biến đổi vế trái :  a. \(\left(a+b\right)^2=a^2+2ab+B^2=VP\)

                          b. \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3=VP\)

                          c. \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=VP\)

                          xem 7 hằng đẳng thức đáng nhớ

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

\(=a^2+2ab+b^2\)

b)\(\left(a-b\right)^3=\left(a-b\right)\left(a-b\right)\left(a-b\right)=\left(a^2-ab-ab+b^2\right)\left(a-b\right)\)

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

\(=a^3-a^2b-2a^2b+2ab^2+ab^2-b^3\)

\(=a^3-3a^2b-3ab^2-b^3\)

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

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

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

biến đổi VT : a. \(\left(a+b\right)^2=a^2+2ab+b^2=VP\)

                    b. \(\left(a-b\right)^3=a^3-3a^2+3ab^2-b^3=VP\)

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

                    xem 7 hằng đẳng thức đáng nhớ

Bài làm:

a) Ta có: \(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2\ge0\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)

luôn đúng

b) \(\left(a+b+c\right)^2\)

\(=\left[\left(a+b\right)+c\right]^2\)

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

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

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

a) Ta có : \(2\left(a^2+b^2\right)-\left(a+b\right)^2=2a^2+2b^2-\left(a^2+2ab+b^2\right)\)

\(=2a^2+2b^2-a^2-2ab-b^2\)

\(=a^2-2ab+b^2\)

\(=\left(a-b\right)^2\ge0\)( đúng với mọi a,b )

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\left(đpcm\right)\)

Dấu " = " xảy ra <=> a = b = 0

b) \(VT=\left(a+b+c\right)^2=\left[\left(a+b\right)+c\right]^2\)

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

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

\(=a^2+b^2+c^2+2ab+2bc+2ac=VP\left(đpcm\right)\)

theo bài ra ta có: \(c^2+2ab-2bc-2ca=0.\)

\(\Rightarrow2\left(c^2+ab-bc-ca\right)=c^2\)

\(\Rightarrow2\left(a-c\right)\left(b-c\right)=c^2\)

Mặt khác: \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{2a^2-2ac+c^2}{2b^2-2bc+c^2}=\frac{2a\left(a-c\right)+2\left(a-c\right)\left(b-c\right)}{2b\left(b-c\right)+2\left(a-c\right)\left(b-c\right)}\)

                                                                               \(=\frac{\left(a-c\right)\left(a+b-c\right)}{\left(b-c\right)\left(b+a-c\right)}=\frac{a-c}{b-c}\) => đpcm

Câu hỏi tương tự: Câu hỏi của Đinh Tuấn Việt - Toán lớp 8 | Học trực tuyến

Cho a = b = c = 1 vào thì đề sai

Để ý phần mẫu \(2bc\le b^2+c^2\)

chắc hướng làm là như vậy @@

ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b 

ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)

M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)

M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)

M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)

M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)

M=-1-1-1=-3

Vậy với a+b+c=0 thì M=-3