Tia Od thuộc nửa mặt phẳng bờ Ob không chứa Oc

=> Tia Ob nằm trong ^cOd

=> ^cOd = ^cOb + ^bOd = ^cOb + ^ aOc = ^aOb = 90 độ.

=> Tia Oc và tia Od vuông góc với nhau.

\(-\frac{1}{3}\cdot\frac{5}{4}\cdot\left(-\frac{1}{3}\right)^3\text{ : }\frac{1}{18}\)

\(=-\frac{1}{3}\cdot\frac{5}{4}\cdot\frac{\left(-1\right)^3}{3^3}\text{ : }\frac{1}{18}\)

\(=\frac{5}{4}\cdot\frac{1}{3}\cdot\frac{-1}{3^3}\cdot18\)

\(=\frac{5}{4}\cdot\frac{-1}{3\cdot3^3}\cdot9\cdot2\)

\(=\frac{5}{4}\cdot\frac{-1}{3^4}\cdot3^2\cdot2\)

\(=\frac{5}{4}\cdot\frac{-1}{3^2}\cdot2\)

\(=\frac{5}{2}\cdot\frac{-1}{9}\)

\(=-\frac{5}{18}\)

Trả lời :

\(-\frac{1}{3}\cdot\frac{5}{4}\cdot\left(-\frac{1}{3}\right)^3\text{ : }\frac{1}{18}\)

\(=-\frac{1}{3}\cdot\frac{5}{4}\cdot\frac{\left(-1\right)^3}{3^3}\text{ : }\frac{1}{18}\)

\(=\frac{5}{4}\cdot\frac{1}{3}\cdot\frac{-1}{3^3}\cdot18\)

\(=\frac{5}{4}\cdot\frac{-1}{3\cdot3^3}\cdot9\cdot2\)

\(=\frac{5}{4}\cdot\frac{-1}{3^4}\cdot3^2\cdot2\)

\(=\frac{5}{4}\cdot\frac{-1}{3^2}\cdot2\)

\(=\frac{5}{2}\cdot\frac{-1}{9}\)

\(=-\frac{5}{18}\)

Vậy câu hỏi là gì?
 

\(7^{1000}=\left(7^4\right)^{250}=\overline{......1}^{250}=\overline{.......1}\)

\(3^{1000}=\left(3^4\right)^{250}=\overline{......1}^{250}=\overline{......1}\)

\(\Rightarrow7^{1000}-3^{1000}=\overline{......1}-\overline{......1}=\overline{......0}⋮10\)

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

\(\Rightarrow\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}\)

\(\Rightarrow\frac{a+2b+3c}{2+6+12}=\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\) mà a + 2b + 3c = -20

\(\Rightarrow\frac{-20}{20}=\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

\(\Rightarrow-1=\frac{a}{2}=\frac{b}{3}=\frac{c}{4}\)

\(\Rightarrow\hept{\begin{cases}a=-1\cdot2=-2\\b=-1\cdot3=-3\\c=-1\cdot4=-4\end{cases}}\)

\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{1a}{1.2}=\frac{2b}{2.3}=\frac{3c}{3.4}=\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{2}=\frac{b}{3}=\frac{c}{4}=\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}=\frac{a+2b+3c}{2+6+12}=-\frac{20}{20}=-1\)
\(\Rightarrow\hept{\begin{cases}a=2.\left(-1\right)=-2\\b=3.\left(-1\right)=-3\\c=4.\left(-1\right)=-4\end{cases}}\)
Vậy a = -2, b = -3, c = -4

\(7^{2x}+7^{2x+2}=2450\)

\(7^{2x}+7^{2x}\times7^2=2450\)

\(7^{2x}\left(1+49\right)=2450\)

\(7^{2x}=2450\div50\)

\(7^{2x}=49\)

\(7^{2x}=7^2\)

\(\Rightarrow2x=2\Leftrightarrow x=1\)

\(7^{2x}+7^{2x+2}=2450\)

\(\Rightarrow7^{2x}+7^{2x}\cdot7^2=2450\)

\(\Rightarrow7^{2x}\cdot\left(1+49\right)=2450\)

\(\Rightarrow7^{2x}=2450:50=49\)

\(\Rightarrow7^{2x}=7^2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(\frac{1}{4}:\left(\frac{5}{3}\right)^{x-1}=0,09\)

\(\frac{1}{4}:\left(\frac{5}{3}\right)^{x-1}=\frac{9}{100}\)

\(\left(\frac{5}{3}\right)^{x-1}=\frac{1}{4}:\frac{9}{100}\)

\(\left(\frac{5}{3}\right)^{x-1}=\frac{1}{4}\cdot\frac{100}{9}\)

\(\left(\frac{5}{3}\right)^{x-1}=\frac{100}{36}\)

\(\left(\frac{5}{3}\right)^{x-1}=\frac{25}{9}\)

\(\left(\frac{5}{3}\right)^{x-1}=\left(\frac{5}{3}\right)^2\)

\(\Rightarrow x-1=2\Rightarrow x=3\)

Vậy x cần tìm bằng 3

=> (5/3)^x-1=25/9

=> (5/3)^x-1=(5/30)²

=> x-1=2

=> x=3