HOC24
Lớp học
Môn học
Chủ đề / Chương
Bài học
\(2480-\dfrac{4710}{3}+\left(200-\left(x-5\right)\right)=1010.\\ \Leftrightarrow910+200-x+5-1010=0.\\ \Leftrightarrow105-x=0.\\ \Leftrightarrow x=105.\)
\(\dfrac{2x}{\left(\dfrac{1309}{11}-19\right)-2}=0.\\ \Leftrightarrow\dfrac{2x}{100-2}=0.\\ \Leftrightarrow\dfrac{2x}{98}=0.\\ \Leftrightarrow2x=0.\Leftrightarrow x=0.\)
a) \(MN//PQ\left(gt\right).\)
\(\Rightarrow\widehat{M}+\widehat{Q}=180^o\) (Trong cùng phía).
\(\Rightarrow\widehat{Q}=180^o-105^o=75^o.\)
Tứ giác MNPQ là hình thang cân (gt).
\(\Rightarrow\widehat{Q}=\widehat{P};\widehat{M}=\widehat{N}.\)
\(\Rightarrow\widehat{P}=75^o;\widehat{N}=105^o.\)
b) \(MN//PQ\left(gt\right).\)
\(\Rightarrow\widehat{N}+\widehat{P}=180^o\) (Trong cùng phía).
\(\Rightarrow\widehat{N}=180^o-40^o=140^o.\)
\(\Rightarrow\widehat{Q}=40^o;\widehat{M}=140^o.\)
\(2x\left(x-2\right)-\left(2-x^2\right)=0.\\ \Leftrightarrow2x\left(x-2\right)+\left(x+2\right)\left(x-2\right)=0.\\ \Leftrightarrow\left(x-2\right)\left(2x+x+2\right)=0.\\ \Leftrightarrow\left(x-2\right)\left(3x+2\right)=0.\\ \Leftrightarrow\left[{}\begin{matrix}x=2.\\x=-\dfrac{2}{3}.\end{matrix}\right.\)
Xét \(\Delta ABH\) vuông tại A:
\(AB^2=AH^2+BH^2\left(Pytago\right).\\ 12^2=6^2+BH^2.\\ BH^2=108.\\ \Rightarrow BH=6\sqrt{3}.\)
\(\sin B=\dfrac{AH}{AB}.\\ \sin B=\dfrac{6}{12}.\\ \Rightarrow\widehat{B}=30^o.\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2.\\ =\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{16}.4.\\ =\dfrac{55}{56}-\dfrac{3}{4}.\\ =\dfrac{13}{56}.\)