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a, \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
=> a=b=c
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thử bài bất :D
Ta có: \(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b+c}{4}\ge5\sqrt[5]{\dfrac{1}{a^3\left(b+c\right)}.\dfrac{a^3}{2^3}.\dfrac{\left(b+c\right)}{4}}=\dfrac{5}{2}\) ( AM-GM cho 5 số ) (*)
Hoàn toàn tương tự:
\(\dfrac{1}{b^3\left(c+a\right)}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{b}{2}+\dfrac{c+a}{4}\ge5\sqrt[5]{\dfrac{1}{b^3\left(c+a\right)}.\dfrac{b^3}{2^3}.\dfrac{\left(c+a\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (**)
\(\dfrac{1}{c^3\left(a+b\right)}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{c}{2}+\dfrac{a+b}{4}\ge5\sqrt[5]{\dfrac{1}{c^3\left(a+b\right)}.\dfrac{c^3}{2^3}.\dfrac{\left(a+b\right)}{4}}=\dfrac{5}{2}\) (AM-GM cho 5 số) (***)
Cộng (*),(**),(***) vế theo vế ta được:
\(P+\dfrac{3}{2}\left(a+b+c\right)+\dfrac{2\left(a+b+c\right)}{4}\ge\dfrac{15}{2}\) \(\Leftrightarrow P+2\left(a+b+c\right)\ge\dfrac{15}{2}\)
Mà: \(a+b+c\ge3\sqrt[3]{abc}=3\) ( AM-GM 3 số )
Từ đây: \(\Rightarrow P\ge\dfrac{15}{2}-2\left(a+b+c\right)=\dfrac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
1. \(a^3+b^3+c^3+d^3=2\left(c^3-d^3\right)+c^3+d^3=3c^3-d^3\) :D
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1) ta có: A= x^3 -8y^3=> A=(x-2y)(x^2 +2xy+4y^2)=>A=5.(29+2xy) (vì x-2y=5 và x^2+4y^2=29) (1)
Mặt khác : x-2y=5(gt)=> (x-2y)^2=25=> x^2-4xy+4y^2=25=>29-4xy=25(vì x^2+4y^2=29)
=> xy=1 (2)
Thay (2) vào (1) ta đc: A= 5.(29+2.1)=155
Vậy gt của bt A là 155
2) theo bài ra ta có: a+b+c=0 => a+b=-c=>(a+b)^2=c^2=> a^2 +b^2+2ab=c^2=>c^2-a^2-b^2=2ab
=> \(\left(c^2-a^2-b^2\right)^2=4a^2b^2\)
=>\(c^4+a^4+b^4-2c^2a^2+2a^2b^2-2b^2c^2=4a^2b^2\)
=>\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)
=>\(2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)
=> \(a^4+b^4+c^4=\frac{1}{2}\left(a^2+b^2+c^2\right)^2\) (đpcm)
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ta co:
a-b=a^3+b^3
a-b-b^3=a^3
Mà một số luôn nhỏ hơn hoặc bằng chính nó lũy thừa 3
Nhưng a-b-b^3=a^3 nên b=0
Mà a=a^3 suy ra a=1
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a) \(a,b>0\Rightarrow a^3-b^3< a^3+b^3\)
Mà \(a^3+b^3=a-b\)
\(\Rightarrow a^3-b^3< a-b\)
\(\Rightarrow\frac{a^3-b^3}{a-b}< 1\)
\(\Leftrightarrow\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}< 1\)
\(\Leftrightarrow a^2+ab+b^2< 1\)
\(\Rightarrow a^2+b^2< 0\)(Vì a,b > 0)
b) Câu hỏi của ta là ai - Toán lớp 7 - Học toán với OnlineMath
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\(\Leftrightarrow\dfrac{2a^2}{b^2}+\dfrac{2b^2}{c^2}+\dfrac{2c^2}{a^2}=\dfrac{2a}{c}+\dfrac{2c}{b}+\dfrac{2b}{a}\)
\(\Leftrightarrow\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}-\dfrac{2a}{c}\right)+\left(\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}-\dfrac{2c}{b}\right)+\left(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}-\dfrac{2b}{a}\right)=0\)
\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{b}{c}\right)^2+\left(\dfrac{a}{b}-\dfrac{c}{a}\right)^2+\left(\dfrac{b}{c}-\dfrac{c}{a}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}-\dfrac{b}{c}=0\\\dfrac{a}{b}-\dfrac{c}{a}=0\\\dfrac{b}{c}-\dfrac{c}{a}=0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}\Leftrightarrow a=b=c\)
\(a>0;b>0\Rightarrow a^2>0;b^2>0\Rightarrow a^2-b^2< =a^2+b^2=1\)
\(\Rightarrow\left(a-b\right)\left(a+b\right)< 1\Rightarrow a-b< \frac{1}{a+b}\Rightarrow\frac{1}{a+b}>a-b\)
với a>0;b>0 ta có: \(a^3b+ab^3>=2\sqrt{a^3bab^3}=2\sqrt{a^4b^4}=2a^2b^2\)(bđt cosi)
\(\Rightarrow a^4+a^3b+ab^3+b^4>=a^4+2a^2b^2+b^4=\left(a^2+b^2\right)^2=1^2=1\)
\(\Rightarrow a^3\left(a+b\right)+b^3\left(a+b\right)=\left(a^3+b^3\right)\left(a+b\right)>=1\Rightarrow a^3+b^3>=\frac{1}{a+b}\)mà \(\frac{1}{a+b}>a-b\)
\(\Rightarrow a^3+b^3>a-b\left(đpcm\right)\)