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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 \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)
Bài 1:
a)\(a^2+b^2+c^2=ab+bc+ca\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Khi \(a=b=c\)
b)\(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)
\(\Rightarrow-2a^2-2b^2-2c^2+2ab+2bc+2ca=0\)
\(\Rightarrow-\left(a^2-2ab+b^2\right)-\left(b^2-2bc+c^2\right)-\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2\le0\)
Khi \(a=b=c\)
c)\(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Khi \(a=b=c\)
Bài 2:
Từ \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow-2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Rightarrow ab+bc+ca=-1\)\(\Rightarrow\left(ab+bc+ca\right)^2=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2\left(a^2bc+b^2ca+c^2ab\right)=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=1\left(vi`....a+b+c=0\right)\)
Khi đó: \(a^2+b^2+c^2=2\Rightarrow\left(a^2+b^2+c^2\right)^2=4\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\)
\(\Rightarrow a^4+b^4+c^4+2=4\Rightarrow a^4+b^4+c^4=2\)
so u cn tk m sl fr u
a2 + b2+ c2 = ab + bc + ca
=> a2 + b2+ c2 -ab - bc - ca = 0
=> 2 ( a2 + b2 + c2 -ab -bc - ca) =0
=> ( a2 - 2ab + b2 ) + ( b2 -2bc + c2 ) + ( c2 - 2ca + a2 ) = 0
<=> ( a-b )2 + ( b -c)2 + ( c- a)2 =0
Do ( a -b)2 \(\ge\)0 ( b-c)2 + \(\ge\)0 ( c -a )2 \(\ge\)0
=> a-b =0 ; b -c = 0 ; c -a = 0
=> a=b ; b = c ; c =a
Vậy a = b = c
bn chép lại đề nhé
\(\Rightarrow2\left(a^2+b^2+c^2-ab-ac-bc\right)=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Rightarrow0=2\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow}\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c\left(đpcm\right)}\)
chúc bn hc tốt
a) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(Vt\ge0\left(\forall a,b,c\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Rightarrow a=b=c\)
Ta có : a2 + b2 + c2 = ab + bc + ca
=> 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0
= (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0
=> (a - b)2 + (b - c)2 + (c - a)2 = 0
=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\left(\text{đpcm}\right)\)
b) Ta có : 2(x2 + t2) + (y + t)(y - t) = 2x(y + t)
=> 2x2 + 2t2 + y2 - t2 = 2xy + 2t
=> 2x2 + t2 + y2 = 2xt + 2xy
=> 2x2 + t2 + y2 - 2xt - 2xy = 0
=> (x2 - 2xy + y2) + (x2 + t2 - 2xt) = 0
=> (x - y)2 + (x - t)2 = 0
=> \(\hept{\begin{cases}x-y=0\\x-t=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y\\x=t\end{cases}}\Rightarrow x=y=t\left(\text{đpcm}\right)\)
c) Ta có a + b + c = 0
=> (a + b + c)2 = 0
=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
=> a2 + b2 + c2 + 2(ab + bc + ca) = 0
=> a2 + b2 + c2 = 0
=> a = b = c = 0
Khi đó A = (0 - 1)2003 + 02004 + (0 + 1)2005
= - 1 + 0 + 1 = 0
Vậy A = 0
\(a)\) 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
Bài 2:
\(A=\left(2ac-a^2-c^2+b^2\right)\left(2ac+a^2+c^2-b^2\right)\)
\(=\left[b^2-\left(a-c\right)^2\right]\left[\left(a+c\right)^2-b^2\right]\)
\(=\left(b-a+c\right)\left(b+a-c\right)\left(a+c-b\right)\left(a+c+b\right)\)>0