cmr
a] neu\(\frac{a+b+c}{a+b-c}=\frac{a-b+c}{a-b-c}\) trong do b khac 0 thi c=0
Những câu hỏi liên quan
cho a, b>0 va c khac 0. cmr neu 1/a+1/b+1/c=0 thi can(a+b)=can(b+c)+can(c+a)
Chon cau tra loi dung Neu a+b/b+c=c+d/d+a thi
a)a phai bang c
b)a=c hay a+b+c+d=0
c)a+b+c+d phai bang 0
d)a+b+c+d khac 0 neu a=c
nếu \(\frac{a+b}{b+c}\)=\(\frac{c+d}{d+a}\)
=> a =c
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Cho 1/c=1/2(1/a+1/b)(voi a,b,ckhac 0; b khac c). CMRa/b=a-c/c-b
cho a,b,c khac 0 ; a++b+c khac 0 thoa man \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
CMR\(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b+c-c}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(\frac{ca+cb+c^2+ab}{abc\left(a+b+c\right)}\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b\left(a+c\right)+c\left(a+c\right)\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Rightarrow a+b=0\Rightarrow a=-b\Rightarrow a^{2009}=-b^{2009}\)
\(\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{c^{2009}}\) (1)
\(\frac{1}{a^{2009}+b^{2009}+c^{2009}}=\frac{1}{c^{2009}}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{1}{a^{2009}}+\frac{1}{b^{2009}}+\frac{1}{c^{2009}}=\frac{1}{a^{2009}+b^{2009}+c^{2009}}\) (đpcm)
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cmr
a+b/c+d=b+c/d+a trong do a+b+c+d khac 0 thi a=c
\(\frac{a+b}{c+d}=\frac{b+c}{d+a}\Rightarrow\frac{a+b}{b+c}=\frac{c+d}{d+a}=\frac{a+b+c+d}{b+c+d+a}=1\Rightarrow a+b=b+c\Rightarrow a=c\)
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biet x=\(\frac{a}{b+c}\)=\(\frac{b}{c+a}\)=\(\frac{c}{a+b}\)
tim x trong 2 truong hop sau
a] a+b+c=0
b] a+b+c khac 0
a] x= a/b+c=b/c+a=c/a+b=a+b+c/b+c+c+a+a+b=0
=> x=0
b]
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cmr neu \(\sqrt{a.a'}+\sqrt{b.b'}+\sqrt{c.c'}=\sqrt{\left(a+b+c\right)\left(a'+b'+c'\right)}\)
voi a,a',b,b',c,c'>0 thi \(\frac{a}{a'}=\frac{b}{b'}=\frac{c}{c'}\)
chun to ran neu \(\frac{a}{b}\)<\(\frac{c}{d}\)(b.0,d>0) thi \(\frac{a}{b}\)<\(\frac{a+c}{b+d}\)<\(\frac{c}{d}\)
P=\(\frac{2a-b}{c+d}\)+\(\frac{2b-c}{d+a}\)+\(\frac{2c-d}{a+b}\)+\(\frac{2d-a}{b+c}\)
Tinh P biet :
\(\frac{a}{b}\)=\(\frac{b}{c}\)=\(\frac{c}{d}\)=\(\frac{d}{a}\)
trong do ( a+b+c+d ) khac 0
Ta có: a/b=b/c=c/d=d/a áp dụng tính chất dãy tỉ số bằng nhau ta được:
a/b=b/c=c/d=d/a=(a+b+c+d)/(a+b+c+d)=1
Do đó: a/b=1 suy ra a=b (1) ; b/c=1 suy ra b=c (2) ; c/d=1 suy ra c=d (3) ; d/a=1 suy ra d=a (4)
Từ (1),(2),(3),(4) ta được: a=b=c=d
Suy ra:P=(2a-a)/(a+a)+(2a-a)/(a+a)+(2a-a)/(a+a)+(2a+a)/(a+a)
=4.a/2a=4.1/2=2
Vậy P=2
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