Ta có: \(3\left(a+b+c\right)^2\ge0\forall a,b,c\)

\(4\left|ab+bc+ca\right|\ge0\forall a,b,c\)

Do đó: \(3\left(a+b+c\right)^2+4\left|ab+bc+ca\right|\ge0\forall a,b,c\)

Dấu '=' xảy ra khi a=b=c=0

Ta có: \(a^{2020}-b^{2021}+c^{2022}\)

\(=0^{2020}-0^{2021}+0^{2022}\)

=0

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)

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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{a}=\dfrac{1}{b}\end{matrix}\right.\) \(\Rightarrow a=b=c\)

\(\Rightarrow M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)

Bài này xuất hiện trong câu cuối đề GKI năm ngoái của mình :v

-Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\left\{{}\begin{matrix}\dfrac{a}{2020}=\dfrac{c}{2022}=\dfrac{a-c}{2020-2022}=\dfrac{a-c}{-2}\\\dfrac{a}{2020}=\dfrac{b}{2021}=\dfrac{a-b}{2020-2021}=\dfrac{a-b}{-1}\\\dfrac{c}{2022}=\dfrac{b}{2021}=\dfrac{c-b}{2022-2021}=c-b\end{matrix}\right.\)

\(\Rightarrow c-b=-\left(a-b\right)=\dfrac{a-c}{-2}\)

\(\Rightarrow\left\{{}\begin{matrix}a-c=-2\left(c-b\right)\\a-b=-\left(c-b\right)\end{matrix}\right.\)

\(\left(a-c\right)^3+8\left(a-b\right)^2.\left(c-b\right)=\left[-2\left(c-b\right)\right]^3+8\left[-\left(c-b\right)\right]^2.\left(c-b\right)=-8\left(c-b\right)^3+8\left(c-b\right)^3=0\left(đpcm\right)\)

Ta có : \(a^2+b^2+c^2=ab+bc+ca\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)

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

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

\(T=\frac{a^{2021}+b^{2021}+c^{2021}}{\left(a+b+c\right)^{2021}}=\frac{b^{2021}+b^{2021}+b^{2021}}{\left(b+b+b\right)^{2021}}=\frac{3b^{2021}}{\left(3b\right)^{2021}}=\frac{3}{3^{2021}}=\frac{1}{3^{2020}}\)

Nhầm là, tính A=(a-1)²⁰¹⁹+(b²-1)²⁰²⁰+(c³-1)²⁰²¹

Ta có : \(a+b+c=3\Rightarrow\left(a+b+c\right)^2=9\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)=9-2\times6=3\)

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

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

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

\(\Rightarrow a=b=c\)

Mà \(a+b+c=3\Rightarrow a=b=c=1\)

\(\Rightarrow A=\left(1-1\right)^{2019}+\left(1^2-1\right)^{2020}+\left(1^3-1\right)^{2021}\)

\(=0^{2019}+0^{2020}+0^{2021}=0\)

Câu hỏi của Đậu Đình Kiên - Toán lớp 7 - Học toán với OnlineMath

a^2+b^2+c^2=ab+bc+ac

=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0

=>a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0

=>(a-b)^2+(b-c)^2+(a-c)^2=0

=>a=b=c

\(T=\dfrac{a^{2022}+a^{2022}+a^{2022}}{\left(3a\right)^{2022}}=\dfrac{3}{3^{2022}}=\dfrac{1}{3^{2021}}\)

Ta có:\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

\(\iff\)\(\frac{abc}{ac+bc}=\frac{abc}{ab+ac}=\frac{abc}{bc+ba}\)

\(\iff\) \(ac+bc=ab+ac=bc+ba\)

+)\(ac+bc=ab+ac\) 

\(\implies\)\(bc=ab\)

\(\implies\) \(c=a\left(1\right)\)

+)\(ab+ac=bc+ba\)

\(\implies\) \(ac=bc\)

\(\implies\) \(a=b\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\implies\) \(a=b=c\)

\(\implies\) \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{aa+bb+cc}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)

Vậy \(M=1\)