Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Rightarrow\dfrac{ayz}{xyz}+\dfrac{bxz}{xyz}+\dfrac{cxy}{xyz}=0\)
\(\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Rightarrow ayz+bxz+cxy=0\)
+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\dfrac{xy}{ab}+2\dfrac{xz}{ac}+2\dfrac{yz}{bc}=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy}{abc}+\dfrac{bxz}{abc}+\dfrac{ayz}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1\)
\(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1\) \(\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2016}\)
\(\Rightarrow\dfrac{bc+ac+bc}{abc}=\dfrac{1}{2016}\)
\(\Rightarrow\dfrac{bc+ac+ab}{abc}=\dfrac{1}{a+b+c}\)
\(\Rightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)
\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc=abc\)
\(\Rightarrow ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow a=-b\) hay \(b=-c\) hay \(c=-a\)
-Vậy trong ba số a,b,c tồn tại 2 số đối nhau.
Theo bất đẳng thức tam giác
\(\Rightarrow\left\{\begin{matrix}a< b+c\\b< c+a\\c< a+b\end{matrix}\right.\Rightarrow\left\{\begin{matrix}b+c-a>0\\c+a-b>0\\a+b-c>0\end{matrix}\right.\)
Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\forall a,b>0\)
\(\Rightarrow\left\{\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{a+c-b}\ge\dfrac{2}{a}\end{matrix}\right.\)
Cộng theo từng vế
\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Rightarrow\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{a+c-b}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ( đpcm )
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow ab+bc+ca=0\)
\(a+b+c=\sqrt{2019}\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=2019\)
\(\Rightarrow a^2+b^2+c^2=2019\) ( vì \(ab+bc+ca=0\))
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow ab+bc+ca=0\\ A=a^2+b^2+c^2\\ \Leftrightarrow A=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\\ \Leftrightarrow A=\left(\sqrt{2019}\right)^2-2\cdot0=2019\)
Bạn chỉ cần để ý điều này thôi: \(\left(x-\frac{1}{x}\right)^2=x^2-2.x.\frac{1}{x}+\frac{1}{x^2}=x^2-2+\frac{1}{x^2}\)
Do đó giả thiết viết lại thành:
\(\left(a^2-2+\frac{1}{a^2}\right)+\left(b^2-2+\frac{1}{b^2}\right)+\left(c^2-2+\frac{1}{c^2}\right)=0\)
\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2+\left(b-\frac{1}{b}\right)^2+\left(c-\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-\frac{1}{a}=0\\b-\frac{1}{b}=0\\c-\frac{1}{c}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=\frac{1}{a}\\b=\frac{1}{b}\\c=\frac{1}{c}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^2=1\\b^2=1\\c^2=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a^2\right)^{1010}=1^{1010}\\\left(b^2\right)^{1010}=1^{1010}\\\left(c^2\right)^{1010}=1^{1010}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{2020}=1\\b^{2020}=1\\c^{2010}=1\end{matrix}\right.\) \(\Leftrightarrow a^{2020}+b^{2020}+c^{2020}=3\)
a: Xét hình thang ABCD có MN//AB//CD
nên AM/MN=BN/NC
=>AM/AD=BN/BC(1)
Xét ΔADC có MO//DC
nên MO/DC=AM/AB(2)
Xét ΔBDC có ON//DC
nên ON/DC=BN/BC(3)
Từ (1), (2) và (3) suy ra MO=ON(đpcm)
b:
Để \(\dfrac{1}{AB}+\dfrac{1}{CD}=\dfrac{2}{MN}\) thì \(\dfrac{MN}{AB}+\dfrac{MN}{CD}=2\)
MN=2ON=2OM
\(\dfrac{2OM}{AB}+\dfrac{2ON}{CD}=2\left(\dfrac{OM}{AB}+\dfrac{ON}{CD}\right)\)
mà OM/AB=DO/DB
và ON/CD=BO/BD
nên \(VT=2\cdot\left(\dfrac{DO}{DB}+\dfrac{BO}{DB}\right)=2\left(đpcm\right)\)
\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(=\dfrac{bxz-cxy}{ax}=\dfrac{cyx-ayz}{by}=\dfrac{azy-bxz}{cz}\)
\(=\dfrac{bxz-cxy+cyx-ayz+azy-bxz}{ax+by+cz}=0\)
\(\Rightarrow bz-cy=0\Rightarrow bz=cy\Rightarrow\dfrac{y}{b}=\dfrac{z}{c}\)
Tương tự...
\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\left(dpcm\right)\)
\(\dfrac{4a^2}{a-1}=\dfrac{a\left(a^2-1\right)+4}{a-1}=4\left(a+1\right)+\dfrac{4}{a-1}+8\ge8+8=16\)
\(\dfrac{5b^2}{b-1}=5\left(b-1\right)+\dfrac{5}{b-1}+10\ge20\)
\(\dfrac{3c^2}{c-1}=3\left(c-1\right)+\dfrac{3}{c-1}+6=12\)
\(\Rightarrow dpcm\)
\(C=\dfrac{-\left(x+1\right)+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2}{1-2x}\)
\(C=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Rightarrow C=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2\left(1-x\right)-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{-\left(1-x\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=-2.\dfrac{-1}{1-2x}\)
\(\Rightarrow C=\dfrac{2}{1-2x}\)
Theo bđt cauchy schwarz dạng engel
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+c}\ge\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\dfrac{a+b+c}{2}\)
Dấu ''='' xảy ra khi a = b = c