Lời giải:
Ta có:
$BH=\frac{AH}{\tan B}=\frac{AH}{\tan 60^0}$
$CH=\frac{AH}{\tan C}=\frac{AH}{\tan 40^0}$
$\Rightarrow BH+CH=\frac{AH}{\tan 60^0}+\frac{AH}{\tan 40^0}$
$\Leftrightarrow 12=AH(\frac{1}{\tan 60^0}+\frac{1}{\tan 40^0})$
$\Rightarrow AH=6,78$ (cm)
$AC=\frac{AH}{\sin C}=\frac{AH}{\sin 40^0}=10,55$ (cm)
b.$S_{ABC}=\frac{AH.BC}{2}=\frac{6,78.12}{2}=40,68$ (cm vuông)
1. rút gọn biểu thức
B=\(\dfrac{3}{\sqrt{6}-2}+\dfrac{2}{\sqrt{6}+2}-\dfrac{5\sqrt{6}}{2}\)
rút gọn
A=\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
cho x<0, y<0. Biểu thức A= \(x\sqrt{16y^2}\)bằng
rút gọn không dùng máy tính \(A=\frac{3+\sqrt{5}}{\sqrt{5}+2}+\frac{\sqrt{5}}{\sqrt{5}-1}-\frac{3\sqrt{5}}{3+\sqrt{5}}\)
RUT GON:
\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)
Rút gọn biểu thức
A=[1+(11-√11):(1-√11)]×[(11+√11):(1+√11)+1]
Cho biểu thức: \(P=\left(\frac{x+2}{\sqrt{x}+1}-\sqrt{x}\right):\left(\frac{\sqrt{x}-4}{1-x}-\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\frac{x\sqrt{x}-1}{x-1}-\frac{x+1}{\sqrt{x}+1}\)
rút gọn biểu thức
1 .
a)\(A=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)
b)\(B=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\)
c)C=\(\frac{2}{\sqrt[3]{4}+\sqrt[3]{2}+2}\)
2 .
a)\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
b)\(B=\frac{\left(5+2\sqrt{6}\right).\left(49-20\sqrt{6}\right).\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
c)C=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7}+4\sqrt{3}}}}\)
d)D=(\(\left(\sqrt{3}-1\right).\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
Toán 9
(\(\frac{6-2\sqrt{2}}{3-2\sqrt{2}}\)-\(\frac{5}{\sqrt{5}}\)):\(\frac{1}{2-\sqrt{5}}\)
