cho a+b+c=0. chứng minh rằng a3+b3+c3=3abc
cho a+b+c=0. chứng minh rằng a3+b3+c3=3abc
Ta cần CM BĐT a3+b3+c3=3abc luôn đúng với a+b+c=0
ta có \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\) \(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\) \(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\) \(\left(\left(a+b\right)+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)\)\(-3ab\left(a+b+c\right)\)=0
\(\Leftrightarrow\) \(\left(a+b+c\right)\left(\left(a+b+c\right)^2-3\left(a+b\right)c-3ab\right)\)=0(đúng vì a+b+c=0)
Vậy \(a^3+b^3+c^3=3abc\) với a+b+c=0
Cần gì phải vất vả thế!!
Giải:
Từ giả thiết \(a+b+c=0\) ta có:
\(\Rightarrow a+b=-c\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)\)
\(=-3ab\left(-c\right)=3abc\)
Vậy \(a^3+b^3+c^3=3abc\) (Đpcm)
a+b+c=0 => a+b=-c => (a+b)^3 = -c^3 => a^3 + b^3 +3ab(a+b) + c^3 = 0 . Vì a+b = -c => 3ab(a+b) = -3abc => a^3 +b^3 +c^3 - 3abc = 0 => a^3 +b^3 +c^3 = 3abc
chứng minh rằng a2+b2+c2\(\ge\)ab+ac+bc với mọi số a,b,c
Lại copy!!!
Giải:
Áp dụng BĐT Bunhiacopski
Xét cặp số \(\left(1,1,1\right)\) và \(\left(a,b,c\right)\) ta có:
\(\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\ge\left(1.a+1.b+1.c\right)^2\)
\(\Rightarrow3\left(a^2+b^2+c^2\right)\ge a^2+b^2+c^2+2\left(ab+bc+ac\right)\)
\(\Rightarrow a^2+b^2+c^2\ge ab+ac+bc\) (Đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Chúng ta có thể dễ dàng bất đức thức này bằng vài bước suy luận cơ bản như sau:
Điều này luôn đúng nên ta có điều phải chứng minh. Đẳng thức xảy ra khi a = b = c.
Ta có \(\left(a-b\right)^2\ge0\Leftrightarrow a^2+b^2-2ab\ge0\)
Cộng cả hai vế của bất phương trình ta được \(a^2+b^2\ge2ab\) (1)
Tương tự ta có:
Cho \(\dfrac{x}{a}\) + \(\dfrac{y}{b}\) + \(\dfrac{z}{c}\) = 1 và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
Tính A = \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)
Có:
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
Có:
\(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1^2\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(ayz+bxz+cxy=0\right)\)
Cho x > 1. Tìm Min
P = \(2x+\dfrac{5}{x-1}\)
cho a,b,c>0 thỏa mãn a+b+c=3. chứng minh rằng \(\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge\dfrac{3}{2}\)
\(VT=a-\dfrac{ab^2}{b^2+1}+b-\dfrac{bc^2}{c^2+1}+c-\dfrac{ca^2}{a^2+1}\)
\(VT=3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}b^2+1\ge2\sqrt{b^2}=2b\\c^2+1\ge2\sqrt{c^2}=2c\\a^2+1\ge2\sqrt{a^2}=2a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab^2}{b^2+1}\le\dfrac{ab^2}{2b}=\dfrac{ab}{2}\\\dfrac{bc^2}{c^2+1}\le\dfrac{bc^2}{2c}=\dfrac{bc}{2}\\\dfrac{ca^2}{a^2+1}\le\dfrac{ca^2}{2a}=\dfrac{ca}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\le\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge3-\dfrac{ab+bc+ca}{2}\) ( 1 )
Theo hệ quả của bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow3\ge ab+bc+ca\)
\(\Rightarrow\dfrac{3}{2}\ge\dfrac{ab+bc+ca}{2}\)
\(\Rightarrow\dfrac{3}{2}\le3-\dfrac{ab+bc+ca}{2}\) ( 2 )
Từ (1) và (2)
\(\Rightarrow3-\left(\dfrac{ab^2}{b^2+1}+\dfrac{bc^2}{c^2+1}+\dfrac{ca^2}{a^2+1}\right)\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}\ge\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=1\)
Cho \(a,b,c\ne0\)thỏa mãn
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)= 1
Tính Q = \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)
Có:
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\)
\(\Rightarrow\left(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right).\left(a+b+c\right)=1.\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{a+c}+\dfrac{c\left(a+b+c\right)}{a+b}=a+b+c\)
\(\Rightarrow\dfrac{a^2+a\left(b+c\right)}{b+c}+\dfrac{b^2+b\left(a+c\right)}{a+c}+\dfrac{c^2+c\left( a+b\right)}{a+b}=a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{a\left(b+c\right)}{b+c}+\dfrac{b^2}{a+c}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c^2}{a+b}+\dfrac{c\left(a+b\right)}{a+b}=a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+a+\dfrac{b^2}{a+c}+b+\dfrac{c^2}{a+b}+c=a+b+c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=a+b+c-a-b-c\)
\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0\)
giải pt a ( x-2)(2x+3)=0
\(\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=0\) và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2\)
Tính \(A=\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)
1.Thực hiện phép tính:
a) ( \(\dfrac{1}{1-x}\)- 1)( x - \(\dfrac{1-2x}{1-x}\) + 1)
b) ( \(\dfrac{1}{x}\)+ \(\dfrac{x-2}{x^2-4}\) - \(\dfrac{2+x}{x^2+2x}\))
c) ( \(\dfrac{2+x}{2-x}\) - \(\dfrac{4x^2}{x^2-4}\) - \(\dfrac{2-x}{2+x}\)): \(\dfrac{x^2-3x}{2x^2-x^3}\)
d) [ \(\dfrac{1}{x^2}\) + \(\dfrac{1}{y^2}\) + \(\dfrac{2}{x+y}\)( \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\))] : \(\dfrac{x^3+y^3}{x^2y^2}\)
Rút gọn biểu thức
a. 2x+2y/a2+2ab+b2 . ax-ay+bx-by/2x2-2y2
b. a+b-c/a2+2ab+b2-c2 . a2+2ab+b2+ac+bc/a2-b2
c.x3+1/x2+2x+1 . x2-1/2x2-2x+2
d. x8-1/x+1 . 1/ (x2+1) (x4+1)
e. x-y/xy+y2 - 3x+y/x2-xy . y-x/x+y
a2 c2... là em viết số mũ đó ạ. anh chị giúp em giải mấy bài này nha
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{1}{a+b}\)
\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{a+b-c}{\left(a+b-c\right)\left(a+b+c\right)}.\dfrac{\left(a+b\right)\left(a+b+c\right)}{\left(a-b\right)\left(a+b\right)}\)
\(=\dfrac{1}{a-b}\)
\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)
\(=\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)^2}.\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x^2-x+1\right)}\) \(=\dfrac{x-1}{2}\) \(d,\dfrac{x^8-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4\right)^2-1}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{x+1}.\dfrac{1}{\left(x^2+1\right)\left(x^4+1\right)}\) \(=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x+1}.\dfrac{1}{x^2+1}\) \(=\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\) \(=x-1\) \(e,\dfrac{x-y}{xy+y^2}-\dfrac{3x+y}{x^2-xy}.\dfrac{y-x}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x\left(x-y\right)}.\dfrac{-\left(x-y\right)}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{3x+y}{x}.\dfrac{-1}{x+y}\) \(=\dfrac{x-y}{y\left(x+y\right)}-\dfrac{-3x-y}{x\left(x+y\right)}\) \(=\dfrac{x\left(x-y\right)+y\left(3x+y\right)}{xy\left(x+y\right)}\) \(=\dfrac{x^2-xy+3xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{x^2+2xy+y^2}{xy\left(x+y\right)}\) \(=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}=\dfrac{x+y}{xy}\)