\(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\Leftrightarrow\frac{a+b}{ab}=-\frac{a+b}{c\left(a+b+c\right)}\)=> a = -b  hoặc ab +c(a+b+c) =0

+ a = -b => thay => dpcm

+ab + c(a+b+c) =0 =>(a+c)(b+c) =0 => a =-c hoạc  b =-c thay => dpcm

Ta có : 1/a+1/b+1/c=1/a+b+c
suy ra:(ab+bc+ac)/abc=1/a+b+c
=>(ab+ac+bc)(a+b+c)=abc
=>a²b+ab²+a²c+ac²+b²c+bc²+3abc=abc
(a+b)(b+c)(a+c)=0
=>a=-b hc b=-c hc a=-c
ta sẽ dễ dàng c/m được với 3 t/h trên thì 1/a¹⁹⁹⁵+1/b¹⁹⁹⁵+1/c¹⁹⁹⁵=1/a¹⁹⁹⁵+b¹⁹⁹⁵+c¹⁹⁹⁵

a) Ta có: \(\frac{a+2}{a-2}=\frac{b+3}{b-3}.\)

\(\Leftrightarrow\frac{a+2}{b+3}=\frac{a-2}{b-3}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{a+2}{b+3}=\frac{a-2}{b-3}=\frac{a+2+a-2}{b+3+b-3}=\frac{2a}{2b}=\frac{a}{b}\) (1)

\(\frac{a+2}{b+3}=\frac{a-2}{b-3}=\frac{a}{b}=\frac{4}{6}=\frac{2}{3}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{a}{b}=\frac{2}{3}\)

\(\Rightarrow\frac{a}{2}=\frac{b}{3}\left(đpcm\right).\)

Chúc bạn học tốt!

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

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

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

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

\(P=0\)

\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)

\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)

\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)

\(B=\frac{3}{4}\)

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)

\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)

\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)

=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)

\(A=\frac{2}{3}-\frac{1}{192}\)

\(A=\frac{127}{192}\)

\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

      \(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)

      \(C=\frac{1990.997}{1994.995}\)

      \(C=\frac{995.2+997}{997.2+995}=1\)

\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)

\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)

B=\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)

B= \(\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)

B=\(\frac{3}{4}\)

Sau mình làm tiếp vội quá! k mình nha