bạn lớp 7 mà học kém quá nhỉ

dễ ot

b,c=1

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\Leftrightarrow abc=\left(a+b+c\right)\left(ab+bc+ca\right)\)

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

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

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

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

\(\Leftrightarrow\left(a+b\right)\left[ab+c^2+c\left(a+b\right)\right]=0\Leftrightarrow\left(a+b\right)\left(ab+c^2+ac+bc\right)=0\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)Chúc bạn học tốt!

Phú Quý Lê Tăng ơi! Hình như bn làm lộn dấu 1 bước phải ko? Chỗ đó hình như phải là +2ab mới đúng.

Bài 1:

a: Sửa đề: \(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}=0\)

Ta có: \(\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)

\(=\frac{1}{\left(a+b\right)^2-2ab-c^2}+\frac{1}{\left(b+c\right)^2-2bc-a^2}+\frac{1}{\left(c+a\right)^2-2ac-b^2}\)

\(=\frac{1}{\left(a+b\right)^2-c^2-2ab}+\frac{1}{\left(b+c\right)^2-a^2-2bc}+\frac{1}{\left(c+a\right)^2-b^2-2ac}\)

\(=\frac{1}{\left(a+b-c\right)\left(a+b+c\right)-2ab}+\frac{1}{\left(b+c+a\right)\left(b+c-a\right)-2bc}+\frac{1}{\left(c+a-b\right)\left(c+a+b\right)-2ac}\)

\(=\frac{-1}{2ab}+\frac{-1}{2bc}+\frac{-1}{2ac}=\frac{-c-a-b}{2abc}=0\)

2:

Sửa đề: \(M=\frac{x^2}{x^4+x^2+1}\)

TH1: x=0

=>\(a=\frac{x}{x^2+x+1}=0\)

\(M=\frac{x^2}{x^4+x^2+1}=0\)

TH2: x<>0

\(\frac{x}{x^2+x+1}=a\)

=>\(\frac{x^2+x+1}{x}=\frac{1}{a}\)

=>\(\frac{x^2-x+1+2x}{x}=\frac{1}{a}\)

=>\(\frac{x^2-x+1}{x}+2=\frac{1}{a}\)

=>\(\frac{x^2-x+1}{x}=\frac{1}{a}-2=\frac{1-2a}{a}\)

=>\(\frac{x}{x^2-x+1}=\frac{a}{1-2a}\)

\(M=\frac{x^2}{x^4+x^2+1}\)

\(=\frac{x^2}{x^4+2x^2+1-x^2}\)

\(=\frac{x^2}{\left(x^2+1\right)^2-x^2}=\frac{x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

\(=\frac{x}{x^2-x+1}\cdot\frac{x}{x^2+x+1}=a\cdot\frac{a}{1-2a}=\frac{a^2}{1-2a}\)

a) b,c=1

còn lại chịu

a, Thay a=1 ta có hệ phương trình:

       1+\(\)1/b=c+\(\)1/1

       Và 1+1/b=b+1/c

<=>c=1/b

      Và1+1/b=b+1/1/b

Giải hệ này ta tìm được b=-1/2 và c=-2

\(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ca\right)\left(a-abc\right)\)

\(\Leftrightarrow a^2b+ab^2c^2-a^3bc-b^2c=b^2a+a^2bc^2-ca^2-ab^3c\)

\(\Leftrightarrow a^2b-ab^2-b^2c+ca^2=a^2bc^2-ab^3c+a^3bc-ab^2c^2\)

\(\Leftrightarrow\left(a-b\right)\left(ab+bc+ca\right)=abc\left(a-b\right)\left(a+b+c\right)\)

\(\Leftrightarrow ab+bc+ca=abc\left(a+b+c\right)\Leftrightarrow a+b+c=\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)

\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)

Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)

\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)