2axby+b^2y^2\)  

                                 \(a^2y^2-2axby+b^2x^2=0\) 

                                 \(\left(ay-bx\right)^2\)  = 0

                                 \(ay-bx=0\)

                                 \(ay=bx\)

                                \(\frac{a}{x}=\frac{b}{y}\)   dpcm

Câu b, c làm tương tự câu a

2. ta co bieu thuc x - ( f-1)

3.

a) sửa đề: \(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)

=\(\frac{-x^2\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-y^2\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(-\frac{\left\lbrace x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

xét tử số:

Tử=\(x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)

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

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

=\(\left(y-z\right)\left\lbrace x^2-x\left(y+z\right)+yz\right\rbrace\)

=\(\left(y-z\right)\left\lbrace x\left(x-y\right)-z\left(x-y\right)\right\rbrace\)

=\(\left(y-z\right)\left(x-y\right)\left(x-z\right)\)

=\(-\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

thay lại vào biểu thức cũ:

\(\Rightarrow-\frac{\left\lbrace-\left(x-y\right)\left(y-z\right)\left(z-x\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(1\)

b) \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)

=\(\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

2) ta có: \(VT=\left(a^2+b^2\right)\left(x^2+y^2\right)\) và \(VP=\left(ax+by\right)^2\)

tính hiệu của cả VT và VP

suy ra: \(\left(ay+bx\right)^2=0\Rightarrow ay=bx\)

vì \(x,y\ne0\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}\left(đpcm\right)\)

3)

biến đổi đẳng thức (1) thành (ay+bx)² + (bz-cy)² +(az-cx)² =0

\(\Rightarrow\) Đpcm