\(AC=\sqrt{AB^2+BC^2-2AB.BC.cosB}=\sqrt{9^2+12^2-2.9.12.cos60^0}=3\sqrt{13}\)

Sửa đề: cosA=3/5

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

=>\(\frac{7^2+5^2-BC^2}{2\cdot7\cdot5}=\frac35\)

=>\(49+25-BC^2=\frac35\cdot14\cdot5=42\)

=>\(BC^2=49+25-42=32\)

=>\(BC=4\sqrt2\)

\(cos^2A+\sin^2A=1\)

=>\(\sin^2A=1-\left(\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25}\)

=>\(\sin A=\frac45\)

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

\(=\frac12\cdot7\cdot5\cdot\frac45=\frac12\cdot7\cdot4=7\cdot2=14\)

\(S_{ABC}=\frac12\cdot AH\cdot BC\)

=>\(AH\cdot\frac{4\sqrt2}{2}=14\)

=>\(AH=\frac{14}{2\sqrt2}=\frac{7}{\sqrt2}\)

=>Chọn B

\(BM=\dfrac{1}{2}BC=3\)

\(AM=\sqrt{AB^2+BM^2-2AB.BM.cos60^0}=\sqrt{19}\)

\(BN=\dfrac{\sqrt{2\left(AB^2+BM^2\right)-AM^2}}{2}=\dfrac{7}{2}\)

ban  ấy biết làm thì đâu có phải đăng

Tính p , S Rồi mới tính R phải ko bn Hoàng

Gọi G là trọng tâm tam giác \(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=0\)

\(\overrightarrow{MA}^2+\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MA}.\overrightarrow{MC}=0\)

\(\Leftrightarrow\overrightarrow{MA}\left(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right)=0\)

\(\Leftrightarrow\overrightarrow{MA}\left(\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\right)=0\)

\(\Leftrightarrow3\overrightarrow{MA}.\overrightarrow{MG}=0\)

\(\Rightarrow\) M thuộc đường tròn đường kính AG

Bán kính: \(R=\dfrac{1}{2}AG=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{6}\)

Xét ΔABC có AD là phân giác

nên \(\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{5}{7}\)

=>\(\dfrac{BD}{5}=\dfrac{DC}{7}\)

mà BD+DC=BC=6

nên \(\dfrac{BD}{5}=\dfrac{CD}{7}=\dfrac{BD+CD}{5+7}=\dfrac{6}{12}=\dfrac{1}{2}\)

=>BD=2,5; CD=3,5

=>\(\dfrac{BD}{BC}=\dfrac{5}{12};\dfrac{CD}{CB}=\dfrac{7}{12}\)

\(\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BD}\)

\(=\overrightarrow{AB}+\dfrac{5}{12}\cdot\overrightarrow{BC}\)

\(=\overrightarrow{AB}+\dfrac{5}{12}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)

\(=\dfrac{7}{12}\cdot\overrightarrow{AB}+\dfrac{5}{12}\cdot\overrightarrow{AC}\)

=>Chọn C

\(R=\dfrac{BC}{2sin\widehat{BAC}}=\dfrac{a}{2sin120^0}=\dfrac{a\sqrt{3}}{3}\)

Giải hệ phương trình:

\(\left\{{}\begin{matrix}y^3-4y^2+4y=\sqrt{x+1}\left(y^2-5y+4+\sqrt{x+1}\right)\\2\sqrt{x^2-3x+3}+6x-7=y^2\left(x-1\right)^2+\left(y^2-1\right)\sqrt{3x-2}\end{matrix}\right.\)

Từ pt trên giải ra \(y=\sqrt{x+1}\) rồi thay vào dưới lại bị bí quá :((

\(C=180^0-\left(A+B\right)=75^0\)

Áp dụng định lý hàm sin:

\(\dfrac{b}{sinB}=\dfrac{c}{sinC}\Rightarrow c=\dfrac{b.sinC}{sinB}=\dfrac{8.sin75^0}{sin45^0}=4+4\sqrt{3}\)

Đặt \(\frac{a}{\sqrt3}=\frac{b}{\sqrt2}=\frac{2c}{\sqrt6-\sqrt2}=k\)

=>\(a=k\cdot\sqrt3;b=k\cdot\sqrt2;2c=k\left(\sqrt6-\sqrt2\right)\)

=>\(a=k\cdot\sqrt3;b=k\cdot\sqrt2;c=\frac{k\left(\sqrt6-\sqrt2\right)}{2}\)

Xét ΔABC có \(cosA=\frac{b^2+c^2-a^2}{2\cdot b\cdot c}\)

\(=\left\lbrack2k^2+k^2\cdot\frac{\left(\sqrt6-\sqrt2\right)^2}{4}-3k^2\right\rbrack:\left\lbrack2\cdot k\cdot\sqrt2\cdot\frac{k\left(\sqrt6-\sqrt2\right)}{2}\right\rbrack\)

\(=\left\lbrack2k^2+k^2\cdot\frac{8-4\sqrt3}{4}-3k^2\right\rbrack:\left\lbrack k^2\cdot\left(\sqrt{12}-2\right)\right\rbrack\)

\(=\frac{\left(2+2-\sqrt3-3\right)}{\sqrt{12}-2}=\frac{1-\sqrt3}{2\left(\sqrt3-1\right)}=-\frac12\)

=>\(\hat{BAC}=120^0\)

Xét ΔABC có \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

=>\(\frac{k\cdot\sqrt3}{\sin120}=\frac{k\cdot\sqrt2}{\sin B}=\frac{\frac{k\left(\sqrt6-\sqrt2\right)}{2}}{\sin C}\)

=>\(\frac{\sqrt2}{\sin B}=\frac{\frac{\left(\sqrt6-\sqrt2\right)}{2}}{\sin C}=\sqrt3:\frac{\sqrt3}{2}=2\)

=>\(\sin B=\frac{\sqrt2}{2};\sin C=\frac{\sqrt6-\sqrt2}{4}\)

=>\(\hat{B}=45^0;\hat{C}=15^0\)

Xét ΔABC có \(\frac{BC}{\sin A}=2R\)

=>\(2R=2\sqrt3:\sin120=2\sqrt3:\frac{\sqrt3}{2}=2\sqrt3\cdot\frac{2}{\sqrt3}=4\)

=>R=2