Đặt \(p=\frac{a+b+c}{2}\)

=>a+b+c=2p

=>a+b-c=2p-2c

=>a+b-c=2(p-c)

Ta có: a+b+c=2p

=>a-b+c=2p-2b

=>a-b+c=2(p-b)

Ta có: a+b+c=2p

=>b+c-a=2p-2a

=>b+c-a=2(p-a)

\(S=\frac14\left(a+b-c\right)\left(a-b+c\right)\)

=>\(S=\frac14\cdot2\left(p-c\right)\cdot2\cdot\left(p-b\right)=\left(p-c\right)\left(p-b\right)\) (1)

Vì S là diện tích tam giác ABC và p là nửa chu vi của tam giác ABC

nên \(S_{ABC}=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)

=>\(\sqrt{p\left(p-a\right)\left(p-c\right)\left(p-b\right)}=\left(p-c\right)\left(p-b\right)\)

=>\(p\left(p-a\right)\left(p-c\right)\left(p-b\right)=\left(p-c\right)^2\left(p-b\right)^2\)

=>p(p-a)=(p-c)(p-b)

=>\(2p\cdot2\left(p-a\right)=2\left(p-c\right)\cdot2\cdot\left(p-b\right)\)

=>(a+b+c)(b+c-a)=(a+b-c)(a-b+c)

=>\(\left(b+c\right)^2-a^2=a^2-\left(b-c\right)^2\)

=>\(\left(b+c\right)^2+\left(b-c\right)^2=2a^2\)

=>\(2a^2=b^2+2bc+c^2+b^2-2bc+c^2=2b^2+2c^2\)

=>\(a^2=b^2+c^2\)

=>ΔABC vuông tại A

=>\(\hat{BAC}=90^0\)

Giả thiết tương đương: 

\(a^4+b^4+c^4+2b^2c^2=2a^2\left(b^2+c^2\right)+2b^2c^2\)

\(\Leftrightarrow a^4+\left(b^2+c^2\right)^2=2a^2\left(b^2+c^2\right)+2b^2c^2\)

\(\Leftrightarrow\left(b^2+c^2-a^2\right)^2=2b^2c^2\)

\(\Leftrightarrow b^2+c^2-a^2=\pm\sqrt{2}bc\)

\(cosA=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{\pm\sqrt{2}bc}{2bc}=\pm\dfrac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}A=45^0\\A=135^0\end{matrix}\right.\)

kết quả ra 60 hoặc 120

Ta có  P = A B → + A C → . B C → = A B → + A C → . B A → + A C → .

= A C → + A B → . A C → − A B → = A C → 2 − A B → 2 = A C 2 − A B 2 = b 2 − c 2 .  

Chọn A.

5 căn 2

Áp dụng định lí cosin trong tam giác ABC ta có:

\({a^2} = {b^2} + {c^2} - 2bc.\cos A\)\( \Rightarrow \cos A = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)

Mà \(\sin A = \sqrt {1 - {{\cos }^2}A} \).

\( \Rightarrow \sin A = \sqrt {1 - {{\left( {\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right)}^2}}  = \sqrt {\frac{{{{(2bc)}^2} - {{({b^2} + {c^2} - {a^2})}^2}}}{{{{(2bc)}^2}}}} \)

\( \Leftrightarrow \sin A = \frac{1}{{2bc}}\sqrt {{{(2bc)}^2} - {{({b^2} + {c^2} - {a^2})}^2}} \)

Đặt \(M = \sqrt {{{(2bc)}^2} - {{({b^2} + {c^2} - {a^2})}^2}} \)

\(\begin{array}{l} \Leftrightarrow M = \sqrt {(2bc + {b^2} + {c^2} - {a^2})(2bc - {b^2} - {c^2} + {a^2})} \\ \Leftrightarrow M = \sqrt {\left[ {{{(b + c)}^2} - {a^2}} \right].\left[ {{a^2} - {{(b - c)}^2}} \right]} \\ \Leftrightarrow M = \sqrt {(b + c - a)(b + c + a)(a - b + c)(a + b - c)} \end{array}\)

Ta có: \(a + b + c = 2p\)\( \Rightarrow \left\{ \begin{array}{l}b + c - a = 2p - 2a = 2(p - a)\\a - b + c = 2p - 2b = 2(p - b)\\a + b - c = 2p - 2c = 2(p - c)\end{array} \right.\)

\(\begin{array}{l} \Leftrightarrow M = \sqrt {2(p - a).2p.2(p - b).2(p - c)} \\ \Leftrightarrow M = 4\sqrt {(p - a).p.(p - b).(p - c)} \\ \Rightarrow \sin A = \frac{1}{{2bc}}.4\sqrt {p(p - a)(p - b)(p - c)} \\ \Leftrightarrow \sin A = \frac{2}{{bc}}.\sqrt {p(p - a)(p - b)(p - c)} \end{array}\)

b) Ta có: \(S = \frac{1}{2}bc\sin A\)

Mà \(\sin A = \frac{2}{{bc}}\sqrt {p(p - a)(p - b)(p - c)} \)

\(\begin{array}{l} \Rightarrow S = \frac{1}{2}bc.\left( {\frac{2}{{bc}}\sqrt {p(p - a)(p - b)(p - c)} } \right)\\ \Leftrightarrow S = \sqrt {p(p - a)(p - b)(p - c)} .\end{array}\)

a: A(5;3); B(-2;-1); C(-1;5)

\(\overrightarrow{AB}=\left(-2-5;-1-3\right)=\left(-7;-4\right)\)

\(\overrightarrow{BC}=\left(-1+2;5+1\right)=\left(1;6\right)\)

\(\overrightarrow{AC}=\left(-1-5;5-3\right)=\left(-6;2\right)\)

\(\overrightarrow{AB}+2\cdot\overrightarrow{BC}=\left(-7+2\cdot1;-4+2\cdot6\right)\)

=>\(\overrightarrow{AB}+2\cdot\overrightarrow{BC}=\left(-5;8\right)\)

=>\(\left(\overrightarrow{AB}+2\cdot\overrightarrow{BC}\right)\cdot\overrightarrow{AC}=\left(-5\right)\cdot\left(-6\right)+2\cdot8=30+16=46\)

\(\overrightarrow{AB}-2\cdot\overrightarrow{BC}=\left(-7-2\cdot1;-4-2\cdot6\right)\)

=>\(\overrightarrow{AB}-2\cdot\overrightarrow{BC}=\left(-9;-16\right)\)

=>\(\left(\overrightarrow{AB}-2\cdot\overrightarrow{BC}\right)\cdot\overrightarrow{BC}=\left(-9\right)\cdot1+\left(-16\right)\cdot6=-9-96=-105\)

b: Tọa độ trọng tâm của ΔABC là:

\(\begin{cases}x_{G}=\frac13\cdot\left(x_{A}+x_{B}+x_{C}\right)=\frac13\left(5-2-1\right)=\frac23\\ y_{G}=\frac13\cdot\left(y_{A}+y_{B}+y_{C}\right)=\frac13\cdot\left(3-1+5\right)=\frac13\cdot7=\frac73\end{cases}\)

c: Gọi H(x;y) là trực tâm của ΔABC

=>AH⊥BC và BH⊥AC

=>\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0;\overrightarrow{BH}\cdot\overrightarrow{AC}=0\)

A(5;3); H(x;y); B(-2;-1); C(-1;5)

\(\overrightarrow{AC}=\left(-6;2\right);\overrightarrow{BH}=\left(x+2;y+1\right)\)

\(\overrightarrow{BH}\cdot\overrightarrow{AC}=0\)

=>-6(x+2)+2(y+1)=0

=>-3(x+2)+y+1=0

=>-3x-6+y+1=0

=>y-3x-5=0

=>y=3x+5

\(\overrightarrow{BC}=\left(1;6\right);\overrightarrow{AH}=\left(x-5;y-3\right)\)

\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0\)

=>1(x-5)+6(y-3)=0

=>x-5+6y-18=0

=>x+6y-23=0

=>x+6y=23

=>x+6(3x+5)=23

=>x+18x+30=23

=>19x=-7

=>x=-7/19

=>\(y=3x+5=3\cdot\frac{-7}{19}+5=-\frac{21}{19}+5=\frac{-21+95}{19}=\frac{74}{19}\)

=>H(-7/19;74/19)

d: AH⊥BC

nên AH sẽ đi qua A và AH nhận \(\overrightarrow{BC}=\left(1;6\right)\) làm vecto pháp tuyến

Phương trình đường cao AH là:

1(x-5)+6(y-3)=0

=>x-5+6y-18=0

=>x+6y-23=0

\(\overrightarrow{BC}=\left(1;6\right)\)

=>vecto pháp tuyến là (-6;1)

Phương trình BC là:

-6(x+2)+1(y+1)=0

=>-6x-12+y+1=0

=>-6x+y-11=0

=>y=6x+11

x+6y-23=0

=>x+6(6x+11)-23=0

=>x+36x+66-23=0

=>37x=-43

=>\(x=-\frac{43}{37}\)

=>\(y=6x+11=6\cdot\frac{-43}{37}+11=\frac{149}{37}\)

=>Tọa độ chân đường cao kẻ từ A xuống BC là K(-43/37;149/37)

e: \(AB=\sqrt{\left(-7\right)^2+\left(-4\right)^2}=\sqrt{49+16}=\sqrt{65}\)

\(BC=\sqrt{1^2+6^2}=\sqrt{37}\)

\(AC=\sqrt{\left(-6\right)^2+2^2}=\sqrt{40}=2\sqrt{10}\)

Xét ΔABC có \(cosBAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

\(=\frac{65+40-37}{2\cdot\sqrt{65}\cdot2\sqrt{10}}=\frac{68}{4\sqrt{650}}=\frac{17}{\sqrt{650}}\)

=>\(\sin BAC=\sqrt{1-\left(\frac{17}{\sqrt{650}}\right)^2}=\sqrt{1-\frac{289}{650}}=\sqrt{\frac{361}{650}}=\frac{19}{\sqrt{650}}\)

Diện tích tam giác BAC là:

\(S_{ABC}=\frac12\cdot AB\cdot AC\cdot\sin BAC\)

\(=\frac12\cdot\sqrt{65}\cdot2\sqrt{10}\cdot\frac{19}{\sqrt{650}}=19\)

a: A(5;3); B(-2;-1); C(-1;5)

\(\overrightarrow{AB}=\left(-2-5;-1-3\right)=\left(-7;-4\right)\)

\(\overrightarrow{BC}=\left(-1+2;5+1\right)=\left(1;6\right)\)

\(\overrightarrow{AC}=\left(-1-5;5-3\right)=\left(-6;2\right)\)

\(\overrightarrow{AB}+2\cdot\overrightarrow{BC}=\left(-7+2\cdot1;-4+2\cdot6\right)\)

=>\(\overrightarrow{AB}+2\cdot\overrightarrow{BC}=\left(-5;8\right)\)

=>\(\left(\overrightarrow{AB}+2\cdot\overrightarrow{BC}\right)\cdot\overrightarrow{AC}=\left(-5\right)\cdot\left(-6\right)+2\cdot8=30+16=46\)

\(\overrightarrow{AB}-2\cdot\overrightarrow{BC}=\left(-7-2\cdot1;-4-2\cdot6\right)\)

=>\(\overrightarrow{AB}-2\cdot\overrightarrow{BC}=\left(-9;-16\right)\)

=>\(\left(\overrightarrow{AB}-2\cdot\overrightarrow{BC}\right)\cdot\overrightarrow{BC}=\left(-9\right)\cdot1+\left(-16\right)\cdot6=-9-96=-105\)

b: Tọa độ trọng tâm của ΔABC là:

\(\begin{cases}x_{G}=\frac13\cdot\left(x_{A}+x_{B}+x_{C}\right)=\frac13\left(5-2-1\right)=\frac23\\ y_{G}=\frac13\cdot\left(y_{A}+y_{B}+y_{C}\right)=\frac13\cdot\left(3-1+5\right)=\frac13\cdot7=\frac73\end{cases}\)

c: Gọi H(x;y) là trực tâm của ΔABC

=>AH⊥BC và BH⊥AC

=>\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0;\overrightarrow{BH}\cdot\overrightarrow{AC}=0\)

A(5;3); H(x;y); B(-2;-1); C(-1;5)

\(\overrightarrow{AC}=\left(-6;2\right);\overrightarrow{BH}=\left(x+2;y+1\right)\)

\(\overrightarrow{BH}\cdot\overrightarrow{AC}=0\)

=>-6(x+2)+2(y+1)=0

=>-3(x+2)+y+1=0

=>-3x-6+y+1=0

=>y-3x-5=0

=>y=3x+5

\(\overrightarrow{BC}=\left(1;6\right);\overrightarrow{AH}=\left(x-5;y-3\right)\)

\(\overrightarrow{AH}\cdot\overrightarrow{BC}=0\)

=>1(x-5)+6(y-3)=0

=>x-5+6y-18=0

=>x+6y-23=0

=>x+6y=23

=>x+6(3x+5)=23

=>x+18x+30=23

=>19x=-7

=>x=-7/19

=>\(y=3x+5=3\cdot\frac{-7}{19}+5=-\frac{21}{19}+5=\frac{-21+95}{19}=\frac{74}{19}\)

=>H(-7/19;74/19)

d: AH⊥BC

nên AH sẽ đi qua A và AH nhận \(\overrightarrow{BC}=\left(1;6\right)\) làm vecto pháp tuyến

Phương trình đường cao AH là:

1(x-5)+6(y-3)=0

=>x-5+6y-18=0

=>x+6y-23=0

\(\overrightarrow{BC}=\left(1;6\right)\)

=>vecto pháp tuyến là (-6;1)

Phương trình BC là:

-6(x+2)+1(y+1)=0

=>-6x-12+y+1=0

=>-6x+y-11=0

=>y=6x+11

x+6y-23=0

=>x+6(6x+11)-23=0

=>x+36x+66-23=0

=>37x=-43

=>\(x=-\frac{43}{37}\)

=>\(y=6x+11=6\cdot\frac{-43}{37}+11=\frac{149}{37}\)

=>Tọa độ chân đường cao kẻ từ A xuống BC là K(-43/37;149/37)

e: \(AB=\sqrt{\left(-7\right)^2+\left(-4\right)^2}=\sqrt{49+16}=\sqrt{65}\)

\(BC=\sqrt{1^2+6^2}=\sqrt{37}\)

\(AC=\sqrt{\left(-6\right)^2+2^2}=\sqrt{40}=2\sqrt{10}\)

Xét ΔABC có \(cosBAC=\frac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

\(=\frac{65+40-37}{2\cdot\sqrt{65}\cdot2\sqrt{10}}=\frac{68}{4\sqrt{650}}=\frac{17}{\sqrt{650}}\)

=>\(\sin BAC=\sqrt{1-\left(\frac{17}{\sqrt{650}}\right)^2}=\sqrt{1-\frac{289}{650}}=\sqrt{\frac{361}{650}}=\frac{19}{\sqrt{650}}\)

Diện tích tam giác BAC là:

\(S_{ABC}=\frac12\cdot AB\cdot AC\cdot\sin BAC\)

\(=\frac12\cdot\sqrt{65}\cdot2\sqrt{10}\cdot\frac{19}{\sqrt{650}}=19\)