Tìm x cho trường hợp x/30 = 35/18 - 86/45
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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, 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#

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#
\(\frac{x}{30}=\frac{35}{18}-\frac{86}{45}\)
\(\Leftrightarrow\frac{x}{30}=\frac{1}{30}\)
\(\Leftrightarrow x=1\)
Vậy x=1