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\(a^3-b^3-ac^2+bc^2=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)-c^2\left(a-b\right)=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2-c^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a^2+b^2-c^2=-ab\end{matrix}\right.\)
TH1: \(a=b\Rightarrow\) chịu thua ko tính được góc C
TH2: \(a^2+b^2-c^2=-ab\Rightarrow cosC=\frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2}\)
\(\Rightarrow C=120^0\)
\(b\left( {{b^2} - {a^2}} \right) = c\left( {{a^2} - {c^2}} \right)\left( {a,b,c \ne 0} \right)\left( * \right)\)
Ta có: \(a^2=b^2+c^2-2bc.cosA\)
\(\Rightarrow\left\{{}\begin{matrix}a^2-c^2=b^2-2bc.cosA\\b^2-a^2=2bc.cosA-c^2\end{matrix}\right.\)
Thay vòa $(*)$ ta được:
\(\begin{array}{l} b\left( {2bc.\cos A - {c^2}} \right) = c\left( {{b^2} - 2bc.\cos A} \right)\\ \Leftrightarrow bc\left( {2b\cos A - c} \right) = bc\left( {b - 2c\cos A} \right)\\ \Leftrightarrow 2bc\cos A - c = b - 2c\cos A\left( {do:a,b,c \ne 0} \right)\\ \Leftrightarrow \cos A = \dfrac{1}{2} \Rightarrow \widehat A = {60^o} \end{array}\)
Ta có
\(a< b+c\left(bđt\Delta\right)\)
\(\Rightarrow2a< a+b+c\)
\(\Rightarrow2a< 2\)
\(\Rightarrow a< 1\)
\(\Rightarrow-a>-1\)
\(\Rightarrow1-a>0\)
Tương tự với b và c
\(\Rightarrow\begin{cases}1-b>0\\1-c>0\end{cases}\)
\(\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)>0\)
\(\Rightarrow1-\left(a+b+c\right)+ab+bc+ca-abc>0\)
\(\Rightarrow1-\left(a+b+c\right)+ab+bc+ca>abc\)
\(\Rightarrow1-2+ab+bc+ca>abc\)
\(\Rightarrow-1+ab+bc+ca>abc\)
\(\Rightarrow-2+2ab+2bc+2ca>2abc\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca-2>2acb+a^2+b^2+c^2\)
Áp dụng hằng đẳng thức \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca\)
\(\Rightarrow\left(a+b+c\right)^2-2>2abc+a^2+b^2+c^2\)
\(\Rightarrow2^2-2>2abc+a^2+b^2+c^2\)
\(\Rightarrow2abc+a^2+b^2+c^2< 2\)
đpcm
a<b+c => 2a<a+b+c=2=>a<1=> b<1,c<1
=> (1-a)(1-b)(1-c)>0. Rút gọn ta được
ab+bc+ca >1+abc
Ta lại có: (a+b+)^2 =a^2+b^2+c^2 +2(ab+bc+ca)
=> 4= a^2+b^2+c^2+2(ab+bc+ca)
=> 4> a^2+b^2+c^2+2(1+abc)=> 4>a^2+b^2+c^2+2+2abc
=> a^2+b^2_c^2+2abc<2
Câu 1: Diện tích tam giác là: \(\frac{h_A.a}{2}=\frac{3.6}{2}=9\)(đvdt)
Câu 2: Diện tích tam giác là: \(\frac{1}{2}ab.\sin C=\frac{1}{2}.4.5.\sin60^o=5\sqrt{3}\)(đvdt)
Câu 2: Ta có: \(\hept{\begin{cases}c^2=a^2+b^2-2ab.\cos C\\a^2+b^2>c^2\end{cases}\Rightarrow c^2>c^2-2ab.\cos C\Leftrightarrow2ab.\cos C>0}\)
\(\Rightarrow\cos C>0\Rightarrow C< 90^o\)
Vậy C là góc nhọn
\(\hept{\begin{cases}a+b+c=4\\a^2+b^2+c^2=6\end{cases}}\)
\(b^2+c^2=6-a^2\Rightarrow\left(b+c\right)^2-2bc=6-a^2\)
\(\Rightarrow2bc=\frac{\left(b+c\right)^2-6+a^2}{2}\)
\(=\frac{\left(4-a\right)^2-6+a^2}{2}\left(Do:a+b+c=4\right)\)
\(=\frac{2a^2-8a+10}{2}=a^2-4a+5\)
\(\Rightarrow P=a^3+bc\left(b+c\right)=a^3+\left(a^2-4a+5\right)\left(4-a\right)\left(Do:a+b+c=4\right)\)
\(=a^3+4a^2-16a+20-a^3+4a^2-5a\)
\(=8a^2-21a+20\)
\(=8\left(a^2-2.\frac{21}{16}a+\frac{441}{256}\right)+\frac{199}{32}\)
\(=8\left(a-\frac{21}{16}\right)^2+\frac{119}{32}\)
.............................................................
\(b^3-a^2b=a^2c-c^3\)
\(\Leftrightarrow b^3+c^3-a^2b-a^2c=0\)
\(\Leftrightarrow\left(b+c\right)\left(b^2+c^2-bc\right)-a^2\left(b+c\right)=0\)
\(\Leftrightarrow b^2+c^2-bc-a^2=0\)
\(\Leftrightarrow b^2+c^2-a^2=bc\)
\(\Rightarrow cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{bc}{2bc}=\frac{1}{2}\)
\(\Rightarrow A=60^0\)
\(sigma\frac{a}{1+b-a}=sigma\frac{a^2}{a+ab-a^2}\ge\frac{\left(a+b+c\right)^2}{a+b+c+\frac{\left(a+b+c\right)^2}{3}-\frac{\left(a+b+c\right)^2}{3}}=1\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
\(\frac{1}{b^2+c^2}=\frac{1}{1-a^2}=1+\frac{a^2}{b^2+c^2}\le1+\frac{a^2}{2bc}\)
Tương tự cộng lại quy đồng ta có đpcm
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
\(b^3-a^2b=c^3-a^2c\)
\(\Leftrightarrow b^3-c^3-a^2b+a^2c=0\)
\(\Leftrightarrow\left(b-c\right)\left(b^2+bc+c^2\right)-a^2\left(b-c\right)=0\)
\(\Leftrightarrow\left(b-c\right)\left(b^2+bc+c^2-a^2\right)=0\)
\(\Leftrightarrow b^2+bc+c^2=a^2\)
\(\Rightarrow cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{b^2+c^2-\left(b^2+bc+c^2\right)}{2bc}=-\frac{1}{2}\)
\(\Rightarrow A=120^0\)