Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)\) Ta có :
\(a+b+c=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)^3=0^3\)
\(\Leftrightarrow\)\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(a+b+c=0\)\(\Rightarrow\)\(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Leftrightarrow\)\(a^3+b^3+c^3+3.\left(-c\right)\left(-a\right)\left(-b\right)=0\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\)\(a^3+b^3+c^3=3abc\) ( đpcm )
Vậy \(a^3+b^3+c^3=3abc\)
Chúc bạn học tốt ~
a, a+b+c=0 => a+b=-c
=>(a+b)3=(-c)3
=>a3+3a2b+3ab2+b3=-c3
=>a3+3ab(a+b)+b3=-c3
Mà a+b=-c
=>a3-3abc+b3=-c3
=>a3+b3+c3=3abc (đpcm)
b, \(P=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3}{abc}+\frac{b^3}{abc}+\frac{c^3}{abc}=\frac{a^3+b^3+c^3}{abc}\)
mà a3+b3+c3=3abc (bài a)
\(\Rightarrow P=\frac{3abc}{abc}=3\)
Vậy P=3
1 ) Ta có :
\(a+b-c=0\Leftrightarrow a+b=c\Leftrightarrow\left(a+b\right)^3=c^3\)
\(\Rightarrow a^3+b^3-c^3=a^3+b^3-\left(a+b\right)^3\)
\(\Rightarrow a^3+b^3-c^3=a^3+b^3-3a^2b-3b^2a-b^3\)
\(\Rightarrow a^3+b^3-c^3=-3a^2b-3b^2a\)
\(\Rightarrow a^3+b^3-c^3=-3ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3-c^3=-3abc\left(đpcm\right)\)
2 ) Ta có :
\(a-b+c=0\Leftrightarrow c=b-a\Leftrightarrow c^3=\left(b-a\right)^3\)
\(\Rightarrow a^3-b^3+c^3=a^3-b^3+\left(b-a\right)^3\)
\(\Rightarrow a^3-b^3+c^3=a^3-b^3+b^3-3a^2b+3b^2a-a^3\)
\(\Rightarrow a^3-b^3+c^3=-3a^2b+3b^2a\)
\(\Rightarrow a^3-b^3+c^3=-3ab\left(a-b\right)\)
\(\Rightarrow a^3-b^3+c^3=3ab\left(b-a\right)\)
\(\Rightarrow a^3-b^3+c^3=3abc\left(đpcm\right)\)
1 ) Bổ sung dấu \(\Rightarrow\) thứ 2 :
\(\Rightarrow...=a^3+b^3-a^3-3a^2b-3b^2a-b^3\)
có a + b + c = 0
\(\Rightarrow\)a + b = -c
\(\Rightarrow\)(a + b)3 = (-c)3
\(\Rightarrow\)a3 + b3 + 3ab(a + b) = -c3
\(\Rightarrow\) a3 + b3 + c3 = 3abc
b) có a + b + c = 0
nên a + b = c
(a + b)2 = c2
nên c2 - a2 - b2 = 2ab
cm tương tự ta có \(a^2-b^2-c^2=2bc\);\(b^2-a^2-c^2=2ac\)
\(P=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-a^2-c^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}\)
\(=\frac{1}{2}\left(\frac{a^3+b^3+c^3}{abc}\right)\)
\(=\frac{1}{2}\cdot3=1,5\)
a) Ta có: (a + b + c + d)(a - b - c +d )=( (a + d) + (b + c) )( (a + d) - (b + c) )
=(a + d )2 - (b +c )2 (1)
(a - b + c - d)(a + b - c - d)=(a - d)2 - (b - c)2 (2)
Từ (1) và (2) => a2 + 2ad + d2 - b2 - 2bc - c2=a2 - 2ad + d2 - b2 + 2bc - c2
4ad=4bc => ad=bc <=> \(\frac{a}{c}=\frac{b}{d}\) (đpcm)
Do \(a+b+c=0\Rightarrow a+b=-c\)
Ta có: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3\Rightarrow a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)=\left(-c\right)^3-3ab\left(-c\right)=\left(-c\right)^3+3abc\)
Do đó: \(a^3+b^3+c^3=\left(-c\right)^3+3abc+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
Kết luận: bạn ghi sai đề.
- Biết a – b = 7 tính : A = a2(a + 1) – b2(b – 1) + ab – 3ab(a – b + 1)
- Cho ba số a, b, c khác 0 thỏa nãm đẳng thức :
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)=0\)
Mà a+b+c=0\(\Rightarrow0.\left[\left(z+b\right)^2-\left(a+b\right)c+c^2\right]-3ab.0=0\Rightarrow0+0=0\)
0+0=0 đúng suy ra \(a^3+b^3+c^3=3abc\)đúng với \(a+b+c=0\)
Bạn học tốt nha
1 cái T I C K nha mình cảm ơn
\(a^3+b^3+c^3=3abc\\ \Rightarrow\left(a^3+3a^2b+3ab^2+b^3\right)-3a^2b-3ab^2-3abc+c^3=0\\ \Rightarrow\left(a+b\right)^3-3a^2b-3ab^2-3abc+c^3=0\\ \Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3a^2b-3ab^2-3abc=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-3ab-ac-bc\right)=0\\ \Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\\ \)
\(\Rightarrow\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)=0\\ \Rightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2=\left(b-c\right)^2=\left(c-a\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\) (DPCM)