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4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm
\(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}=\frac{2}{a+2b+2c}\)
\(\Leftrightarrow\frac{2b+a}{ab}=\frac{2c-\left(a+2b+2c\right)}{c\left(a+2b+2c\right)}\)
\(\Leftrightarrow\frac{a+2b}{ab}=\frac{-\left(2b+a\right)}{ac+2ab+2c^2}\)
\(\Leftrightarrow\left(a+2b\right)\left(ac+2bc+2c^2\right)+\left(2b+a\right)ab=0\)
\(\Leftrightarrow\left(a+2b\right)\left(ac+2bc+2c^2+ab\right)=0\)
\(\Leftrightarrow\left(a+2b\right)\left[a\left(b+c\right)+2c\left(b+c\right)\right]=0\)
\(\Rightarrow\left(a+2b\right)\left(b+c\right)\left(2c+a\right)=0\) (đpcm)
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=-\dfrac{1}{c^3}\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}=0\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=0\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{ab}.\left(-\dfrac{1}{c}\right)=0\)
\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}-\dfrac{3}{abc}=0\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)
Ta có: Điều cần chứng minh là \(A=3abc\) hay \(\dfrac{A}{3abc}=1\)
Thật vậy:
\(\dfrac{A}{3abc}=\left(\dfrac{b^2c^2}{a}+\dfrac{c^2a^2}{b}+\dfrac{a^2b^2}{c}\right).\dfrac{1}{3abc}\)
\(\dfrac{A}{3abc}=\dfrac{b^2c^2}{3a^2bc}+\dfrac{c^2a^2}{3ab^2c}+\dfrac{a^2b^2}{3abc^2}\)
\(\dfrac{A}{3abc}=\dfrac{bc}{3a^2}+\dfrac{ac}{3b^2}+\dfrac{ab}{3c^2}\)
\(\dfrac{A}{3abc}=\dfrac{abc}{3a^3}+\dfrac{abc}{3b^3}+\dfrac{abc}{3c^3}\)
\(\dfrac{A}{3abc}=\dfrac{abc}{3}\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{abc}{3}.\dfrac{3}{abc}=1\)
\(\dfrac{A}{3abc}=1\Leftrightarrow A=3abc\left(đpcm\right)\)
Lớp 8 nên chắc biết Bunhiacopxki chứ. Nếu ko biết thì google.
Dùng Bunhiacopxki để chứng minh cái này: \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2+\left(\frac{c}{\sqrt{z}}\right)^2\right]\)
\(\ge\left(\sqrt{x}.\frac{a}{\sqrt{x}}+\sqrt{y}.\frac{b}{\sqrt{y}}+\sqrt{z}.\frac{c}{\sqrt{z}}\right)^2=\left(a+b+c\right)^2\)
hay\(\left(x+y+z\right)\left(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\right)\ge\left(a+b+c\right)^2\)
\(\Rightarrow\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)
Áp dụng BĐT trên ta có:
\(VT=\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}\)
\(=\left(a^2+b^2+c^2\right).\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\)
Áp dụng BĐT Bunhiacopxki, ta có: \(\left(1.a+1.b+1.c\right)^2\le\left(1^2+1^2+1^2\right)\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}\ge\frac{1}{3}\)
Vậy BĐT được chứng minh
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
a. \(a^3+a^2c-abc+b^2c+b^3\)
<=> \(\left(a^3+b^3\right)+c\left(a^2-ab+b^2\right)\)
<=> (\(\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)\)
<=> \(\left(a+b+c\right)\left(a^2-ab+b^2\right)\)
vì a+b+c =0 => đpcm
b. 2(a+1)(b+1)=(a+b)(a+b+2)
<=> \(2\left(ab+a+b+1\right)=\)\(a^2+ab+2a+ab+b^2+2b\)
<=> \(2ab+2a+2b+2=a^2ab+2a+ab+b^2+2b\)
<=> \(a^2+b^2=2\)=> đpcm