Question

[Trần Văn Tú]

a,Cho a,b,c thỏa mãn a+b+c=0 và ab+bc+ca=0.Tính giá trị của M=(a-1)¹⁹⁹⁹+b²⁰⁰⁰+(c+1)²⁰⁰¹

b,Cho a²+b²+c²=1 và \(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)=\(\dfrac{1}{abc}\).Tính giá trị của biểu thức P=\(\dfrac{a+b}{b+c}\)+\(\dfrac{b+c}{c+a}\)+\(\dfrac{c+a}{a+b}\)

c,Cho a+b+c=3.Chứng minh rằng (a-1)³+(b-1)³+(c-1)³=3(a-1)(b-1)(c-1)

Answer 1

[Khôi Bùi]

a ) \(a+b+c=0\)

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

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

\(\Leftrightarrow a^2+b^2+c^2+2.0=0\)

\(\Leftrightarrow a^2+b^2+c^2=0\)

Do \(a^2\ge0;b^2\ge0;c^2\ge0\)

\(\Rightarrow a^2+b^2+c^2\ge0\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=0\) ( * )

Thay * vào biểu thức M , ta được :

\(M=\left(0-1\right)^{1999}+0^{2000}+\left(0+1\right)^{2001}\)

\(=-1^{1999}+0+1^{2001}\)

\(=-1+0+1\)

\(=0\)

Vậy \(M=0\)

Answer 2

[Khôi Bùi]

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{abc}\)

\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=\dfrac{1}{abc}\)

\(\Leftrightarrow\dfrac{bc+ac+ab-1}{abc}=0\)

\(\Leftrightarrow bc+ac+ab-1=0\)

\(\Leftrightarrow bc+ac+ab=1\)

Mà \(a^2+b^2+c^2=1\)

\(\Rightarrow bc+ac+ab=a^2+b^2+c^2\)

\(\Rightarrow2bc+2ac+2ab=2a^2+2b^2+2c^2\)

\(\Rightarrow2a^2+2b^2+2c^2-2bc-2ac-2ab=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)

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

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)

Mà \(P=\dfrac{a+b}{b+c}+\dfrac{b+c}{c+a}+\dfrac{c+a}{a+b}\)

\(\Rightarrow P=\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\)

\(\Rightarrow P=1+1+1=3\)

Vậy \(P=3\)