Câu 4
Câu a, b, c xem lại đề
d) \(\left(\dfrac{-10}{3}\right).\left(-2,5\right)=\left(\dfrac{-10}{3}\right).\left(\dfrac{-5}{2}\right)=\dfrac{50}{6}=\dfrac{25}{3}\)
e) \(2\dfrac{1}{7}:1\dfrac{1}{14}=\dfrac{15}{7}:\dfrac{15}{14}=\dfrac{15}{7}.\dfrac{14}{15}=2\)
Câu 5:
a) \(\dfrac{3}{5}+x=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}-\dfrac{3}{5}\)
\(x=\dfrac{5}{15}-\dfrac{9}{15}\)
\(x=-\dfrac{4}{15}\)
Vậy \(x=-\dfrac{4}{15}\)
b) \(x-\dfrac{3}{4}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{3}{4}\)
\(x=\dfrac{6}{4}+\dfrac{3}{4}\)
\(x=\dfrac{9}{4}\)
Vậy \(x=\dfrac{9}{4}\)
Câu 6:
Ta có: \(\widehat{MON}+\widehat{MOQ}=180^0\) (kề bù)
\(\Rightarrow\widehat{MON}=180^0-\widehat{MOQ}=180^0-35^0=145^0\)
Lại có: \(\widehat{NOK};\widehat{MOQ}\) là hai góc đối đỉnh
\(\Rightarrow\widehat{NOK}=\widehat{MOQ}=35^0\)
ai giúp mình với ạ cảm ơi