Những câu hỏi liên quan
a). Cho a/b=c/d( với b+d khác 0)
CM: a/b=a+c/b+c
b). Cho a/b+c/d( a,b,c,d khác 0)
CM: a-b/a=c-d/c
CM: 16^a +16^b +16^c >= 2^a+ 2^b +2^c, biết a+b+c= 0
cho a,b,c>0. CM: a/b + b/a + a/c>= căn a/b + căn b/a+ căn a/c
cho(a+b+c):(a+b-c)=(a-b+c):(a-b-c) và b khác 0.CM c=0
Từ \(\left(a+b+c\right):\left(a+b-c\right)=\left(a-b+c\right):\left(a-b-c\right)\)
\(\Rightarrow\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}=\frac{\left(a+b+c\right)-\left(a-b+c\right)}{\left(a+b-c\right)-\left(a-b-c\right)}\)
\(=\frac{a+b+c-a+b-c}{a+b-c-a+b+c}=\frac{2b}{2b}=1\)
\(\Rightarrow a+b+c=a+b-c\)\(\Rightarrow\left(a+b+c\right)-\left(a+b-c\right)=0\)
\(\Rightarrow a+b+c-a-b+c=0\)\(\Rightarrow2c=0\)\(\Rightarrow c=0\)( đpcm )
CM: 16^a +cho a,b,c>0.1 6^b +16^c >= 2^a+ 2^b +2^c, biết a+b+c= 0
CM: a/b + b/a + a/c>= căn a/b + căn b/a+ căn a/c
bài 1: CM a+c=2b và 2bd=c (b+d)
(b,d khác 0) thì a/b=c/d
bài 2: cho 2/a=1/b+1/c (a,b,c khác 0)
CM: b/c=b-a/a-c
Bài 1:
2bd=c(b+d)
=>d(a+c)=c(b+d)
=>ad+cd=cb+cd
=>ad=cb
=>a/b=c/d
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Cho a/b-c+b/c-a+c/a-b=0
CM: a/(b-c)2+b/(c-a)2+c/(a-b)2=0
cho a b c khác 0 thoa mãn a+b+c=0
a/ CM a+b=-c
b/Tính A= (1+a/b).(1+b/c).(1+c/a)
cho a + b + c = 0 . CM : M = N = P M = a ( a + b ) ( a + c ) N = b ( b + c ) ( a + b ) P = c ( c + b ) ( a + c )
\(a+b+c=0\)
\(\Rightarrow\)\(\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)
\(M=a\left(a+b\right)\left(a+c\right)=a.\left(-c\right).\left(-b\right)=abc\)
\(N=b\left(b+c\right)\left(a+b\right)=b.\left(-a\right).\left(-c\right)=abc\)
\(P=c\left(b+c\right)\left(a+c\right)=c.\left(-a\right).\left(-b\right)=abc\)
\(\Rightarrow\)\(M=N=P\)
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Cho a khác +-b và a(a+b)(a+c)=b(b+c)(b+a). Cm a+b+c=0
\(a\left(a+b\right)\left(a+c\right)=b\left(b+c\right)\left(b+a\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a^2+ac-b^2-bc\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[\left(a+b\right)\left(a-b\right)+c\left(a-b\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a+b+c\right)=0\)
=>a+b+c=0
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cho a/b+c +b/c+a +c/a+b =1.Cm a^2/b+c +b^2/c+a +c^2/a+b =0
Ta có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1\)
\(\Rightarrow\left(a+b+c\right)\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)=a+b+c\)
\(\Rightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+\left(a+b+c\right)=a+b+c\)
\(\Rightarrow\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)=0\)(đpcm)