Ta có:

\(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}=\frac{a}{a\left(a+1\right)}+\frac{1}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}=\frac{1}{a}=y\)

Đúng 100%

Ta có : \(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}=\frac{a}{a\left(a+1\right)}+\frac{1}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}=\frac{1}{a}\)  Với mọi a

=> \(\frac{1}{a}=\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}\)

\(\frac{1}{a+1}+\frac{1}{a\left(a+1\right)}=\frac{a}{a\left(a+1\right)}+\frac{1}{a\left(a+1\right)}=\frac{a+1}{a\left(a+1\right)}=\frac{1}{a}\)

Vậy....

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)

\(\frac{yz+xz+xy}{xyz}=0\)

yz + xz + xy = 0

\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2\times\left(xy+xz+yz\right)=x^2+y^2+z^2+2\times0=x^2+y^2+z^2\left(\text{đ}pcm\right)\)

a) Từ giả thiết suy ra: xy + yz + zx = 0

Do đó:

\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=x^2+y^2+z^2\)

b) Đặt \(\frac{1}{a-b}=x\); \(\frac{1}{b-c}=y\); \(\frac{1}{c-a}=z\)

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=a-b+b-c+c-a=0\)

Theo câu a ta có: \(x^2+y^2+z^2=\left(x+y+z\right)^2\)

Suy ra điều phải chứng minh

a)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

\(\Rightarrow\frac{xy+yz+xz}{xyz}=0\)

\(\Rightarrow xy+yz+xz=0\)

\(x^2+y^2+z^2=\left(x+y+z\right)^2\)

\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2+2xy+2yz+2xz\)

\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)

Do \(xy+yz+xz=0\)

\(\Rightarrow x^2+y^2+z^2=x^2+y^2+z^2\) ( đpcm )

b)

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

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

\(\Rightarrow\frac{2}{\left(a-b\right)\left(b-c\right)}+\frac{2}{\left(b-c\right)\left(c-a\right)}+\frac{2}{\left(a-b\right)\left(c-a\right)}=0\)

\(\Rightarrow2\left(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)}\right)=0\)

\(\Rightarrow\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)}=0\)

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

\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right)\left[\left(a-b\right)+\left(b-c\right)+\left(c-a\right)\right]}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)

\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right)\left[\left(-a+a\right)+\left(-b+b\right)+\left(-c+c\right)\right]}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)

\(\Rightarrow\frac{\left(c-a\right)\left(b-c\right)\left(a-b\right).0}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=0\)

\(\Rightarrow0=0\) ( đpcm )

a)(a+b+c)(ab+bc+ac)-abc=a(ab+bc+ac)+b(ab+bc+ac)+c(ab+bc+ac)-abc

=a²b+abc+a²c+ab²+b²c+abc+abc+bc²+ac²-abc

=(abc+a²b)+(a²c+ac²)+(b²c+ab²)+(bc²+abc)+(abc-abc)

=ab(c+a)+ac(c+a)+b²(c+a)+bc(c+a)

=(ab+ac+b²+bc)(c+a)

=(a+b)(b+c)(c+a)

a) \(\left(a+b+c\right)\left(ab+bc+ac\right)-abc=a^2b+abc+a^2c+ab^2+b^2c+abc+abc+c^2b+c^2a-abc\)

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

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

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

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

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)(áp dụng từ câu a) )

\(\Rightarrow a+b=0\)hoặc \(b+c=0\)hoặc \(c+a=0\)

Đặt \(a^{2n+1}=x;b^{2n+1}=y;c^{2n+1}=z\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\left(xy+yz+xz\right)\left(x+y+z\right)-xyz=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)( áp dụng câu a) )

\(\Rightarrow x+y=0\)hoặc \(y+z=0\)hoặc \(z+x=0\)

Mà ta lại có \(a+b=0\left(cmt\right)\)\(\Rightarrow\)\(\frac{1}{a^{2n+1}}+\frac{1}{b^{2n+1}}=0\)\(\Rightarrow\frac{1}{c^{2n+1}}=\frac{1}{c^{2n+1}}\)(luôn đúng)

Tương tự với các trường hợp còn lại, ta có điều phải chứng minh.

\(\)

Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)

\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)

\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)

\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)

\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)

Dấu "=" xảy ra khi x=y=z

Biến đổi vế trái ta có :

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

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

 \(=\frac{a^2+3a+2-2a^2-4a+a^2+a}{a\left(a+1\right)\left(a+2\right)}=\frac{2}{a\left(a+1\right)\left(a+2\right)}\)

Vậy Vế trái = Vế phải 

2) 1/x - 1/y - 1/z = 1

=> (1/x - 1/y - 1/z)^2 = 1

<=> 1/x^2 + 1/y^2 + 1/z^2 - 2/xy - 2/xz + 2/yz = 1

<=> 1/x^2 + 1/y^2 + 1/z^2 - 2.(1/xy + 1/xz - 1/yz) = 1

<=> 1/x^2 + 1/y^2 + 1/z^2 - 2.(z+y-x/xyz) = 1

<=> 1/x^2 + 1/y^2 + 1/z^2 - 2.0 = 1

<=> 1/x^2 + 1/y^2 + 1/z^2 = 1 (đpcm)