Vòng 3
Bài 1:
\(\left\{{}\begin{matrix}a\left(a+b+c\right)=-\dfrac{1}{24}\\b\left(a+b+c\right)=\dfrac{1}{16}\\c\left(a+b+c\right)=-\dfrac{1}{72}\end{matrix}\right.\)
Cộng theo vế 3 đẳng thức trên ta có:
\(a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=-\dfrac{1}{24}+\dfrac{1}{16}+\left(-\dfrac{1}{72}\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2=\dfrac{1}{144}\Leftrightarrow a+b+c=\pm\dfrac{1}{12}\)
Xét \(a+b+c=\dfrac{1}{12}\)\(\Rightarrow\left\{{}\begin{matrix}a\cdot\dfrac{1}{12}=-\dfrac{1}{24}\\b\cdot\dfrac{1}{12}=\dfrac{1}{16}\\c\cdot\dfrac{1}{12}=-\dfrac{1}{72}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=-\dfrac{1}{2}\\b=\dfrac{3}{4}\\c=-\dfrac{1}{6}\end{matrix}\right.\)
Xét \(a+b+c=-\dfrac{1}{12}\)\(\Rightarrow\left\{{}\begin{matrix}a\cdot\dfrac{-1}{12}=-\dfrac{1}{24}\\b\cdot\dfrac{-1}{12}=\dfrac{1}{16}\\c\cdot\dfrac{-1}{12}=-\dfrac{1}{72}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{4}\\c=\dfrac{1}{6}\end{matrix}\right.\)
Bài 2:
a)Từ \(a^2+b^2+c^2=\dfrac{b^2-c^2}{a^2+3}+\dfrac{c^2-a^2}{b^2+4}+\dfrac{a^2-b^2}{c^2+5}\)
\(\Rightarrow a^2-\dfrac{a^2-b^2}{c^2+5}+b^2-\dfrac{b^2-c^2}{a^2+3}+c^2-\dfrac{c^2-a^2}{b^2+4}=0\)
\(\Rightarrow\dfrac{a^2c^2+4a^2+b^2}{c^2+5}+\dfrac{a^2b^2+2b^2+c^2}{a^2+3}+\dfrac{b^2c^2+3c^2+a^2}{b^2+4}=0\)
Dễ thấy: \(\dfrac{a^2c^2+4a^2+b^2}{c^2+5}+\dfrac{a^2b^2+2b^2+c^2}{a^2+3}+\dfrac{b^2c^2+3c^2+a^2}{b^2+4}\ge0\forall a,b,c\)
Dấu "=" xảy ra khi \(a=b=c=0\)
Tức là \(2012ab+2013c=0\)
b)Áp dụng t/c dãy tỉ số "=" nhau ta có:
\(\dfrac{x-y}{z}=\dfrac{3y}{x-z}=\dfrac{x}{y}=\dfrac{x-y+3y+x}{z+x-z+y}=\dfrac{2\left(x+y\right)}{x+y}=2\)
\(\Rightarrow\dfrac{x}{y}=2\Rightarrow x=2y\). Và \(\dfrac{x-y}{z}=2\Rightarrow x-y=2z\)
\(\Rightarrow2y-y=2z\Rightarrow y=2z\)
Vậy \(y=2z;x=2y\)
Bài 3:
a)\(A\left(x\right)=x^2+10x+36=x^2+10x+25+11\)
\(=x^2+5x+5x+25+11\)
\(=x\left(x+5\right)+5\left(x+5\right)+11=\left(x+5\right)^2+11\ge11\forall x\)
Dấu "=" xảy ra khi \(\left(x+5\right)^2=0\Rightarrow x=-5\)
b)Xét c=0 suy ra ta có pt
\(\left(a+b\right)\left(a-b\right)=11=11\cdot1=1\cdot11=-1\cdot\left(-11\right)=-11\cdot\left(-1\right)\)
Suy ra \(\left\{{}\begin{matrix}a+b=11\\a-b=1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a+b=1\\a-b=11\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a+b=-11\\a-b=-1\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a+b=-1\\a-b=-11\end{matrix}\right.\)
Giải các hệ trên ta thu được \(\left\{{}\begin{matrix}a=6\\b=5\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a=6\\b=-5\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a=-6\\b=-5\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a=-6\\b=5\end{matrix}\right.\)
*)Xét \(c< 0\) thì 8c không là số nguyên=>8c+10 ko là số nguyên nên VT cũng ko tồn tại a,b nguyên (làm gì có tích 2 số nguyên nào mà ko dc số nguyên)
*)Xét \(c>0\) ta có: \(8^c+10\equiv2\left(mod4\right)\)
Mà \(a^2\equiv\)0 hoặc 1 (mod 4) và \(b^2\equiv0\) hoặc 1 (mod 4)
Nên ko thể \(a^2-b^2\equiv2\left(mod4\right)\) như VP
Vậy c>0 thì vô nghiệm
Bài 4:
Đặt \(AB=AC=a;CD=2b\)\(\Rightarrow MD=b\) (M trung điểm CD)
\(\Delta MED;\Delta ACD\) có:
\(\widehat{ACD}=\widehat{MED}=\left(90^o\right)\)
\(\widehat{ADC}\) chung
\(\Rightarrow\Delta MED=\Delta ACD\left(g-g\right)\)
\(\Rightarrow\dfrac{ED}{CD}=\dfrac{MD}{AD}\left(1\right)\).\(\Delta ACD\left(\widehat{ACD}=90^o\right)\) có:
\(AD^2=AC^2+CD^2=a^2+\left(2b\right)^2=a^2+4b^2\) (ĐL py-ta-go)
\(\left(1\right)\Leftrightarrow\dfrac{ED}{2b}=\dfrac{b}{\sqrt{a^2+4b^2}}\Rightarrow ED=\dfrac{2b^2}{\sqrt{a^2+4b^2}}\)
Có: \(AE+ED=AD\) (E nằm giữa A và D)
Vì thế \(AE=AD-ED=\sqrt{a^2+4b^2}-\dfrac{2b^2}{\sqrt{a^2+4b^2}}=\dfrac{a^2+2b^2}{\sqrt{a^2+4b^2}}\)
\(\Delta KED\left(\widehat{KED}=90^o\right)\) theo ĐL py-ta-go có:
\(KD^2=KE^2+ED^2=AK^2-AE^2+ED^2=BK^2+AB^2-AE^2+ED^2\)
\(=BK^2+a^2-\left(\dfrac{a^2+2b^2}{\sqrt{a^2+4b^2}}\right)^2+\left(\dfrac{2b^2}{\sqrt{a^2+4b^2}}\right)^2=BK^2\)
\(\Rightarrow KD=KB\) suy ra tam giác KBD cân tại K
Bài 5:
Đặt \(\,\widehat{MBA}=\widehat{MCB}=\widehat{MAC}=\alpha\), khi đó ta có:
\( \widehat{AMB}= \pi - \widehat{MBA}-\widehat{MAB}= \pi - \widehat{MBA}-\left(\frac{\pi}{2}-\widehat{MAC}\right)\)
\(=\pi-{\alpha}-\left(\frac{\pi}{2}-{\alpha}\right)=\frac{\pi}{2}\). Tương tự cũng có:
\(\,\widehat{BMC}=\pi-\left(\cfrac{\pi}{4}-{\alpha}\right)-{\alpha}=\cfrac{3\pi}{4}\,\), vì vậy \(\,\widehat{CMA}=2 \pi - \widehat{AMB} - \widehat{BMC}=\cfrac{3\pi}{4}\,\)
Sau đó, theo định luật Sin có:
\(\Delta MAB:\dfrac{MA}{\sin\left(\alpha\right)}=\dfrac{AB}{\sin\left(\dfrac{\pi}{2}\right)}=AB\)
\(\Delta MBC:\dfrac{MB}{\sin\left(\alpha\right)}=\dfrac{BC}{\sin\left(\dfrac{3\pi}{4}\right)}=\sqrt{2}BC=2AB\)
\(\Delta MCA=\dfrac{MC}{\sin\left(\alpha\right)}=\dfrac{AC}{\sin\left(\dfrac{3\pi}{4}\right)}=\sqrt{2}AC=\sqrt{2}AB\)
Vậy \(MA:MB:MC=1:2:\sqrt{2}\)