ta co : (1/a +1/b+1/c)^2=1/a^2+1/b^2+1/c^2+2/ab+2/ac+2bc =1/a^2+1/b^2+1/c^2+2(1/ab+1/ac+1/bc) =>2^2=2+2(1/ab+1/ac+1/bc) 2=2(1/ab+1/ac+1/bc) =>1/ab+1/ac+1/bc=1 =>c/abc+b/abc+a/abc=abc/abc =>a+b+c/abc=abc/abc =>.a+b+c=abc(dpcm) (minh nghi la dung)
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
cho :frac{1}{a}+frac{1}{b}+frac{1}{c}2 ;frac{1}{a^2}+frac{1}{b^2}+frac{1}{c^2}2 .CMR:a+b+cabc2.Cho: frac{x}{a}+frac{y}{b}+frac{z}{c}0;frac{a}{x}+frac{b}{y}+frac{c}{z}2.Tính Afrac{a^2}{x}+frac{b^2}{y^2}+frac{c^2}{z^2}
Đọc tiếp
cho :
\(\frac{1}{a}+\frac{1}{b}\)+\(\frac{1}{c}\)=2 ;\(\frac{1}{a^2}\)+\(\frac{1}{b^2}\)+\(\frac{1}{c^2}\)=2 .CMR:a+b+c=abc
2.Cho: \(\frac{x}{a}\)+\(\frac{y}{b}\)+\(\frac{z}{c}\)=0;\(\frac{a}{x}\)+\(\frac{b}{y}\)+\(\frac{c}{z}\)=2.Tính A=\(\frac{a^2}{x}\)+\(\frac{b^2}{y^2}\)+\(\frac{c^2}{z^2}\)
Cho a,b,c >0
Chứng minh: \(\frac{1}{a^2+b^2+abc}+\frac{1}{b^2+c^2+abc}+\frac{1}{c^2+a^2+abc}\ge\frac{1}{abc}\)
Cho: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) và a+b+c=abc. Tính: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\);\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
cmr a+b+c=abc
cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) va\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2.\)
Chung minh rang a+b+c=abc
Cho các số a,b,c khác 0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\) và a+b+c=abc
Tính B=\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
Cho
\(\hept{\begin{cases}a+b+c=abc\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\\a,b,c\ne0\end{cases}}\)
Tính \(\frac{1}{a^{2\:}}+\frac{1}{b^2}+\frac{1}{c^2}\)
giúp mình với:
Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
và a+b+c=abc
C/m \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
1) Cho \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)
CM: \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1\)
2) Cho \(abc\ne1\)và \(\frac{ab+1}{b}=\frac{bc+1}{c}=\frac{ac+1}{a}\)
CM: a=b=c