trả lời

e nhân ra cx ko đc hả

?????????? sao vậy

\(\frac{52}{33}-x:\frac{11}{84}=\frac{10}{33}\)

\(\Leftrightarrow x:\frac{11}{84}=\frac{52}{33}-\frac{10}{33}=\frac{14}{11}\)

\(\Leftrightarrow x=\frac{14}{11}.\frac{11}{84}=\frac{1}{6}\)

~ Hok tốt ~

\(\frac{x}{30}=\frac{35}{18}-\frac{86}{45}\)

\(\Leftrightarrow\frac{x}{30}=\frac{1}{30}\)

\(\Leftrightarrow x=1\)

Vậy x=1

a, \(\sin25^0\)< \(\sin70^0\)

b, \(\cos40^0\)> \(\cos75^0\)

c, \(\sin35^0\)= \(\cos55^0\)

\(\cos55^0\)< \(\cos35^0\)

\(\Rightarrow\)\(\sin35^0\)< \(\cos35^0\)

#mã mã#

A:104

B:

\(\frac{-21}{5}\)

C:\(\frac{11}{10}\)

a, ta có \(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}\)

                  \(\frac{1}{3}\)= \(\frac{\sin\alpha}{\cos\alpha}\)

                    \(\cos\alpha\)= 3 \(\sin\alpha\)

ta có \(\frac{\cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}\)= \(\frac{3\sin\alpha+\sin\alpha}{3\sin\alpha-\sin\alpha}\)= \(\frac{4\sin\alpha}{2\sin\alpha}\)= \(2\)

#mã mã#

giúp mình với mình đang cần gấp 

ai giải được thì mình

3ab+56=5ab

56 =5ab - 3 ab

56 =ab.(5 - 3 )

56 = 2ab

=>ab=56:2

ab = 28

Vậy ab =28

#Chém

\(D=2.4+4.6+...+98.100\)

\(\Rightarrow6D=2.4.6+4.6.6+...+98.100.6\)

\(=2.4.6+4.6.\left(8-2\right)+...+98.100.\left(102-96\right)\)

\(=2.4.6-2.4.6+4.6.8+...+98.100.102-96.98.100\)

\(=98.100.102\)

\(=999600\)

\(\Rightarrow D=\frac{999600}{6}=166600\)

\(D=2.4+4.6+...+98.100\)

\(6D=2.4.6+4.6.6+...+98.100.6\)

\(=2.4.6+4.6.\left(8-2\right)+...+98.100.\left(102-96\right)\)

\(=2.4.6-2.4.6+4.6.8+...+98.100.102\)

\(=98.100.102\)

\(6D=999600\)

\(D=999600:6\)

\(D=166600\)

a, ta có \(\cos^2\alpha\)+  \(\sin^2\alpha\)= 1

                  1/5 + \(\cos^2\alpha\)= 1

                               \(\cos^2\alpha\)= 4/5

\(4\cos^2\alpha\)+6 \(\sin^2\alpha\)= 4 . 4/5 + 6.1/5=22/5

b, \(\sin\alpha\)= 2/3 

\(\sin^2\alpha\)= 4/9

\(\cos^2\alpha=\frac{5}{9}\)

\(5\cos^2\alpha+2\sin^2=\frac{5.5}{9}+\frac{2.4}{9}=\frac{33}{9}\)

#mã mã#