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Bài 1: - \(\dfrac{5}{7}\) x \(\dfrac{31}{33}\) + \(\dfrac{-5}{7}\) x \(\dfrac{2}{33}\) + 2\(\dfrac{5}{7}\)
= - \(\dfrac{5}{7}\) \(\times\) ( \(\dfrac{31}{33}\) + \(\dfrac{2}{33}\)) + 2 + \(\dfrac{5}{7}\)
= - \(\dfrac{5}{7}\) + 2 + \(\dfrac{5}{7}\)
= 2
2, \(\dfrac{3}{14}\): \(\dfrac{1}{28}\) - \(\dfrac{13}{21}\): \(\dfrac{1}{28}\) + \(\dfrac{29}{42}\): \(\dfrac{1}{28}\) - 8
= (\(\dfrac{3}{14}\) - \(\dfrac{13}{21}\) + \(\dfrac{29}{42}\)) : \(\dfrac{1}{28}\) - 8
= \(\dfrac{2}{7}\) x 28 - 8
= 8 - 8
= 0
Bài 1:
b: Ta có: \(18^n:2^n=\left(\sqrt{81}\right)^2\)
\(\Leftrightarrow9^n=81\)
hay n=2
\(\left\{{}\begin{matrix}\widehat{AOD}+\widehat{BOC}=180^0\\\widehat{AOD}=\widehat{BOC}\left(đối.đỉnh\right)\end{matrix}\right.\Rightarrow\widehat{AOD}=\widehat{BOC}=90^0\\ \Rightarrow AB\perp CD\\ \Rightarrow\widehat{AOD}=\widehat{BOD}=\widehat{BOC}=\widehat{AOC}=90^0\)
Ta có: \(\widehat{AOD}=\widehat{BOC}\)(đối đỉnh)
Mà \(\widehat{AOD}+\widehat{BOC}=100^0\)
\(\Rightarrow\widehat{AOD}=\widehat{BOC}=50^0\)
Ta có: \(\widehat{AOD}+\widehat{AOC}=180^0\)(kề bù)
\(\Rightarrow\widehat{AOC}=180^0-\widehat{AOD}=130^0\)
\(\Rightarrow\widehat{DOB}=\widehat{AOC}=130^0\)(đối đỉnh)
a: Xét ΔADB và ΔADC có
AD chung
DB=DC
AB=AC
Do đó: ΔABD=ΔACD
Câu 6
a) Ta có: \(\widehat{A}=90^0\) ⇒a⊥c
a//b, a⊥c ⇒b⊥c
b) Ta lại có: M1+N1=1800(trong cùng phía)
1200+N1=1800
N1=1800-1200=600
Bài 3:
1, Áp dụng t/c dtsbn:
\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{z-x}{3-6}=\dfrac{-21}{-3}=7\\ \Rightarrow\left\{{}\begin{matrix}x=42\\y=28\\z=21\end{matrix}\right.\)
2, Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)
Bài 4:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+y+z}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}=\dfrac{130}{\dfrac{13}{12}}=120\)
Do đó: x=60; y=40; z=30
câu đầu:
\(2^{3n}:2^{n+2}=\left[\left(\sqrt{2}\right)^2\right]^2\\ \Leftrightarrow2^{3n-\left(n+2\right)}=2^2\\ \Leftrightarrow3n-\left(n+2\right)=2\\ \Leftrightarrow2n=4\\ \Leftrightarrow n=2\)
câu sau:
\(8^n:2^n.4=4\\ \Leftrightarrow8^n:2^n=4:4=1\\ \Leftrightarrow2^{3n}:2^n=2^0\\ \Leftrightarrow2^{3n-n}=2^0\\ \Leftrightarrow3n-n=0\\ \Leftrightarrow n=0\)