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Ta có: \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
\(\Rightarrow a^3+b^3+c^3=-3ab.\left(-c\right)=3abc\)
Mặt khác: \(a+b+c=0\Rightarrow a^2=\left(-b-c\right)^2=\left(b+c\right)^2\)
\(\Rightarrow a^2-b^2-c^2=\left(b+c\right)^2-b^2-c^2=2bc\)
Tương tự ta có: \(b^2-a^2-c^2=2ca\)
\(c^2-a^2-b^2=2ab\)
\(\Rightarrow B=\frac{a^2}{2ab}+\frac{b^2}{2ca}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)
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) 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
Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\) \(\Rightarrow xyz=1\)
\(P=\frac{1}{\frac{1}{x^3}\left(\frac{1}{y}+\frac{1}{z}\right)}+\frac{1}{\frac{1}{y^3}\left(\frac{1}{z}+\frac{1}{x}\right)}+\frac{1}{\frac{1}{z^3}\left(\frac{1}{x}+\frac{1}{y}\right)}\)
\(P=\frac{x^3yz}{y+z}+\frac{y^3xz}{x+z}+\frac{z^3xy}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{1}{2}\left(x+y+z\right)\ge\frac{1}{2}.3\sqrt[3]{xyz}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(x;y;z\right)=\left(1;1;1\right)\)
a) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2\right)^2-\left(2ab\right)^2\)
\(=\left(a^2+b^2-2ab\right)\left(a^2+b^2+2ab\right)=\left(a-b\right)^2.\left(a+b\right)^2\)( đpcm )
b) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-b+b-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-b+b-c\right)+\left(c-a\right)^3\)
\(-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-c\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-c\right)^3+\left(c-a\right)^3-3\left(a-b\right)\left(b-c\right)\left(a-c\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
\(=\left(a-c\right)^3-\left(a-c\right)^3+3\left(a-b\right)\left(b-c\right)\left(c-a\right)-3\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\)
\(\Rightarrow\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)( đpcm )
1) Ta có: \(\left(a^2+b^2\right)^2-4a^2b^2\)
\(=a^4+2a^2b^2+b^4-4a^2b^2\)
\(=a^4-2a^2b^2+b^4\)
\(=\left(a^2-b^2\right)^2\)
\(=\left[\left(a-b\right)\left(a+b\right)\right]^2\)
\(=\left(a-b\right)^2\left(a+b\right)^2\)
2) Ta có: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(=\left(a-b+b-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2\right]+\left(c-a\right)^3\)
\(=\left(a-c\right)\left(a^2-2ab+b^2-ab+ac+b^2-bc+b^2-2bc+c^2\right)+\left(c-a\right)^3\)
\(=-\left(c-a\right)\left(a^2+3b^2+c^2-3ab+ac-3bc\right)+\left(c-a\right)\left(c^2-2ca+a^2\right)\)
\(=\left(c-a\right)\left(c^2-2ca+a^2-a^2-3b^2-c^2+3ab-ac+3bc\right)\)
\(=\left(c-a\right)\left(3ab+3bc-3b^2-3ac\right)\)
\(=3\left(c-a\right)\left(ab-b^2-ac+bc\right)\)
\(=3\left(c-a\right)\left[b\left(a-b\right)-c\left(a-b\right)\right]\)
\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
(a+b+c)3=[(a+b)+c]3=(a+b)3+c3+3(a+b)c(a+b+c)
=a3+b3+3ab(a+b)+c3+3(a+b)c(a+b+c)
=a3+b3+c3+3(a+b)[ab+c(a+b+c)]
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
==a3+b3+c3+3(a+b)[(ab+ac)+(bc+c2)]
=a3+b3+c3+3(a+b)(a+c)(b+c)
#)Giải :
\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+ca+c^2\right)\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)\)
\(=\left(a+b^3\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)\)
\(=\left(a+b+c\right)^3\)
\(\Rightarrowđpcm\)
Bài 1
a, Ta có
A = x2 + 6x + 13
⇒ A = (x2 + 6x + 9) + 4
⇒ A = (x + 3)2 + 4
Vì (x + 3)2 ≥ 0 với ∀ x ∈ R
⇒ (x + 3)2 + 4 ≥ 4 > 0 với ∀ x ∈ R
⇒ A > 0 với ∀ x ∈ R (đpcm)
b, B = 2x2 + 4y2 - 4x + 4xy + 13
⇒ B = (2x2 - 4x + 2) + (4y2 + 4xy + 1) + 8
⇒ B = 2 (x2 - 2x + 1) + (2y + 1)2 + 8
⇒ B = 2 (x - 1)2 + (2y + 1)2 + 8
Vì \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\text{ với ∀ x ∈ R}\\\left(2y+1\right)^2\ge0\text{ với ∀ y ∈ R}\end{matrix}\right.\)
⇒ 2 (x - 1)2 + (2y + 1)2 ≥ 0 với ∀ x, y ∈ R
⇒ 2 (x - 1)2 + (2y + 1)2 + 8 ≥ 8 với ∀ x, y ∈ R
⇒ B ≥ 8 với ∀ x, y ∈ R
Dấu " = " xảy ra
⇒ 2 (x - 1)2 + (2y + 1)2 = 0
Mà \(\left\{{}\begin{matrix}2\left(x-1\right)^2\ge0\text{ với ∀ x ∈ R}\\\left(2y+1\right)^2\ge0\text{ với ∀ y ∈ R}\end{matrix}\right.\)
nên : Để 2 (x - 1)2 + (2y + 1)2 = 0
⇒ \(\left\{{}\begin{matrix}2\left(x-1\right)^2=0\text{ }\\\left(2y+1\right)^2=0\text{ }\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x-1=0\\2y+1=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=0+1\\2y=0-1\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=1\\2y=-1\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=1\\y=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy giá trị nhỏ nhất của B là 8 tại \(\left\{{}\begin{matrix}x=1\\y=\dfrac{-1}{2}\end{matrix}\right.\)
Chúc bạn học tốt!!!
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 \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\))
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)\(-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ab-ac\right)\)\(-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)