Áp dụng BĐT Bunhiacopxki :
\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
Dấu đẳng thức xảy ra \(\Leftrightarrow\frac{a}{x}=\frac{b}{y}\)
\(\Leftrightarrow ay=bx\)
\(\Leftrightarrow ay-bx=0\)
Ta có đpcm.
Chứng minh rằng nếu \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\) với x,y khác 0 thì \(\dfrac{a}{x}=\dfrac{b}{y}\)
a) \(\left(a+b\right)^2=\left(a-b\right)+4ab
\)
b) \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)
c) \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax-by\right)^2+\left(ay+bx\right)^2\)
Chứng minh rằng nếu:
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2;x,y\ne0\) thì \(\dfrac{a}{x}=\dfrac{b}{y}\)
Chứng minh các hằng đẳng thức sau:
a) \(\left(ax+yy+cz\right)^2+\left(bx-ay\right)^2+\left(cy-bz\right)^2+\left(az-cx\right)^2=\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)
b) \(\left(ab+bc+ac\right)^2+\left(a^2-bc\right)+\left(b^2-ca\right)^2+\left(c^2-ab\right)^2=\left(a^2+b^2+c^2\right)^2\)
Chứng minh:
a) \(x\ne0,y\ne0\) và \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)\) thì \(\dfrac{a}{x}=\dfrac{b}{y}\)
b) \(x\ne0,y\ne0,z\ne0\) và \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\) thì \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
1) Cho \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
CMR: \(a=b=c=1\)
2) CMR: nếu \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\) thì \(\dfrac{a}{x}=\dfrac{b}{y}\)
3) Cho \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
CMR: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)
Chứng minh rằng nếu:\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=\left(y+z-2x\right)^2+\left(z+x-2y\right)^2+\left(x+y-2z\right)^2\)thì x=y=z
Chứng minh rằng:
Nếu \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cx\right)^2\) thì \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\).
Chứng minh nếu \(x^2=b^2+c^2;y^2=c^2+a^2;z^2=a^2+b^2\)thì \(\left(x+y+z\right)\left(-x+y+z\right)\left(x-y+z\right)\left(x+y-z\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)
