Đường thẳng BC nhận \(\overrightarrow{n}=\left(\sqrt{3};-3\right)\) là 1 vtpt

Gọi \(\overrightarrow{n_1}=\left(a;b\right)\) là 1 vtpt của AB (với a;b không đồng thời bằng 0)

Do tam giác ABC đều \(\Rightarrow\widehat{\left(n_1;\overrightarrow{n}\right)}=60^0\)

\(\Rightarrow cos\left(\overrightarrow{n_1};\overrightarrow{n}\right)=\dfrac{\left|a\sqrt{3}-3b\right|}{\sqrt{a^2+b^2}.\sqrt{3+9}}=\dfrac{1}{2}\)

\(\Leftrightarrow\left(a-\sqrt{3}b\right)^2=a^2+b^2\)

\(\Leftrightarrow a^2+3b^2-2\sqrt{3}ab=a^2+b^2\)

\(\Leftrightarrow b^2=\sqrt{3}ab\Rightarrow\left[{}\begin{matrix}b=0\\b=\sqrt{3}a\end{matrix}\right.\)

\(\Rightarrow\) Phương trình 2 cạnh còn lại có dạng:

\(\left\{{}\begin{matrix}a\left(x-2\right)+0\left(y-0\right)=0\\a\left(x-2\right)+\sqrt{3}a\left(y-0\right)=0\end{matrix}\right.\)

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

B

Đáp án B nha

Chọn B nhé bạn

Tam giác ABC là tam giác đều?

Nếu ABC đều thì \(\left|\overrightarrow{BM}\right|=BM=\dfrac{a\sqrt{3}}{2}\)

Chọn C

bạn ơi giúp mình với C/M: (ax^2 - bx^2)^4 + (2ab+bx^2)^4 + (2ab+a^2)^4 = 2(a^2+ab+b^2)

D là điểm nào bạn?

1.

\(\Leftrightarrow x^2-3x+1+\dfrac{\sqrt{3}}{3}\sqrt{\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2b^2-a^2+\dfrac{\sqrt{3}}{3}ab=0\)

\(\Leftrightarrow\left(\sqrt{3}b-a\right)\left(2b+\sqrt{3}a\right)=0\)

\(\Leftrightarrow a=\sqrt{3}b\)

\(\Leftrightarrow\sqrt{x^2+x+1}=\sqrt{3}.\sqrt{x^2-x+1}\)

\(\Leftrightarrow x^2+x+1=3x^2-3x+3\)

\(\Leftrightarrow2x^2-4x+2=0\)

\(\Leftrightarrow x=1\)

Bài 2:

Đặt \(AB=x>0\) 

\(AG=\dfrac{1}{2}BC=\dfrac{1}{2}\sqrt{a^2+x^2}\)

\(CG=\dfrac{2}{3}\sqrt{\left(\dfrac{AB}{2}\right)^2+AC^2}=\dfrac{2}{3}\sqrt{\dfrac{x^2}{4}+a^2}\)

\(BG=\dfrac{2}{3}\sqrt{AB^2+\left(\dfrac{AC}{2}\right)^2}=\dfrac{2}{3}\sqrt{x^2+\dfrac{a^2}{4}}\)

Ta có:

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\Leftrightarrow\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{AG}\)

\(\Leftrightarrow GB^2+GC^2+2GB.GC.cos\left(\overrightarrow{GB};\overrightarrow{GC}\right)=AG^2\)

\(\Leftrightarrow cos\left(\overrightarrow{GB};\overrightarrow{GC}\right)=\dfrac{AG^2-BG^2-CG^2}{2GB.GC}\)

\(=\dfrac{\dfrac{a^2+x^2}{4}-\left[\dfrac{4}{9}\left(\dfrac{x^2}{4}+a^2\right)+\dfrac{4}{9}\left(\dfrac{a^2}{4}+x^2\right)\right]}{\dfrac{2}{9}\sqrt{\left(a^2+4x^2\right)\left(x^2+4a^2\right)}}\)

\(=-\dfrac{11}{4}.\dfrac{x^2+a^2}{2\sqrt{\left(a^2+4x^2\right)\left(x^2+4a^2\right)}}\le-\dfrac{11}{4}.\dfrac{x^2+a^2}{5\left(x^2+a^2\right)}=-\dfrac{11}{20}\)

Dấu "=" xảy ra khi \(a=x\Leftrightarrow AB=a\)

\(tanB=\sqrt{2}\Rightarrow\dfrac{AC}{AB}=\sqrt{2}\Rightarrow\dfrac{AC^2}{AB^2}=2\)

\(\Rightarrow\dfrac{AC^2}{AB^2}+1=3\Rightarrow\dfrac{AC^2+AB^2}{AB^2}=3\Rightarrow\dfrac{BC^2}{AB^2}=3\)

\(\Rightarrow\dfrac{AB}{BC}=\dfrac{1}{\sqrt{3}}\)

Mà \(sinC=\dfrac{AB}{BC}\Rightarrow sinC=\dfrac{1}{\sqrt{3}}\)

\(sin^2C+cos^2C=1\Rightarrow\dfrac{1}{3}+cos^2C=1\Rightarrow cosC=\dfrac{\sqrt{6}}{3}\)

\(tanC=\dfrac{sinC}{cosC}=\dfrac{\sqrt{2}}{2}\)

b.

Trong tam giác vuông ACH:

\(sinC=\dfrac{AH}{AC}\Rightarrow AC=\dfrac{AH}{sinC}=\dfrac{2\sqrt{3}}{\dfrac{1}{\sqrt{3}}}=6\left(cm\right)\)

Trong tam giác vuông ABC:

\(tanB=\dfrac{AC}{AB}\Rightarrow AB=\dfrac{AC}{tanB}=\dfrac{6}{\sqrt{2}}=3\sqrt{2}\)

Áp dụng Pitago:

\(BC=\sqrt{AB^2+AC^2}=3\sqrt{6}\left(cm\right)\)

a: Xét ΔABC vuông tại A có 

\(\tan\widehat{B}=\sqrt{2}\)

\(\Leftrightarrow AC=AB\cdot\sqrt{2}\)

Áp dụng định lí Pytago vào ΔABC vuông tại A, ta được:

\(BC^2=AB^2+AC^2\)

\(\Leftrightarrow BC^2=3\cdot AB^2\)

hay \(BC=AB\cdot\sqrt{3}\)

Xét ΔABC vuông tại A có 

\(\sin\widehat{C}=\dfrac{AB}{BC}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3}\)

\(\cos\widehat{C}=\dfrac{AC}{BC}=\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{\sqrt{6}}{3}\)

\(\tan\widehat{C}=\dfrac{AB}{AC}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

\(\cot\widehat{C}=\sqrt{2}\)