Bài toán 3:
\(a,\dfrac{x+0,5}{-1}=\dfrac{-1}{x+0,5}\\ \Rightarrow\left(x+0,5\right)^2=-1\cdot-1=1\\ \Rightarrow\left(x+0,5\right)^2=1^2\\ TH1:x+0,5=1\\ \Rightarrow x=1-0,5\\ \Rightarrow x=0,5\\ TH2:x+0,5=-1\\ \Rightarrow x=-1-0,5\\ \Rightarrow x=-1,5\\ b,5^{x+1}-5^{x+2}=-20\\ \Rightarrow5^x\cdot5-5^x\cdot5^2=-20\\ \Rightarrow5^x\cdot\left(5-5^2\right)=-20\\ \Rightarrow5^x\cdot\left(-20\right)=-20\\ \Rightarrow5^x=-\dfrac{20}{-20}\\ \Rightarrow5^x=1\\ \Rightarrow5^x=5^0\\ \Rightarrow x=0\)
Bài 3:
\(a,\dfrac{3}{4}-5x=\dfrac{-1}{4}-3x\)
\(5x-3x=\dfrac{3}{4}-\dfrac{-1}{4}\)
\(x\left(5-3\right)=\dfrac{4}{4}\)
\(x2=1\)
\(x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
\(b,\dfrac{3}{2}x-\dfrac{2}{5}=\dfrac{1}{3}x-\dfrac{1}{4}\)
\(\dfrac{3}{2}x-\dfrac{1}{3}x=\dfrac{2}{5}-\dfrac{1}{4}\)
\(x\left(\dfrac{3}{2}-\dfrac{1}{3}\right)=\dfrac{8}{20}-\dfrac{5}{20}\)
\(x\left(\dfrac{9}{6}-\dfrac{2}{6}\right)=\dfrac{3}{20}\)
\(x.\dfrac{7}{6}=\dfrac{3}{20}\)
\(x=\dfrac{3}{20}:\dfrac{7}{6}\)
\(x=\dfrac{3}{20}.\dfrac{6}{7}\)
\(x=\dfrac{3.3}{10.7}\)
\(x=\dfrac{9}{70}\)
Vậy \(x=\dfrac{9}{70}\)
\(d,\dfrac{2}{3}-\dfrac{5}{3}x=\dfrac{7}{10}x+\dfrac{5}{6}\)
\(-\dfrac{5}{3}x-\dfrac{7}{10}x=\dfrac{5}{6}-\dfrac{2}{3}\)
\(\left(\dfrac{-5}{3}-\dfrac{7}{10}\right)x=\dfrac{5}{6}-\dfrac{4}{6}\)
\(\left(\dfrac{-50}{30}-\dfrac{21}{30}\right)x=\dfrac{1}{6}\)
\(\dfrac{-71}{30}.x=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}:\dfrac{-71}{30}\)
\(x=\dfrac{1}{6}.\dfrac{-30}{71}\)
\(x=\dfrac{1.\left(-5\right)}{1.71}\)
\(x=\dfrac{-5}{71}\)
Vậy \(x=\dfrac{-5}{71}\)
Bài 4 :
Vì \(\widehat{xOm} \) và \(\widehat{yOm}\) là `2` góc kề bù nên
Ta có :
\(\widehat{xOm} + \widehat{yOm} = \)\(180^\circ\)
\(\widehat{xOm} + 70^\circ = 180^\circ\)
\(\widehat{xOm} = 180^\circ - 70^\circ\)
\(\widehat{xOm} = 110^\circ\)
Bài 5 :
Vì \(\widehat{mAt} ; \widehat{nAt}\) là `2` góc kề bù nên :
Ta có :
\(\widehat{mAt} + \widehat{nAt} = 180^\circ\)
\(50 ^\circ+ \widehat{nAt} = 180^\circ\)
\(\widehat{nAt} = 180^\circ - 50^\circ\)
\(\widehat{nAt} = 130^\circ\)
