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a: Xét ΔBAH vuông tại A và ΔBDH vuông tại D có
BH chung
\(\widehat{ABH}=\widehat{DBH}\)
Do đó: ΔBAH=ΔBDH
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Gọi số xe đội 1,2,3 lần lượt là a,b,c(xe;a,b,c∈N*)
Vì số hàng bằng nhau nên số xe tỉ lệ nghịch với số ngày
Do đó \(3a=4b=5c\Rightarrow\dfrac{a}{\dfrac{1}{3}}=\dfrac{b}{\dfrac{1}{4}}=\dfrac{c}{\dfrac{1}{5}}\)
Áp dụng tc dtsbn:
\(\dfrac{a}{\dfrac{1}{3}}=\dfrac{b}{\dfrac{1}{4}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b+c}{\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}}=\dfrac{94}{\dfrac{47}{60}}=120\\ \Rightarrow\left\{{}\begin{matrix}a=40\\b=30\\c=24\end{matrix}\right.\)
Vậy ...
Bài 2:
Gọi số máy 3 đội lần lượt là a,b,c(máy;a,b,c∈N*)
Vì số máy tỉ lệ nghịch với thời gian nên \(3a=5b=6c\Rightarrow\dfrac{3a}{30}=\dfrac{5b}{30}=\dfrac{6c}{30}\Rightarrow\dfrac{a}{10}=\dfrac{b}{6}=\dfrac{c}{5}\)
Áp dụng tc dtsbn:
\(\dfrac{a}{10}=\dfrac{b}{6}=\dfrac{c}{5}=\dfrac{a+b+c}{10+6+5}=\dfrac{42}{21}=2\\ \Rightarrow\left\{{}\begin{matrix}a=20\\b=12\\c=10\end{matrix}\right.\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(\dfrac{17}{14}-\dfrac{5}{7}\right)+\left(\dfrac{4}{9}-\dfrac{17}{18}\right)+\dfrac{11}{125}=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{11}{125}=\dfrac{11}{125}\\ b,=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}=1-1+\dfrac{1}{2}=\dfrac{1}{2}\\ c,=\left(-12\right)\left(-\dfrac{1}{12}\right)^2=\dfrac{12}{12^2}=\dfrac{1}{12}\\ d,=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)=-\dfrac{891}{25}:4=-\dfrac{891}{100}\\ e,\dfrac{17}{12}\cdot\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\\ f,=\dfrac{3}{8}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{3}{8}\cdot\left(-14\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{3}{8}\cdot\dfrac{43}{3}+\dfrac{1}{4}=\dfrac{43}{8}+\dfrac{1}{4}=\dfrac{45}{8}\\ g,=\dfrac{5}{3}\left(-16\dfrac{2}{7}+28\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12\dfrac{2}{7}=\dfrac{5}{3}\cdot\dfrac{86}{7}=\dfrac{430}{21}\)
\(h,=\dfrac{7}{2}\left(-\dfrac{3}{4}+\dfrac{5}{13}-\dfrac{9}{4}-\dfrac{8}{13}\right)=\dfrac{7}{2}\left(-3-\dfrac{3}{13}\right)=\dfrac{7}{2}\left(-\dfrac{42}{13}\right)=-\dfrac{147}{13}\\ i,=9^2-25^2+8^2=81-625+64=-480\\ k,=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}-\dfrac{1}{4}\right)+4\\ =-1-1-1+4=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\widehat{KDA}+\widehat{KAD}=90^0\left(\Delta AKD.vuông.tại.K\right)\\ \widehat{KAD}+\widehat{KAB}=\widehat{BAC}=90^0\\ \Rightarrow\widehat{KDA}=\widehat{KAB}\\ \left\{{}\begin{matrix}\widehat{KDA}=\widehat{KAB}\\\widehat{AKD}=\widehat{AHB}\left(=90^0\right)\\AD=AB\left(gt\right)\end{matrix}\right.\Rightarrow\Delta HAB=\Delta KDA\left(ch-gn\right)\\ 2,DE//KH\Rightarrow\widehat{EDH}=\widehat{KHD}\left(so.le.trong\right)\\ \left\{{}\begin{matrix}\widehat{EDH}=\widehat{KHD}\\\widehat{DKH}=\widehat{DHE}\left(=90^0\right)\\DH.chung\end{matrix}\right.\Rightarrow\Delta KDH=\Delta EDH\left(ch-gn\right)\)
\(\Rightarrow\widehat{KDH}=\widehat{EHD}\\ 3,\left\{{}\begin{matrix}HA=DK\left(\Delta HAB=\Delta KDA\right)\\DK=EH\left(\Delta KDH=\Delta EDH\right)\end{matrix}\right.\Rightarrow AH=EH\left(=DK\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1,2 làm rồi
\(3,\\ \Rightarrow\left|3x+2\right|=\dfrac{9}{4}+\dfrac{15}{4}=6\\ \Rightarrow\left[{}\begin{matrix}3x+2=6\left(x\ge-\dfrac{2}{3}\right)\\3x+2=-6\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\left(tm\right)\\x=-\dfrac{8}{3}\left(tm\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}\)
Áp dụng TCDTSBN ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2z^2}{32}=\dfrac{x^2+y^2-2z^2}{4+9-32}=\dfrac{-19}{-19}=1\)
\(\dfrac{x^2}{4}=1\Rightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\\ \dfrac{y^2}{9}=1\Rightarrow\left[{}\begin{matrix}y=-3\\y=3\end{matrix}\right.\\ \dfrac{2z^2}{32}=1\Rightarrow\left[{}\begin{matrix}z=-4\\z=4\end{matrix}\right.\)
Vậy \(\left(x,y,z\right)\in\left\{\left(-2;-3;-4\right);\left(2;3;4\right)\right\}\)
viết cụ thể đi bạn
Bài 4:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{8}=\dfrac{b}{7}=\dfrac{c}{6}=\dfrac{d}{5}=\dfrac{a-c}{8-6}=40\)
Do đó: a=320; b=280; c=240; d=200