a) Ta có: \(\widehat{cNb}+\widehat{MNb}=180^{\circ}\) (hai góc kề bù)

\(\Rightarrow\widehat{MNb}=180^{\circ}-\widehat{cNb}=180^{\circ}-55^{\circ}=125^{\circ}\)

b) Ta có: \(\widehat{MNb}=\widehat{aMN}\left(=125^{\circ}\right)\)

Mà hai góc này đều nằm ở vị trí so le trong

Nên \(Ma//Nb\)

\(7+\left(\dfrac{7}{12}-\dfrac{1}{2}+3\right)-\left(\dfrac{1}{12}+5\right)\)

\(=7+\left(\dfrac{7}{12}-\dfrac{6}{12}+3\right)-\left(\dfrac{1}{12}+5\right)\)

\(=7+3+\dfrac{1}{12}-\dfrac{1}{12}-5\)

=10-5=5

= 5

\(-\dfrac{1}{12}-\left(-\dfrac{1}{10}\right)\\ =-\dfrac{1}{12}+\dfrac{1}{10}\\ =\dfrac{-5}{60}+\dfrac{6}{60}\\ =\dfrac{-5+6}{60}\\ =\dfrac{1}{60}\)

-1/12--1/10

=-5/60--6/60

=-11/60

\(\left(7-\dfrac{1}{5}+\dfrac{1}{3}\right)-\left(6+\dfrac{9}{5}+\dfrac{4}{3}\right)\\ =7-\dfrac{1}{5}+\dfrac{1}{3}-6-\dfrac{9}{5}-\dfrac{4}{3}\\ =\left(7-6\right)-\left(\dfrac{1}{5}+\dfrac{9}{5}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)\\ =1-2-1\\ =-2\)

\(\left(7-\dfrac{1}{5} +\dfrac{1}{3}\right)-\left(6+\dfrac{9}{5}+\dfrac{4}{3}\right)\)

\(=7-\dfrac{1}{5}+\dfrac{1}{3}-6-\dfrac{9}{5}-\dfrac{4}{3}\)

\(=\left(7-6\right)-\left(\dfrac{1}{5}+\dfrac{9}{5}\right)+\left(\dfrac{1}{3}-\dfrac{4}{3}\right)-\dfrac{1}{3}\)

\(=1-2+\left(-1\right)-\dfrac{1}{3}\)

\(=\left[1+\left(-1\right)\right]-2-\dfrac{1}{3}\)

\(=0-2-\dfrac{1}{3}\)

\(=-2-\dfrac{1}{3}\)

\(=-\dfrac{6}{3}-\dfrac{1}{3}\)

\(=-\dfrac{7}{3}\)

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.\left(2^2.5\right)}\)

\(=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\\ =\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\\ =\dfrac{-2}{6}\\ =-\dfrac{1}{3}\)

\(M=\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right).....\left(100-50^2\right)\\ =\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right).....\left(100-10^2\right).....\left(100-50^2\right)\\ =\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right).....\left(100-100\right).....\left(100-50^2\right)\\ =\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right)....0....\left(100-50^2\right)=0\)

dạng toán này được gọi là gì v ạ?

a: Xét ΔABD vuông tại A và ΔEBD vuông tại E có

BD chung

\(\widehat{ABD}=\widehat{EBD}\)

Do đó: ΔBAD=ΔBED

=>DA=DE

b: DA=DE

=>D nằm trên đường trung trực của AE(1)

Ta có: BA=BE

=>B nằm trên đường trung trực của AE(2)

Từ (1),(2) suy ra BD là đường trung trực của AE

mà BD cắt AE tại F

nên F là trung điểm của AE

=>CF là đường trung tuyến của ΔAEC

nhanh giúp với ạ