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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
Với mọi a,b,c ta đều có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0.\)Dấu "=" chỉ xảy ra khi a = b = c.
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac\ge0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)(1)
a) \(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)nên \(a^2+b^2+c^2=ab+bc+ac\Leftrightarrow a=b=c\)đpcm (a)
b) \(\left(1\right)\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2ab+2ba+2ac=\left(a+b+c\right)^2\)
nên \(3\left(a^2+b^2+c^2\right)=\left(a+b+c\right)^2\Leftrightarrow a=b=c\)đpcm (b)
c) Từ \(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ac\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\ge3\left(ab+bc+ac\right)\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\)
nên \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\Leftrightarrow a=b=c\)đpcm (c).
2 bao gạo cân nặng 237 kg nếu gấp bao thứ nhất lên 3 lần gấp bao thứ 2 lên 2 lần thì được 611 hỏi mỗi bao gạo cân nặng bao nhiêu kg
đặt A= \(\left(a-b\right)^2\)+(b-c)^2+(c-a)^2=a^2-2ab+b^2+b^2-2bc+c^2+
-2ca+a^2
=2(a^2+b^2+c^2-ab-bc-ca)
=>A=2A
=>A=0
=>(a-b)^2+(b-c)^2+(c-a)^2=0
=>\(\left[{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
=>a=b=c (ĐPCM)
Hình như bạn viết đề hơi ngược mình nghĩ là :
Cho a,b,c khác 0 Chứng minh rằng : \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
Áp dụng BĐT AM - GM ta có :
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2}{b^2}\cdot\frac{b^2}{c^2}}=2.\frac{a}{c}\)
Tương tự có : \(\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge2\cdot\frac{b}{a}\), \(\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge2\cdot\frac{c}{b}\)
Khi đó : \(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\)
Hay : \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3c\left(a+b\right)\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
=> ĐPCM
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)^2+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-\left(3a^2b+3abc+3ab^2\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2-3ab\right]\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
=> ĐPCM
P/s: Có sao sót xin bỏ qua
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2\cdot c+3\left(a+b\right)c^2+c^3\)\(-a^3-b^3-c^3\)
\(=a^3+b^3+c^3+3a^2b+3ab^2+3\left(a^2+2ab+b^2\right)c\)\(+3ac^2+3bc^2-a^3-b^3-c^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=\left(3abc+3a^2c+3b^2c+3bc^2\right)\)\(+\left(3a^2b+3a^2c+3ab^2+3abc\right)\)
\(=c\left(3ab+3ac+3b^2+3bc\right)\)\(+a\left(3ab+3ac+3b^2+3bc\right)\)
\(=\left(a+c\right)\left[\left(3ab+3b^2\right)+\left(3ac+3bc\right)\right]\)
\(=\left(a+c\right)\left[3b\left(a+b\right)+3c\left(a+b\right)\right]\)
\(=3\left(a+c\right)\left(a+b\right)\left(b+c\right)\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)( do \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\))
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)\(-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ab-ac\right)\)\(-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
a) a2 + b2 + c2 = ab + bc + ac
\(\Rightarrow\) a2 + b2 + c2 - ab - bc - ac = 0
\(\Rightarrow\) 2(a2 + b2 + c2 - ab - bc - ac) = 0
\(\Rightarrow\) a2 + a2 + b2 + b2 + c2 + c2 - 2ab - 2bc - 2ac = 0
\(\Rightarrow\) (a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 - 2bc + c2) = 0
\(\Rightarrow\) (a - b)2 + (a - c)2 + (b - c)2 = 0
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0\\a=c\\b=c\end{matrix}\right.\)
\(\Rightarrow\) a = b = c
b) (a + b + c)2 = 3(a2 + b2 + c2)
a2 + b2 + c2 + 2ab + 2bc + 2ac = 3a2 + 3b2 + 3c2
\(\Rightarrow\) 2ab + 2ac + 2bc = 2a2 + 2b2 + 2c2
\(\Rightarrow\) 0 = a2 + a2 + b2 + b2 + c2 + c2 - 2ab - 2bc - 2ac
Hay: a2 + a2 + b2 + b2 + c2 + c2 - 2ab - 2bc - 2ac = 0
\(\Rightarrow\)(a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 - 2bc + c2) = 0
\(\Rightarrow\) (a - b)2 + (a - c)2 + (b - c)2 = 0
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0\\a=c\\b=c\end{matrix}\right.\)
\(\Rightarrow\) a = b = c
c) (a + b + c)2 = 3(ab + bc + ac)
a2 + b2 + c2 + 2ab + 2ac + 2bc = 3ab + 3bc + 3ac
\(\Rightarrow\) a2 + b2 + c2 = ab + ac + bc2
\(\Rightarrow\) 2(a2 + b2 + c2) = 2(ab + ac + bc)
\(\Rightarrow\) 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ac
\(\Rightarrow\) a2 + a2 + b2 + b2 + c2 + c2 - 2ab - 2bc - 2ac = 0
\(\Rightarrow\) (a2 - 2ab + b2) + (a2 - 2ac + c2) + (b2 - 2bc + c2) = 0
\(\Rightarrow\) (a - b)2 + (a - c)2 + (b - c)2 = 0
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0\\a=c\\b=c\end{matrix}\right.\)
\(\Rightarrow\) a = b = c
CHÚC BN HOK TỐT(nhớ tik mik nha
)
a)Cmr : Nếu : a2 + b2 + c2 = ab + bc + ac thì a = b =c
Bài làm
2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ac = 0
=> ( a2 - 2ab + b2) + ( a2 - 2ac + c2) + ( b2 - 2bc + c2) =0
= > ( a - b)2 + ( a - c)2 + ( b -c)2 = 0
Vậy :
* ( a - b)2 = 0
* ( a - c)2 =0
* (b -c)2 =0
Suy ra :
* a =b
* a =c
* b = c
Suy ra : a = b =c ( đpcm)
\(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\)
\(\Rightarrow a^2+b^2-c^2=-2ab\)
+ Tương tự ta cm đc : \(b^2+c^2-a^2=-2bc\)
\(c^2+a^2-b^2=-2ca\)
Do đó : \(B=-\frac{1}{2}-\frac{1}{2}-\frac{1}{2}=-\frac{3}{2}\)