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\(VT=\frac{c-b}{\left(a-b\right)\left(c-a\right)}+\frac{a-c}{\left(a-b\right)\left(b-c\right)}+\frac{b-a}{\left(b-c\right)\left(c-a\right)}\)
\(=\frac{-\left(b-c\right)^2-\left(c-a\right)^2-\left(a-b\right)^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{-2a^2-2b^2-2c^2+2ab+2ac+2bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{2ab-2ac+2bc-2b^2+2ab+2ac-2bc-2a^2-2ab+2ac+2bc-2c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{2\left(a-b\right)\left(b-c\right)+2\left(a-b\right)\left(c-a\right)+2\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{2}{c-a}+\frac{2}{b-c}+\frac{2}{a-b}\)
Ta có \(\dfrac{2}{a-b}\)+\(\dfrac{2}{b-c}\)+\(\dfrac{2}{c-a}\)
= (\(\dfrac{1}{a-b}\)+\(\dfrac{1}{c-a}\))+(\(\dfrac{1}{b-c}\)+\(\dfrac{1}{a-b}\))+(\(\dfrac{1}{c-a}\)+\(\dfrac{1}{b-c}\))
=(\(\dfrac{1}{a-b}\)- \(\dfrac{1}{a-c}\))+(\(\dfrac{1}{b-c}\)- \(\dfrac{1}{b-a}\))+(\(\dfrac{1}{c-a}\) - \(\dfrac{1}{c-b}\))
=\(\dfrac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right).\left(a-c\right)}\)+\(\dfrac{\left(b-a\right)-\left(b-c\right)}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{\left(c-b\right)-\left(c-a\right)}{\left(c-b\right).\left(c-a\right)}\)
= \(\dfrac{a-c-a+b}{\left(a-b\right).\left(a-c\right)}\)+\(\dfrac{b-a-b+c}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{c-b-c+a}{\left(c-b\right).\left(c-a\right)}\)
= \(\dfrac{-c+b}{\left(a-b\right).\left(a-c\right)}\)+ \(\dfrac{-a+c}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{-b+a}{\left(c-b\right).\left(c-a\right)}\)
= \(\dfrac{b-c}{\left(a-b\right).\left(a-c\right)}\)+\(\dfrac{c-a}{\left(b-a\right).\left(b-c\right)}\)+\(\dfrac{a-b}{\left(c-b\right).\left(c-a\right)}\)
Chúc bạn học tốt.
a,
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=2\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
b,
\(a+b+c=2p\Leftrightarrow p=\dfrac{a+b+c}{2}\)
\(\Leftrightarrow\left(p-a\right)^2+\left(p-b\right)^2+\left(p-c\right)^2=3p^2-2pa-2pb-2pc+a^2+b^2+c^2\)
\(=3\left(\dfrac{a+b+c}{2}\right)^2-2\cdot\dfrac{a+b+c}{2}\cdot a-2\cdot\dfrac{a+b+c}{2}\cdot b-2\cdot\dfrac{a+b+c}{2}\cdot c+a^2+b^2+c^2\)
\(=3p^2-\left(a+b+c\right)^2+a^2+b^2+c^2=3p^2-4p^2+a^2+b^2+c^2=a^2+b^2+c^2-p^2\)
a, (a + b + c)^2 + (a - b - c)^2 +( b - c - a) ^2 + (c - a - b)^2
= (a + b + c)^2 + (a + b - c)^2 + (a - b - c)^2 + (a - b + c)^2
= (a + b)^2 + 2c(a + b) + c^2 + (a + b)^2 - 2c(a + b) + c^2 +
(a - b)^2 - 2c(a - b) + c^2 + (a - b)^2 + 2c(a - b) +c^2
= 2(a + b)^2 + 2c^2 + 2(a - b)^2 + 2c^2
= 2[(a + b)^2 + (a - b)^2] + 4c^2
=2(2a^2 + 2b^2) + 4c^2
= 4(a^2 + b^2 + c^2)
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Leftrightarrow\left(\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\right).\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a}{\left(a-b\right)\left(b-c\right)}+\frac{a}{\left(c-a\right)\left(b-c\right)}+\frac{b}{\left(c-a\right)\left(a-b\right)}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a\left(c-a\right)+a.\left(a-b\right)+b.\left(a-b\right)+b.\left(b-c\right)+c.\left(b-c\right)+c.\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{ac-a^2+ab-ac+ba-b^2+b^2-bc+bc-c^2+c^2-ac}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+0=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
đpcm
Ta biến đổi VT:
\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
Đặt b + c = x khi đó:
\(\left(a+x\right)^2+a^2+b^2+c^2\)
\(=a^2+2ax+x^2+a^2+b^2+c^2\)
Thay b + c vào x ta được:
\(=a^2+2a\left(b+c\right)+\left(b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+2ab+2ac+\left(b^2+c^2+2bc\right)+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(c^2+2ca+a^2\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=VP\)
\(\Rightarrowđpcm\)