bạn lớp 7 mà học kém quá nhỉ

dễ ot

b,c=1

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac})=4 \\<=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2.\dfrac{a+b+c}{abc}=4 \\<=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4(do\ a+b+c=abc) \\<=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2 (đpcm)\)

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

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

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

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

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

\(\Leftrightarrow ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+c\right)=0\)

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

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

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

Câu b :

Ta có :

\(a+b+c=abc\)

\(\Leftrightarrow1=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\)

\(\Leftrightarrow2=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(\Leftrightarrow4=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\) \(+2\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)

\(\)\(\Rightarrow\) \(\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+2=4\)

\(\Rightarrow\) \(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\left(đpcm\right)\)

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

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

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

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

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

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

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

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

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

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

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

Bài 2 :

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

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\frac{a+b+c}{abc}=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot1=4\)

( Do \(a+b+c=abc\) )

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

P/s : Cho hỏi bài 1 có a,b,c > 0 không ?

Khuyến mãi thêm bài 1 :))

Áp dụng BĐT AM-GM ta có :

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{b^2}\cdot\frac{b^2}{c^2}}=\frac{2a}{c}\) (1)

Tương tự ta có :

\(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{2b}{a}\)(2), \(\frac{c^2}{a^2}+\frac{a^2}{b^2}\ge\frac{2c}{b}\) (3)

Cộng các vế của BĐT (1) (2) và (3) và chia 2 ta có :

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

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