CMR
nếu \(a^2+b^2+c^2=ab+ac+bc\)thì a=b=c
nếu\(a^2+b^2+c^{^2}+3=2\left(a.b.c\right)\)thì a=b=c=1
Những câu hỏi liên quan
Afrac{a^2+bc}{b+ac}+frac{b^2+ca}{c+ab}+frac{c^2+ab}{a+bc}frac{3left(a^2+bcright)}{left(a+b+cright)b+3ac}+frac{3left(b^2+caright)}{left(a+b+cright)c+3ab}+frac{3left(c^2+abright)}{left(a+b+cright)a+3bc}gefrac{3left(a^2+bcright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(b^2+caright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(c^2+abright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}3
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\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
Câu 1: CMR : Nếu a^3+b^3+c^33abc thì a+b+c0 hoặc abc Câu 2: Cho frac{1}{a}+frac{1}{b}+frac{1}{c}0 . Tính frac{bc}{a^2}+frac{ca}{b^2}+frac{ab}{c^2}Câu 3 : Cho a^3+b^3+c^33abcleft(a.b.cne0right). TínhAleft(1+frac{a}{b}right)left(1+frac{b}{c}right)left(1+frac{c}{a}right)
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Câu 1: CMR : Nếu \(a^3+b^3+c^3=3abc\) thì \(a+b+c=0\) hoặc \(a=b=c\)
Câu 2: Cho \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) . Tính \(\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}\)
Câu 3 : Cho \(a^3+b^3+c^3=3abc\left(a.b.c\ne0\right)\). Tính\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
Câu 1:
Chứng minh a3+b3+c3=3abc thì a+b+c=0\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3abc\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow0=0\) Đúng (Đpcm)
Chứng minh a3+b3+c3=3abc thì a=b=cÁp dụng Bđt Cô si 3 số ta có:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu = khi a=b=c (Đpcm)
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Câu 2
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\cdot\frac{1}{abc}\)
Ta có:
\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\cdot3\cdot\frac{1}{abc}=3\)
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\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{array}\right.\)
Xét \(a+b+c=0\)\(\Rightarrow\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}\)\(\Rightarrow A=\left(1-\frac{b+c}{b}\right)\left(1-\frac{a+c}{c}\right)\left(1-\frac{a+b}{a}\right)\)
\(=\left(1-1-\frac{c}{b}\right)\left(1-1-\frac{a}{c}\right)\left(1-1-\frac{b}{a}\right)\)
\(=\left(-\frac{c}{b}\right)\left(-\frac{a}{c}\right)\left(-\frac{b}{a}\right)=-1\)
Xét \(a^2+b^2+c^2-ab-bc-ca=0\)\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a-b=b-c=c-a=0\Leftrightarrow a=b=c\)
\(\Leftrightarrow A=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
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CMR nếu
\(c^2+2\left(ab-ac-bc\right)=0,b\ne c,a+b\ne c\) thì \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
Vì \(c^2+2\left(ab-ac-bc\right)=0\) nên :
\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a^2+\left(a-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}{b^2+\left(b-c\right)^2+\left(c^2+2ab-2ac-2bc\right)}\)
\(=\frac{2a^2+2c^2-4ac+2ab-2bc}{2b^2+2c^2-4bc+2ab-2ac}=\frac{\left(a-c\right)^2+b\left(a-c\right)}{\left(b-c\right)^2+a\left(b-c\right)}\)
\(=\frac{\left(a-c\right)\left(a-c+b\right)}{\left(b-c\right)\left(b-c+a\right)}=\frac{a-c}{b-c}\) \(\left(b\ne c,a+b\ne0\right)\)
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CM rằng nếu \(c^2=2\cdot\left(ac+bc-ab\right)\) và b#c , a+b#c thì\(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{a-c}{b-c}\)
We Have \(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ac\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ac\right)\)
\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2or\sqrt{3\left(a^2+b^2+c^2\right)}\ge a+b+c.\left(Q.E.D\right)\)
Chứng minh rằng nếu:
a) \(a^2+b^2+c^2=ab+ac+bc\)thì a = b = c
b) \(a^3+b^3+c^3=3abc\)thì a = b = c hoặc a+ b +c = 0
c) a + b +c = 0 thì \(a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)
a) a2 + b2 + c2 = ab + ac + bc
=> 2a2 + 2b2 + 2c2 = 2ab + 2ac + 2bc
=> 2a2 + 2b2 + 2c2 - 2ab - 2ac - 2bc = 0
=> (a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 - 2bc + c2) = 0
=> (a - b)2 + (a - c)2 + (b - c)2 = 0
Do 3 hạng tử trên đều có giá trị lớn hơn hoặc bằng 0 nên a - b = a - c = b - c = 0
=> a = b = c
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b) a3 + b3 + c3 = 3abc
=> a3 + b3 + c3 - 3abc = 0
=> a3 + 3a2b + 3ab2 + b3 + c3 - 3abc - 3a2b - 3ab2 = 0
=> (a + b)3 + c3 - 3ab(a + b + c) = 0
=> (a + b + c)(a2 + 2ab + b2 - bc - ac + c2) - 3ab(a + b + c) = 0
=> (a + b + c)(a2 + b2 + c2 - ab - bc - ac) = 0
=> a + b + c = 0
hoặc a2 + b2 + c2 = ab + bc + ac => a = b = c
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a)\(a^2+b^2+c^2=ab+bc+ca\)\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow a=b=c}\)
b)\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\hept{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
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frac{a^2b+bc^2-1}{acleft(a+cright)}+frac{b^2c+ca^2-1}{ableft(a+bright)}+frac{c^2a+ab^2-1}{bcleft(b+cright)}frac{a^2b^2+b^2c^2-b}{a+c}+frac{b^2c^2+c^2a^2-c}{a+b}+frac{c^2a^2+a^2b^2-a}{b+c}frac{frac{1}{a^2}-frac{1}{ac}+frac{1}{c^2}}{a+c}+frac{frac{1}{b^2}-frac{1}{ab}+frac{1}{a^2}}{a+b}+frac{frac{1}{c^2}-frac{1}{bc}+frac{1}{b^2}}{b+c}gefrac{1}{acleft(a+cright)}+frac{1}{bcleft(b+cright)}+frac{1}{ableft(b+aright)}frac{a}{b+c}+frac{b}{c+a}+frac{c}{a+b}gefrac{3}{2}
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\(\frac{a^2b+bc^2-1}{ac\left(a+c\right)}+\frac{b^2c+ca^2-1}{ab\left(a+b\right)}+\frac{c^2a+ab^2-1}{bc\left(b+c\right)}\)
\(=\frac{a^2b^2+b^2c^2-b}{a+c}+\frac{b^2c^2+c^2a^2-c}{a+b}+\frac{c^2a^2+a^2b^2-a}{b+c}\)
\(=\frac{\frac{1}{a^2}-\frac{1}{ac}+\frac{1}{c^2}}{a+c}+\frac{\frac{1}{b^2}-\frac{1}{ab}+\frac{1}{a^2}}{a+b}+\frac{\frac{1}{c^2}-\frac{1}{bc}+\frac{1}{b^2}}{b+c}\ge\frac{1}{ac\left(a+c\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ab\left(b+a\right)}\)
\(=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
CMR: Với các số thực dương a;b;c thì\(\dfrac{a^3+2abc+b^3}{c^2+ab}+\dfrac{a^3+2abc+c^3}{b^2+ac}+\dfrac{b^3+2abc+c^3}{a^2+bc}\ge2\left(a+b+c\right)\)
Tính (phân thức)
a)\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)