Cho ac=b^2.cmr a(b^2+c^2)=c(a^2+b^2)
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cho b^2=ac cmr a^2+b^2/b^2+c^2=a/c
Ta có :
\(VT=\dfrac{a^2+b^2}{b^2+c^2}\)
Mà \(b^2=ac\)
\(\Leftrightarrow VT=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}=VP\left(đpcm\right)\)
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Ta có: \(b^2=ac\)
\(\Leftrightarrow ac=b\cdot b\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a^2}{b^2}=\dfrac{b^2}{c^2}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Leftrightarrow\dfrac{a^2}{b^2}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(\Leftrightarrow\dfrac{a^2}{ac}=\dfrac{a^2+b^2}{b^2+c^2}\)
hay \(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)(đpcm)
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a) Cho a2 + b2 + c2+3 = 2.(a + b + c). Cmr: a = b = c =1
b) Cho (a + b + c)2 = 3.(ab + bc + ac). Cmr: a = b = c
a) \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Leftrightarrow a=b=c=1\)
b) \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ac\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2+b^2-2ab\right)+\left(b^2+c^2-2bc\right)+\left(c^2+a^2-2ac\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
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MỌI NGƯỜI LM ĐC CÂU NÀO THÌ LM NHA !
Cho x>y TM: x+y<=1 CMR: 1/x^2+y^2 + 1/xy>=6
Cho a,b,c >0 TM: a+b+c<=1 CMR: (1/a^2+bc) + (1/b^2+ac)+ 1/c^2+2ab >=9
Cho a,b>0 TM: a+b<=1 ;CMR: (1/a^b^2)+ 4b + 1/ab>=7
Cho a,b>0 TM:a+b<=1. CMR: 1/1+a^2+b^2 + 1/2ab >=8/3
Cho a,b,c>0 TM: a+b+c<=3.CMR: 1/a^2+b^2+c^2 + 2009/ab+bc+ac >=670
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CMR:(a^2+bc)/(b^2+ac)+(b+ac)/(c+ab)+(c^2+ac)/(a+ab)>=3