P = \(\frac{a^2c}{a^2c+c^2b+b^2a+}+\frac{b^2a}{b^2a+a^2c+c^2b}+\frac{c^2b}{c^2b+b^2a+a^2c}\)

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

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

Câu hỏi của Chi Chi - Toán lớp 8 - Học toán với OnlineMath

A = (2a + 2b +2c - 3c)^2 + (2b + 2c +2a - 3a)^2 + (2c + 2a +2b -3b)^2

Đặt a + b + c = x thì 

A = (2x - 3c)^2 + (2x - 3a)^2 + (2x - 3b)^2

    =4x^2 - 12cx + 9c^2 + 4x^2 - 12ax + 9x^2 + 4x^2 - 12bx + 9b^2

    =12x^2 - 12x(a + b + c) + 9(a^2 + b^2 + c^2)

    =12x^2 - 12x^2 + 9(a^2 + b^2 + c^2) =9(a^2 + b^2 + c^2) =9m

Áp dụng t/c dtsbn ta có:

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

\(\dfrac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b=3a-c\)\(\dfrac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c=3b-a\)

\(\dfrac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a=3c-b\)

\(P=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{2a.2b.2c}=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{8abc}\)

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

\(=\left(4a^2+4b^2+c^2+8ab-4ac+4bc\right)+\left(4b^2+4c^2+a^2+8bc-4ba-4ac\right)\)\(+\left(4c^2+4a^2+b^2+8ac-4cb-4ab\right)\)

\(=9a^2+9b^2+9c^2\)

\(=9\left(a^2+b^2+c^2\right)\)

\(=9m\)

phân tích nó cho trong có a2+b2+c2 thay vào rsut gọn

\(2a=3b=4c\\ \Leftrightarrow\dfrac{a}{6}=\dfrac{b}{4}=\dfrac{c}{3}=\dfrac{2b}{8}=\dfrac{2c}{6}=\dfrac{a+b-c}{7}=\dfrac{a+2b-2c}{8}\\ \Leftrightarrow A=\dfrac{a+b-c}{a+2b-2c}=\dfrac{7}{8}\)

làm ơn trả lời hộ mk với ah mai mk phải nộp bài r