\(a^2-14a-\left(b^2+c^2+2bc\right)=a^2-14a-\left(b+c\right)^2\)

\(=a^2-14a-\left(7-a\right)^2=a^2-14a-\left(a^2-14a+49\right)\)

\(=-49\)

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\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=a^2+b^2+c^2\Leftrightarrow ab+bc+ca=0\)

-Ta có hằng đẳng thức: \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(P=\dfrac{2bc}{a^2}+\dfrac{2ca}{b^2}+\dfrac{2ab}{c^2}+2bc+2ca+2ab\)

\(=\dfrac{2bc}{a^2}+\dfrac{2ca}{b^2}+\dfrac{2ab}{c^2}=\dfrac{2\left(b^3c^3+c^3a^3+a^3b^3\right)}{a^2b^2c^2}=\dfrac{2.\left(ab+bc+ca\right)\left(b^2c^2+c^2a^2+a^2b^2-ab^2c-abc^2-a^2bc\right)}{a^2b^2c^2}=\dfrac{2.0.\left(b^2c^2+c^2a^2+a^2b^2-ab^2c-abc^2-a^2bc\right)}{a^2b^2c^2}=0\)

Ta có: P = (a^2+b^2+c^2-ab-bc-ca)/(a^2-c^2-2ab+2bc)

=1/2.(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca)/(a^2 - 2ab + b^2 - b^2 +2bc  - c^2)

=1/2.[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]/[(a-b)^2-(b^2-2bc+c^2)]

=1/2.[(a-b)^2 + (b-c)^2 + (a-c)^2]/[(a-b)^2 - (b-c)^2

Lại có: a – b = 7; b – c = 3 ó a – b + b – c = 7 + 3 ó a – c = 10

Thay a - b = 7 ; b – c = 3; a - c  = 10 vào P, ta được:

P = 1/2 .(7^2 + 3^2 + 10^2)/(7^2 – 3^2)

= 1/2.(49 + 9 + 100)/(49 – 9)

= 1/2.158/40

= 158/80

= 79/40

# Chúc bạn học tốt!

\(a-b=7;b-c=3\text{ nên: }\left(a-b\right)+\left(b-c\right)=a-c=10\)

\(\text{tử P}=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]=\frac{1}{2}\left(3^2+7^2+10^2\right)=\frac{1}{2}.158=79\)

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

bạn ktra lại đề :)

\(P=a^2-b^2-c^2-2bc-20a\)

\(=a^2-\left(b+c\right)^2-20a\)

\(=a^2-\left(a-10\right)^2-20a\)(vì a + b + c = 10)

\(=a^2-a^2+20a-100-20a\)

\(=-100\)

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

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

\(=\frac{a+b}{c}-1+2=\frac{a+c}{b}-1+2=\frac{c+b}{a}-1+2\)

\(=\frac{a+b}{c}+1=\frac{a+c}{b}+1=\frac{c+b}{a}+1\)

\(=\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}\)

\(\Rightarrow a=b=c\)Thay vào \(P\)ta được :

\(P=\frac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a^3}=\frac{2a\cdot2a\cdot2a}{a^3}=\frac{8a^3}{a^3}=8\)