Đặt cạnh BC=a=8; AB=c; AC=b

Kẻ đường cao AH. Xét tg vuông ABH có ^BAH=90-^B=90-60=30

=> BH=AB/2=c/2 (trong tg vuông cạnh đối diện góc 30 =1/2 cạnh huyền)

\(\Rightarrow AH=\sqrt{AB^2-BH^2}=\sqrt{c^2-\frac{c^2}{4}}=\frac{c\sqrt{3}}{2}.\)

\(S_{ABC}=\frac{1}{2}.BC.AH=\frac{1}{2}.8.\frac{c\sqrt{3}}{2}=2c\sqrt{3}\)

Nửa chu vi p=(a+b+c)/2=(8+12)/2=10

Áp dụng công thức he rông

\(S_{ABC}=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=\sqrt{10\left(10-8\right)\left(10-b\right)\left(10-c\right)}\)

\(=\sqrt{20\left(100-10c-10b+bc\right)}=\sqrt{20\left(100-10\left(c+b\right)+bc\right)}\)

\(=\sqrt{20\left(100-10.12+bc\right)}=\sqrt{20\left(bc-20\right)}=2c\sqrt{3}\)

Bình phương 2 vê \(20\left(bc-20\right)=12c^2\) (*)

Thay b=12-c vào (*) rồi giải PT bậc 2 tìm c từ đó suy ra b. Bạn tự làm nốt nhé, chúc học tốt!

T

\(ĐKXĐ:a\ge3\)

\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)

\(\Leftrightarrow25.\sqrt{\frac{1}{25}.\left(a-3\right)}-7\sqrt{\frac{4}{9}.\left(a-3\right)}-7\sqrt{a^2-9}+18\sqrt{\frac{9}{81}.\left(a^2-9\right)}=0\)

\(\Leftrightarrow25.\sqrt{\frac{1}{25}}.\sqrt{a-3}-7.\sqrt{\frac{4}{9}}.\sqrt{a-3}-7\sqrt{a^2-9}+18.\sqrt{\frac{9}{81}}.\sqrt{a^2-9}=0\)

\(\Leftrightarrow25.\frac{1}{5}.\sqrt{a-3}-7.\frac{2}{3}.\sqrt{a-3}-7\sqrt{a^2-9}+18.\frac{1}{3}.\sqrt{a^2-9}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}.\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}.\sqrt{a-3}-\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{\left(a-3\right)\left(a+3\right)}=0\)

\(\Leftrightarrow\sqrt{a-3}.\left(\frac{1}{3}-\sqrt{a+3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\frac{1}{3}-\sqrt{a+3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a-3=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\\a+3=\frac{1}{9}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\\a=\frac{-26}{9}\end{cases}}\)

mà \(a\ge3\)\(\Rightarrow a=\frac{-26}{9}\)không thỏa mãn

Vậy \(a=3\)

Bài làm:

đk: \(a\ge3\)

Ta có: \(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}+\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)

\(\Leftrightarrow\sqrt{a^2-9}=\sqrt{a-3}\)

\(\Leftrightarrow\left|a^2-9\right|=\left|a-3\right|\)

\(\Leftrightarrow\orbr{\begin{cases}a^2-9=a-3\\a^2-9=3-a\end{cases}}\Leftrightarrow\orbr{\begin{cases}a^2-a-6=0\\a^2+a-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\left(a-3\right)\left(a+2\right)=0\\\left(a-3\right)\left(a+4\right)=0\end{cases}}\)

=> \(a\in\left\{-4;-2;3\right\}\)

Mà theo đk thì \(a\ge3\) => a = 3 (thỏa mãn)

Vậy a = 3

Thu gọn B-.-?

Ta có: \(B=\frac{1}{3}\sqrt{9+6v+v^2}+\frac{4v}{3}+5\)

\(B=\frac{1}{3}\sqrt{\left(3+v\right)^2}+\frac{4v}{3}+5\)

\(B=\frac{1}{3}\cdot\left|3+v\right|+\frac{4v}{3}+5\)

Vì v < - 3 

=> \(B=\frac{1}{3}\cdot\left[-\left(3+v\right)\right]+\frac{4v}{3}+5\)

\(B=\frac{-3-v}{3}+\frac{4v}{3}+5\)

\(B=\frac{3v-3}{3}+5=v-1+5=v+4\)

Vậy \(B=v+4\)

\(B=\frac{1}{3}\sqrt{9+6v+v^2}+\frac{4v}{3}+5\)

\(B=\frac{1}{3}\sqrt{3^2+3\cdot2\cdot v+v^2}+\frac{4v}{3}+5\)

\(B=\frac{1}{3}\sqrt{\left(3+v\right)^2}+\frac{4v}{3}+5\)

\(B=\frac{1}{3}\left|3+v\right|+\frac{4v}{3}+5\)

Với v < -3

\(B=\frac{1}{3}\cdot\left[-\left(3+v\right)\right]+\frac{4v}{3}+5\)

\(B=\frac{1}{3}\left(-3-v\right)+\frac{4v}{3}+5\)

\(B=-1-\frac{v}{3}+\frac{4v}{3}+5\)

\(B=-1+\frac{-v+4v}{3}+5\)

\(B=4+\frac{3v}{3}=4+v\)

Bài làm:

Ta có: \(\frac{x^2+4x^2}{\left(x+2\right)^2}=12\)

\(\Leftrightarrow5x^2=12\left(x^2+4x+4\right)\)

\(\Leftrightarrow7x^2+48x+48=0\)

\(\Leftrightarrow7\left(x^2+\frac{48}{7}x+\frac{576}{49}\right)-\frac{240}{7}=0\)

\(\Leftrightarrow\left(x+\frac{24}{7}\right)^2-\left(\frac{4\sqrt{15}}{7}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{24-4\sqrt{15}}{7}\right)\left(x+\frac{24+4\sqrt{15}}{7}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4\sqrt{15}-24}{7}\\x=-\frac{24+4\sqrt{15}}{7}\end{cases}}\)

*Bổ sung : Thiếu ĐK rồi : \(x\ne-2\)

Ta có:

\(\Delta_1+\Delta_2+\Delta_3=a^2-4b+b^2-4c+c^2-4a=a^2+b^2+c^2-48\)

Dễ thấy:\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=48\Rightarrow\Delta_1+\Delta_2+\Delta_3\ge0\)

Khi đó có ít nhất một phương trình có nghiệm

còn c/m vô nghiệm thế nào z

Vì x < 0, y < 0 nên\(\frac{\sqrt{50x^4y^2}}{\sqrt{200x^2}y^4}=\frac{\sqrt{50}\left|x^2y\right|}{\sqrt{200}\left|x\right|y^4}=\frac{-x^2y}{-2xy^4}=\frac{x}{2y^3}\)

Bài làm:

Ta có:

\(\frac{\sqrt{50x^4y^2}}{\sqrt{200x^2}y^4}=\frac{5x^2y\sqrt{2}}{10xy^4\sqrt{2}}=\frac{x}{2y^3}\)