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H
28 tháng 10 2018
1. \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(abc\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2+c^2-ac-bc\right)-3ab\left(a+b+c\right)\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
2. \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
3.Còn có a + b + c = 0 nữa mà bn.
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)
+ \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\ \left(c-a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow a=b=c\)
M
1
AH
Akai Haruma
Giáo viên
19 tháng 7 2019
Lời giải:
$a+b-c=0\Rightarro a+b=c$. Kết hợp sử dụng đẳng thức quen thuộc \(a^3+b^3=(a+b)^3-3ab(a+b)\) ta có:
\(a^3+b^3-c^3+3abc=(a+b)^3-3ab(a+b)-c^3+3abc\)
\(=c^3-3ab.c-c^3+3abc=0\) (đpcm)
17 tháng 1 2020
a) \(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)
\(\Leftrightarrow2a^2+2b^2+2-2ab-2a-2b\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-1\right)^2+\left(a-1\right)^2\ge0\left(1\right)\)
Ta thấy \(\hept{\begin{cases}\left(a-b\right)^2\ge0;\forall a,b\\\left(a-1\right)^2\ge0;\forall a,b\\\left(b-1\right)^2\ge0;\forall a,b\end{cases}}\)\(\Rightarrow\left(a-b\right)^2+\left(b-1\right)^2+\left(a-1\right)^2\ge0;\forall a,b\)
\(\Rightarrow\left(1\right)\)luôn đúng
Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(a-1\right)^2=0\\\left(b-1\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=b\\a=1\\b=1\end{cases}\Leftrightarrow}a=b=1\)
Vậy... ( bạn ko cần phải ghi dấu bằng xảy ra cũng đúng nhé )
b) Xét hieuj \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=0\)( vì a+b+c=0 )
\(\Rightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)
13 tháng 11 2017
câu 2
a^4 + b^4 + c^4 + d^4 = 4abcd
<=> \(a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4+2a^2b^2-4abcd+2b^2d^2=0\)
<=> \(\left(a^2-b^2\right)^2+\left(c^2-d^2\right)^2+2\left(ab-cd\right)^2=0\)
<=> \(\left\{{}\begin{matrix}a^2=b^2\\c^2=d^2\\ab=cd\end{matrix}\right.\Leftrightarrow a=b=c=d\)
CQ
2
21 tháng 11 2017
\(\text{Ta có : }a+b+c=0\\ \Rightarrow c=-\left(a+b\right)\text{ }\text{ }\text{ }\left(\text{*}\right)\\ \Rightarrow c^3=-\left(a+b\right)^3\\ \Rightarrow a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3\\ =\left(a^3+b^3\right)-\left(a^3+3a^2b+3ab^2+b^3\right)\\ =-\left(3a^2b+3ab^2\right)\\ =-3ab\left(a+b\right)\text{ }\text{ }\text{ }\text{ }\left(1\right)\\ Thay\text{ }\left(\text{*}\right)\text{ }vào\text{ }\left(1\right),ta\text{ }được:\\ \left(1\right)=\left(-3ab\right)\cdot\left(-c\right)=3abc\left(đpcm\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\)
Vậy \(a^3+b^3+c^3=3abc\)
21 tháng 11 2017
\(a+b+c=0\Rightarrow a=-\left(b+c\right)\)
\(\Rightarrow a^3+b^3+c^3=[-\left(b+c\right)]^3+b^3+c^3=-3b^2c-3bc^2=-3bc\left(b+c\right)=3abc\)
12 tháng 8 2017
Câu a : Không hiểu
Câu b :
\(2x^2-x-1=0\)
\(\Leftrightarrow2x^2-2x+x-1=0\)
\(\Leftrightarrow2x\left(x-1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\Rightarrow x=1\\2x+1=0\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
12 tháng 8 2017
a,\(\left(x+5\right)^2-\left(x+5\right)\left(x-5\right)=20\)
\(\Leftrightarrow\left(x+5\right)\left(x+5-x+5\right)=20\)
\(\Leftrightarrow10x+50=20\)\(\Leftrightarrow x=-3\)
b,\(2x^2-x-1=2x^2-2x+x-1\)
\(=2x\left(x-1\right)+\left(x-1\right)\)\(=\left(x-1\right)\left(2x+1\right)\)\(=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{2}\end{matrix}\right.\)
\(a+b+c=0\Rightarrow c=-\left(a+b\right)\)
\(\Rightarrow a^3+b^3+c^3=a^3+b^3+[-\left(a+b\right)]^3=\)\(a^3+b^3-a^3-3a^2b-3ab^2-b^3\)
\(=3ab[-\left(a+b\right)]=3abc\left(đpcm\right)\)
Q
27 tháng 4 2017
Thay \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) và \(a+b=-c\), ta được:
\(a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3\\ =-c^3-3ab\cdot\left(-c\right)+c^3=3abc\)
Vậy \(a^3+b^3+c^3=3abc\)
9 tháng 8 2018
Câu a : Ta có : \(x^3+x^2z+y^2z-xyz+y^3=0\)
\(\Leftrightarrow\left(x^3+y^3\right)+\left(x^2z+y^2z-xyz\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow\left(x^2-xy+y^2\right)\left(x+y+z\right)=0\)
\(\Leftrightarrow x+y+z=0\) ( đpcm )
Câu b : \(VT=\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP\)
Câu c : Ta có : \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a+b+c=0\) ( đúng )
Ta có: a + b + c = 0
⇒ a + b = -c ⇒ (a + b)3 = (-c)3
⇒ a3 + b3 + 3ab(a + b) = -c3 ⇒ a3 + b3 + 3ab(-c) + c3 = 0
⇒ a3 + b3 + c3 = 3abc
a+b+c=0
\(\Rightarrow\)(a+b+c)\(^3\)=0
\(\Rightarrow\)a\(^3\)+b\(^3\)+c\(^3\)+3a\(^2\)b+3ab\(^2\)+3b\(^2\)c+3bc\(^2\)+3a\(^2\)c+3ac\(^2\)+6abc=0
\(\Rightarrow\)a\(^3\)+b\(^3\)+c\(^3\)+(3a\(^2\)b+3ab\(^2\)+3abc)+(3b\(^2\)c+3bc\(^2\)+3abc)+(3a\(^2\)c+3ac\(^2\)+3abc)-3abc=0
=>a\(^3\)+b\(^3\)+c\(^3\)+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc
Do a+b+c=0
=>a\(^3\)+b\(^3\)+c\(^3\)=3abc(ĐPCM)