ta có :

\(A=\frac{67^{2016}}{67^{2016}-11}=1+\frac{11}{67^{2016}-11}\)

\(B=\frac{67^{2016}+13}{67^{2016}+2}=1+\frac{11}{67^{2016}+2}\)

Vì \(67^{2016}-11< 67^{2016}+2\Rightarrow A>B\)

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

\(\Rightarrow b=c=4,5\)

\(\frac{a}{sinA}=\frac{b}{sinB}\Rightarrow a=\frac{b.sinA}{sinB}=\frac{9}{4}\left(\sqrt{6}-\sqrt{2}\right)\)

e/ \(cosA=\frac{b^2+c^2-a^2}{2bc}\Rightarrow a=\sqrt{b^2+c^2-2bc.cosA}\approx23\)

\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{433}{460}\Rightarrow B\approx19^043'\)

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

f/ \(cosA=\frac{b^2+c^2-a^2}{2bc}=\frac{11}{15}\Rightarrow A\approx42^050'\)

\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{17}{35}\Rightarrow B\approx60^056'\)

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

a/ \(cosA=\frac{b^2+c^2-a^2}{2bc}=-\frac{1}{2}\Rightarrow A=120^0\)

\(cosB=\frac{a^2+c^2-b^2}{2ac}=\frac{\sqrt{2}}{2}\Rightarrow B=45^0\)

\(C=180^0-\left(A+B\right)=15^0\)

b/\(A=180^0-\left(B+C\right)=79^037'\)

\(\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}\Rightarrow\left\{{}\begin{matrix}b=\frac{sinB}{sinA}.a\approx61\\c=\frac{sinC}{sinA}.a\approx102\end{matrix}\right.\)

c/\(\frac{a}{sinA}=\frac{b}{sinB}\Rightarrow sinB=\frac{bsinA}{a}\approx0,6\Rightarrow B\approx36^052'\)

\(\Rightarrow C=180^0-\left(A+B\right)=75^045'\)

\(\frac{a}{sinA}=\frac{c}{sinC}\Rightarrow c=\frac{a.sinC}{sinA}\approx21\)

\(\left(C\right):\left(x-2\right)^2+\left(y-1\right)^2=25\)

\(\Rightarrow\left(C\right)\) có tâm \(I\left(2;1\right)\) ; Bán kính \(R=5\)

\( \left(C\right)//d:5x-12y+67=0\)

nên \(\Delta:5x-12y+m=0\left(m\ne67\right)\)

Vì \(d\) có \(VTPT\overrightarrow{n}=\left(5;-12\right)\) cũng là \(VTPT\) của \(\Delta\)

\(R=d\left(I,\Delta\right)=\dfrac{\left|5x_I-12y_I+m\right|}{\sqrt{5^2+\left(-12\right)^2}}\Leftrightarrow\dfrac{\left|5.2-12.1+m\right|}{13}=5\)

\(\Leftrightarrow\left|-2+m\right|=65\)

\(\Leftrightarrow\left[{}\begin{matrix}-2+m=65\\-2+m=-65\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}m=67\left(ktm\right)\\m=-63\left(tm\right)\end{matrix}\right.\)

Vậy pt tiếp tuyến là \(5x-12y-63=0\)

(x-2)^2+(y-1)^2=25

=>R=5; I(2;1)

(d')//(d) nên (d'): 5x-12y+c=0

Theo đề, ta có; d(I;(d'))=5

=>\(\dfrac{\left|5\cdot2+\left(-12\right)\cdot1+c\right|}{\sqrt{5^2+12^2}}=5\)

=>|c-2|=65

=>c=67 hoặc c=-63

Có:
\(DC=AC.tan43^o=\left(AB+BC\right).tan43^o\).
\(DC=BC.tan67^o\).
Vì vây:
\(\left(AB+BC\right).tan43^o=BC.tan67^o\)
\(\Leftrightarrow BC=\dfrac{AB.tan43^o}{tan67^o-tan43^o}=26,55m\).
Suy ra: \(DC=BC.tan67^o=26,55.tan67^o=62,55m\).
Vậy chiều cao DC của chân tháp là 62,55m.

a, 3⁸ : 3⁴ = 3^8-4 = 3⁴

b, 10⁸ : 10² = 10^8-2 = 10⁶

c, a⁶ : a = a^6-1 = a⁵

a) 3 mũ 15 : 3 mũ 15 = 3 mũ 10

b) 4 mũ 6 : 4 mũ 6 = 1

c) 9 mũ 8 : 3 mũ 2 = 9 mũ 8 : 9 = 9 mũ 7

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