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phân tích a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0
=>a=b=c(vì a+b+c khác 0)
thay a=b=c vào P
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 \(a^2-b^2-c^2=2bc\);\(b^2-a^2-c^2=2ac\)
\(P=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-a^2-c^2}+\frac{c^2}{c^2-a^2-b^2}\)
\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}\)
\(=\frac{1}{2}\left(\frac{a^3+b^3+c^3}{abc}\right)\)
\(=\frac{1}{2}\cdot3=1,5\)
Bài 1:
ta có: a + b + c = 0 => a + b = - c => (a+b)2 = (-c)2 => a2 + 2ab + b2 = c2 => a2 + b2 - c2 = -2ab
chứng minh tương tự, ta có: b2 + c2 -a2 = -2bc; c2 + a2 - b2 = -2ac
\(A=\frac{ab}{a^2+b^2-c^2}+\frac{bc}{b^2+c^2-a^2}+\frac{ca}{c^2+a^2-b^2}\)
\(A=\frac{ab}{-2ab}+\frac{bc}{-2bc}+\frac{ca}{-2ac}=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)
=> A là số hữu tỉ
...
\(1)\)
\(a)\)\(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=100+99+98+97+...+2+1\)
\(A=\frac{100\left(100+1\right)}{2}\)
\(A=5050\)
\(b)\)\(B=3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(B=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)
\(B=\left(2^8+1\right).....\left(2^{64}+1\right)+1\)
\(............\)
\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)
\(B=2^{128}-1+1\)
\(B=2^{128}\)
Chúc bạn học tốt ~
\(1)\)
\(c)\)\(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(C=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2-2\left(a+b\right)^2\)
\(C=2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2\)
\(C=2c^2\)
\(2)\)
\(a)\)\(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(VP=a^3+3a^2b+3ab^2+b^3-3ab\left(a+b\right)\)
\(VP=a^3+3ab\left(a+b\right)+b^3-3ab\left(a+b\right)\)
\(VP=a^3+b^3=VT\) ( đpcm )
\(b)\)\(VT=a^3+b^3+c^3-3abc\)
\(VT=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(VT=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(VT=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(VT=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\) ( đpcm )
Từ đó suy ra :
\(i)\)\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)\(\Rightarrow\)\(a+b+c=0\)
Hoặc \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)
Chúc bạn học tốt ~
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\)\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right]=0\)
Do \(a+b+c\ne0\) nên \(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-bc+c^2\right)+\left(c^2-ca+a^2\right)=0\)
\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c}\)
\(\Rightarrow\)\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)
...
Nhận xét:\(\left(a+b\right)^3=a^3+b^3+3a^2b+3ab^2\)
=> \(a^3+b^3=\left(a+b\right)^3-3a^2b-3ab^2\)
ta có \(a^3+b^3+c^3-3abc\)
Thay vào biểu thức trên ta có:
\(\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
= \(\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
=\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
= \(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
=\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Vay \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
Do \(a^3+b^3+c^3=3abc\)và theo đầu bài \(a+b+c\ne0\)nen \(a^2+b^2+c^2-ac-bc-ab=0\)
=> \(a=b=c\)
Vay N = \(\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)
Câu 9.
a) Ta có: \(\left(a-1\right)^2\ge0\)(điều hiển nhiên)
\(\Leftrightarrow a^2-2a+1\ge0\)
\(\Leftrightarrow a^2+2a+1\ge4a\)
\(\Leftrightarrow\left(a+1\right)^2\ge4a\left(đpcm\right)\)
b) Áp dụng BĐT Cauchy cho 2 số không âm:
\(a+1\ge2\sqrt{a}\)
\(b+1\ge2\sqrt{b}\)
\(c+1\ge2\sqrt{c}\)
\(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\sqrt{abc}=8\)(Vì abc = 1)
Câu 10.
a) Ta có: \(-\left(a-b\right)^2\le0\)(điều hiển nhiên)
\(\Leftrightarrow-a^2+2ab-b^2\le0\)
\(\Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\)
\(\Leftrightarrow\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
b) \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
Có: \(2ab\le a^2+b^2;2bc\le b^2+c^2;2ac\le a^2+c^2\)(BĐT Cauchy)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac\le3\left(a^2+b^2+c^2\right)\)
Vậy \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
\(a)\) 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