cho 3 số a,b,c thỏa mãn 0<=a,b,c<=1 và a+b+c=2. cm: a^2+b^2+c^2<=2
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b: (3x-2)^5+(5-x)^5+(-2x-3)^5=0
Đặt a=3x-2; b=-2x-3
Pt sẽ trở thành:
a^5+b^5-(a+b)^5=0
=>a^5+b^5-(a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5)=0
=>-5a^4b-10a^3b^2-10a^2b^3-5ab^4=0
=>-5a^4b-5ab^4-10a^3b^2-10a^2b^3=0
=>-5ab(a^3+b^3)-10a^2b^2(a+b)=0
=>-5ab(a+b)(a^2-ab+b^2)-10a^2b^2(a+b)=0
=>-5ab(a+b)(a^2-ab+b^2+2ab)=0
=>-5ab(a+b)(a^2+b^2+ab)=0
=>ab(a+b)=0
=>(3x-2)(-2x-3)(5-x)=0
=>\(x\in\left\{\dfrac{2}{3};-\dfrac{3}{2};5\right\}\)
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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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Từ a+b+c=0 => b+c=-a
Theo đề ra ta có a3 + b3 + c3 = 0
=> a3 + (b+c)(b2 - bc + c2 )=0
<=> a3- a[(b + c )2 -3bc]= 0
<=> a3- [( -a )2 - 3bc] = 0
<=> a3 - a3 +3bc = 0
<=> 3bc= 0
<=> a =0 hoặc b=0 hoặc c=0 ( đpcm)
cho mik điểm nha bạn ơiii
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\(a,\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}=\dfrac{2\left(a+b+c\right)}{6}=\dfrac{a+b+c}{3}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{a+b+c}{3}\\ \Rightarrow3\left(a+b+c\right)=3\left(a+b\right)\\ \Rightarrow3\left(a+b\right)+3c=3\left(a+b\right)\\ \Rightarrow3c=0\\ \Rightarrow c=0\)
Vậy \(P=\dfrac{a+b-2019c}{a+b+2018c}=\dfrac{a+b}{a+b}=1\)
Cho các số a, b, c thỏa mãn a^3+ b^3+ c^3= 3abc với a, b, c khác 0. Chứng minh a+ b+c = 0 hoặc a=b=c
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a3 + b3 + c3 = 3abc
⇒ a3 + b3 + c3 - 3abc = 0
⇒ ( a3 + b3 ) + c3 - 3abc = 0
⇒ ( a + b )3 - 3ab( a + b ) + c3 - 3abc = 0
⇒ [ ( a + b )3 + c3 ] - [ 3ab( a + b ) + 3abc ] = 0
⇒ ( a + b + c )[ ( a + b )2 - ( a + b ).c + c2 ] - 3ab( a + b + c ) = 0
⇒ ( a + b + c )( a2 + b2 + c2 - ab - bc - ac ) = 0
⇒ \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)
+) a2 + b2 + c2 - ab - bc - ac = 0
⇒ 2( a2 + b2 + c2 - ab - bc - ac ) = 2.0
⇒ 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac = 0
⇒ ( a2 - 2ab + b2 ) + ( b2 - 2bc + c2 ) + ( a2 - 2ac + c2 ) = 0
⇒ ( a - b )2 + ( b - c )2 + ( a - c )2 = 0
VT ≥ 0 ∀ a,b,c . Dấu "=" xảy ra khi a = b = c
⇒ a + b + c = 0 hoặc a = b = c ( đpcm )
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Ta có: \(a+b+c=0\Rightarrow a+b=-c\)
\(a^3+b^3+c^3=0\)
\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)+c^3=0\)
\(\Rightarrow-c.\left(a^2+2ab+b^2-3ab\right)+c^3=0\)
\(\Rightarrow-c\left[\left(a+b\right)^2-3ab\right]+c^3=0\)
\(\Rightarrow-c\left(c^2-3ab\right)+c^3=0\)
\(\Rightarrow-c^3+3abc+c^3=0\Rightarrow3abc=0\Rightarrow abc=0\)
\(\Rightarrow\)\(a=0\) hoặc \(b=0\) hoặc \(c=0\)
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\2a^2+2b^2+2c^2-2ab-2bc-2ca=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
bạn thay vào M giải tiếp nha
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a^3+b^3\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Nếu \(a^2+b^2+c^2-ab-bc-ca\)
\(=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\left(\forall a,b,c\right)\)
Dấu "=" xảy ra khi: a = b = c
Khi đó: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)^3=8\)
Nếu \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{-abc}{abc}=-1\)
Cho a = 1; b =0,5; c = 0,5
1^2+0,5^2+0,5^2=1+0,25+0,25=1,5