Đường tròn \(\left(C\right)\) có\(\left\{{}\begin{matrix}I\left(1;2\right)\\R=\sqrt{10}\end{matrix}\right.\)

Trong tam giác đều,trọng tâm và trực tâm, tâm đường tròn ngoại tiếp trùng nhau

Có phương trình đường thẳng AI: \(\left\{{}\begin{matrix}quaA\left(0;-1\right)\\vtcp\overrightarrow{AI}\left(1;3\right)\Rightarrow vtpt\overrightarrow{n}\left(3;-1\right)\end{matrix}\right.\)

\(\Rightarrow\left(AI\right):3x-y-1=0\)

Gọi \(H=AI\cap BC\) \(\Rightarrow\) H là trung điểm của BC

Có \(\overrightarrow{AI}=\dfrac{2}{3}\overrightarrow{AH}\) \(\Rightarrow\left\{{}\begin{matrix}1=\dfrac{2}{3}\left(x_H-0\right)\\3=\dfrac{2}{3}\left(y_H+1\right)\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x_H=\dfrac{3}{2}\\y_H=\dfrac{7}{2}\end{matrix}\right.\)\(\Rightarrow H\left(\dfrac{3}{2};\dfrac{7}{2}\right)\)

Có pt đường thẳng BC \(\left\{{}\begin{matrix}quaH\left(\dfrac{3}{2};\dfrac{7}{2}\right)\\\perp AI\Rightarrow vtpt\overrightarrow{AI}\left(1;3\right)\end{matrix}\right.\) \(\Rightarrow\left(BC\right):x+3y-12=0\)

Tọa độ B, C là nghiệm của hệ: \(\left\{{}\begin{matrix}x^2+y^2-2x-4y-5=0\\x+3y-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-2x-4y-5=0\\x=12-3y\end{matrix}\right.\)

\(\Rightarrow\left(12-3y\right)^2+y^2-2\left(12-3y\right)-4y-5=0\)

\(\Leftrightarrow10y^2-70y+115=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{7+\sqrt{3}}{2}\Rightarrow x=\dfrac{3-3\sqrt{3}}{2}\\y=\dfrac{7-\sqrt{3}}{2}\Rightarrow x=\dfrac{3+3\sqrt{3}}{2}\end{matrix}\right.\)

\(\Rightarrow B\left(\dfrac{3-3\sqrt{3}}{2};\dfrac{7+\sqrt{3}}{2}\right)\),\(C\left(\dfrac{3+3\sqrt{3}}{2};\dfrac{7-\sqrt{3}}{2}\right)\) 

hoặc \(B\left(\dfrac{3+3\sqrt{3}}{2};\dfrac{7-\sqrt{3}}{2}\right)\) , \(C\left(\dfrac{3-3\sqrt{3}}{2};\dfrac{7+\sqrt{3}}{2}\right)\)

a)Thay m=-1 vào pt có:

\(x^2+2x-8=0\) \(\Leftrightarrow x^2-2x+4x-8=0\)\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

b) Để pt có hai nghiệm pb\(\Leftrightarrow\Delta>0\Leftrightarrow4m^2-4\left(m-1\right)^3>0\)\(\Leftrightarrow m^2>\left(m-1\right)^3\)

Theo định lí viet có:\(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=\left(m-1\right)^3\end{matrix}\right.\)

Giả sử \(x_1=x_2^2\) .Ta có hệ: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1=x_2^2\\x_1x_2=\left(m-1\right)^3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2m\\x_1=x_2^2\\x_2^3=\left(m-1\right)^3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=2m\\x_1=\left(m-1\right)^2\\x_2=m-1\end{matrix}\right.\)

\(\Rightarrow\left(m-1\right)^2+m-1=2m\)

\(\Leftrightarrow m^2-3m=0\) \(\Leftrightarrow\left[{}\begin{matrix}m=3\\m=0\end{matrix}\right.\)(tmđk)

Vậy...

\(sin18^0=cos72^0=2cos^236^0-1=2\left(1-2.sin^218^0\right)^2-1\)\(=1-8.sin^218^0+8.sin^418^0\)

Có \(8.sin^318^0+m.sin^218^0=1\)

\(\Rightarrow\left(sin18^0-1\right)\left(8.sin^318^0+m.sin^218^0-1\right)=0\) ( vì sin18⁰-1\(\ne\)0)

\(\Leftrightarrow8.sin^418^0+sin^318^0\left(m-8\right)-m.sin^218^0-sin18^0+1=0\)

\(\Leftrightarrow8.sin^418^0+sin^318^0\left(m-8\right)-m.sin^218^0-\left(1-8.sin^218^0+8.sin^418^0\right)+1=0\)

\(\Leftrightarrow sin^318^0\left(m-8\right)-sin^218^0\left(m-8\right)=0\)

\(\Leftrightarrow\left(m-8\right)\left(sin^318^0-sin^218^0\right)=0\)

\(\Rightarrow m=8\)

Giả sử (d) cắt tia Ox, Oy tại A(a;0), B(0;b); a,b>0

=>Phương trình đt \(\left(d\right):\dfrac{x}{a}+\dfrac{y}{b}=1\)

\(M\in\left(d\right)\Rightarrow\dfrac{1}{a}+\dfrac{4}{b}=1\)

Có \(\dfrac{1}{a}+\dfrac{4}{b}\ge\dfrac{\left(1+2\right)^2}{a+b}\)\(\Leftrightarrow1\ge\dfrac{9}{a+b}\) \(\Leftrightarrow a+b\ge9\)

\(\Leftrightarrow OA+OB\ge9\)

Dấu = xảy ra\(\Leftrightarrow\dfrac{1}{a}=\dfrac{2}{b}\) và \(a+b=9\) \(\Leftrightarrow a=3,b=6\)

\(\Rightarrow\left(d\right):\dfrac{x}{3}+\dfrac{y}{6}=1\) \(\Leftrightarrow\left(d\right):2x+y-6=0\)

Để toi giải thích: Dấu = bđt |A|+|B|≥|A+B| xảy ra khi AB≥0

Nên trong bài dấu bằng xảy ra khi (x-2017)(2015-x)≥0 và x-2016=0

<=> 2017≥x≥2015 và x=2016 

=>x=2016 ( 2017≥x≥2015 chỉ là một điều kiện thôi,với cả x không nguyên nên trong khoảng này có rất nhiều x thỏa mãn)

Còn bài bạn dưới, x=2015 hoặc 2017 làm P=3 >2 => không phải giá trị của x để P nhỏ nhất

Akai Haruma cô lớp e dậy như thế 

 Gọi pt cạnh AC có dạng: ax+by+c=0 (a²+b²>0)

\(E\in\left(AC\right)\Rightarrow2a+2b+c=0\) \(\Leftrightarrow c=-2a-2b\)

Có \(cos\widehat{CAB}=cos\widehat{CBA}\) \(\Leftrightarrow\left|cos\left(\overrightarrow{n_{AC}},\overrightarrow{n_{AB}}\right)\right|=\left|cos\left(\overrightarrow{n_{AC}},\overrightarrow{n_{CB}}\right)\right|\)

\(\Leftrightarrow\dfrac{\left|a-3b\right|}{\sqrt{10}\sqrt{a^2+b^2}}=\dfrac{\left|7a-b\right|}{5\sqrt{2}.\sqrt{a^2+b^2}}\)

\(\Leftrightarrow\left|a-3b\right|=\dfrac{\sqrt{5}}{5}\left|7a-b\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}a-3b=\dfrac{\sqrt{5}}{5}\left(7a-b\right)\\a-3b=-\dfrac{\sqrt{5}}{5}\left(7a-b\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a\left(1-\dfrac{7\sqrt{5}}{5}\right)=b\left(3-\dfrac{\sqrt{5}}{5}\right)\\a\left(1+\dfrac{7\sqrt{5}}{5}\right)=b\left(3+\dfrac{\sqrt{5}}{5}\right)\end{matrix}\right.\)

TH1: \(a\left(1-\dfrac{7\sqrt{5}}{5}\right)=b\left(3-\dfrac{\sqrt{5}}{5}\right)\) 

Chọn \(a=3-\dfrac{\sqrt{5}}{5}\),\(b=1-\dfrac{7\sqrt{5}}{5}\) \(\Rightarrow c=\dfrac{-40+16\sqrt{5}}{5}\)

=> \(AC:\left(3-\dfrac{\sqrt{5}}{5}\right)x+\left(1-\dfrac{7\sqrt{5}}{5}\right)y+\dfrac{-40+16\sqrt{5}}{5}=0\)

TH2: \(a\left(1+\dfrac{7\sqrt{5}}{5}\right)=b\left(3+\dfrac{\sqrt{5}}{5}\right)\) 

Chọn \(a=3+\dfrac{\sqrt{5}}{5},b=1+\dfrac{7\sqrt{5}}{5}\) \(\Rightarrow c=-\dfrac{40+16\sqrt{5}}{5}\)

\(AC:\left(3+\dfrac{\sqrt{5}}{5}\right)x+\left(1+\dfrac{7\sqrt{5}}{5}\right)y-\dfrac{40+16\sqrt{5}}{5}=0\)

a)\(y=\sqrt{3}sinx+cosx=2\left(\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx\right)\)\(=2\left(sinx.cos\dfrac{\pi}{6}+cosx.sin\dfrac{\pi}{6}\right)\)\(=2sin\left(x+\dfrac{\pi}{6}\right)\)

Có \(-1\le sin\left(x+\dfrac{\pi}{6}\right)\le1\) \(\Leftrightarrow-2\le2sin\left(x+\dfrac{\pi}{6}\right)\le2\)

\(\Leftrightarrow-2\le y\le2\)

miny=-2 \(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=-1\)  \(\Leftrightarrow x+\dfrac{\pi}{6}=-\dfrac{\pi}{2}+2k\pi\left(k\in Z\right)\) \(\Leftrightarrow x=-\dfrac{2\pi}{3}+k2\pi\left(k\in Z\right)\)

maxy=2\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=1\) \(\Leftrightarrow x+\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)\(\Leftrightarrow x=\dfrac{\pi}{3}+k2\pi\left(k\in Z\right)\)

b) \(y=sin2x-cos2x=\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)\)

Có \(\sqrt{2}\ge\sqrt{2}sin\left(2x-\dfrac{\pi}{4}\right)\ge-\sqrt{2}\)

\(\Leftrightarrow\sqrt{2}\ge y\ge-\sqrt{2}\)

miny=\(-\sqrt{2}\) \(\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=-1\)\(\Leftrightarrow2x-\dfrac{\pi}{4}=-\dfrac{\pi}{2}+k2\pi\left(k\in Z\right)\)\(\Leftrightarrow x=-\dfrac{\pi}{8}+k\pi\left(k\in Z\right)\)

maxy=\(\sqrt{2}\Leftrightarrow sin\left(2x-\dfrac{\pi}{4}\right)=1\)\(\Leftrightarrow x=\dfrac{3\pi}{8}+k\pi\left(k\in Z\right)\)

c) \(y=3sinx+4cosx=5\left(\dfrac{3}{5}sinx+\dfrac{4}{5}cosx\right)\)

Đặt \(cosa=\dfrac{3}{5}\) và \(sina=\dfrac{4}{5}\)(vì cos²a+sin²a=1)

\(y=5\left(sinx.cosa+cosx.sina\right)\)\(=5sin\left(x+a\right)\)

\(\Rightarrow-5\le y\le5\)

miny=-5 <=> \(sin\left(x+a\right)=-1\)\(\Leftrightarrow x=-\dfrac{\pi}{2}-arc.sina+k2\pi\left(k\in Z\right)\)

maxy=5 <=> \(sin\left(x+a\right)=1\)\(\Leftrightarrow x=\dfrac{\pi}{2}-arc.sina+k2\pi\left(k\in Z\right)\)

(P/s1:cái x ở câu c ấy trông nó ngu ngu??
 P/s2:sau khi load lại câu hỏi ở 1 tab khác ,thấy 1 câu trả lời nhưng vẫn đăng vì cảm thấy bỏ đi hơi phí :?)

\(x+y=1\Rightarrow x=1-y\) 

\(C=x^2+y^2+xy=\left(1-y\right)^2+y^2+\left(1-y\right)y\)

\(=y^2-y+1\)\(=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall y\)

=>minC=\(\dfrac{3}{4}\) \(\Leftrightarrow y=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{2}\)