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Ta có:
\(\dfrac{x}{10}=\dfrac{y}{5}\)
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}\) \(\left(1\right)\)
\(\dfrac{y}{2}=\dfrac{z}{3}\)
\(\Rightarrow\dfrac{y}{10}=\dfrac{z}{15}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{z}{15}\)
Lại có:
\(\dfrac{z}{15}=\dfrac{4z}{60}\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{4z}{60}=\dfrac{x+4z}{20+60}=\dfrac{240}{80}=3\)
\(\Rightarrow x=3\cdot20=60\)
\(y=3\cdot10=30\)
\(z=3\cdot15=45\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xet tam giac BDC va tam giac CEB ta co
^BDC = ^CEB = 900
BC _ chung
^BCD = ^CBE ( gt )
=> tam giac BDC = tam giac CEB ( ch - gn )
=> ^DBC = ^ECB ( 2 goc tuong ung )
Ta co ^B - ^DBC = ^ABD
^C - ^ECB = ^ACE
=> ^ABD = ^ACE
Xet tam giac IBE va tam giac ICD
^ABD = ^ACE ( cmt )
^BIE = ^CID ( doi dinh )
^BEI = ^IDC = 900
Vay tam giac IBE = tam giac ICD (g.g.g)
c, Do BD vuong AC => BD la duong cao
CE vuong BA => CE la duong cao
ma BD giao CE = I => I la truc tam
=> AI la duong cao thu 3
=> AI vuong BC
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow2x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(b,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3+x>0\\2x-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\2x-5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{5}{2}\\x< -3\end{matrix}\right.\)
\(c,\Leftrightarrow x\left(x+3\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-3< x< 0\)
\(d,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\x+5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-3\\x< -5\end{matrix}\right.\)
\(e,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-2x\ge0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-2x\le0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x< 1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< x\le\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\\ =\left(\dfrac{1}{2}+\dfrac{5}{18}\right)+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)-\dfrac{4}{9}\\ =\left(\dfrac{9}{18}+\dfrac{5}{18}\right)+\dfrac{19}{19}-\dfrac{4}{9}\\ =\dfrac{14}{18}+1-\dfrac{4}{9}\\ =\dfrac{7}{9}+1-\dfrac{4}{9}\\ =\left(\dfrac{7}{9}-\dfrac{4}{9}\right)+1\\ =\dfrac{3}{9}+1\\ =\dfrac{1}{3}+1\\ =\dfrac{4}{3}\)
\(c,\dfrac{-20}{23}+\dfrac{2}{3}-\dfrac{3}{23}+\dfrac{2}{5}+\dfrac{7}{15}\\ =\left(-\dfrac{20}{23}-\dfrac{3}{23}\right)+\left(\dfrac{2}{5}+\dfrac{7}{15}\right)+\dfrac{2}{3}\\ =-\dfrac{23}{23}+\left(\dfrac{6}{15}+\dfrac{7}{15}\right)+\dfrac{2}{3}\\ =-1+\dfrac{13}{15}+\dfrac{2}{3}\\ =-\dfrac{15}{15}+\dfrac{13}{15}+\dfrac{10}{15}\\ =\dfrac{8}{15}\)
\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\\ =\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}.\dfrac{-7}{11}\\ =-\dfrac{35}{77}\\ =-\dfrac{5}{11}\)
\(f,\dfrac{2}{11}.\dfrac{-5}{4}+\dfrac{-9}{11}.\dfrac{5}{4}+1\dfrac{3}{4}\\ =-\dfrac{2}{11}.\dfrac{5}{4}+\dfrac{-9}{11}.\dfrac{5}{4}+\dfrac{7}{4}\\=\dfrac{5}{4}.\left(-\dfrac{2}{11}+\dfrac{-9}{11}\right)+\dfrac{7}{4}\\ =\dfrac{5}{4}.1+\dfrac{7}{4}\\ =\dfrac{5}{4}+\dfrac{7}{4}\\=\dfrac{12}{4}\\ =3\)
\(h,\dfrac{7}{4}\cdot\dfrac{29}{5}-\dfrac{7}{5}\cdot\dfrac{9}{4}+3\dfrac{2}{13}\\ =\dfrac{7}{4}\cdot\dfrac{29}{5}-\dfrac{7}{4}\cdot\dfrac{9}{5}+\dfrac{41}{13}\\ =\dfrac{7}{4}\cdot\left(\dfrac{29}{5}-\dfrac{9}{5}\right)+\dfrac{41}{13}\\ =\dfrac{7}{4}\cdot\dfrac{20}{5}+\dfrac{41}{13}\\ =\dfrac{7}{4}.4+\dfrac{41}{13}\\ =\dfrac{28}{4}+\dfrac{41}{13}\\ =7+\dfrac{41}{13}\\ =\dfrac{132}{13}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(c,\Rightarrow\left|x-\dfrac{1}{9}\right|=-\dfrac{4}{5}\\ \Rightarrow x\in\varnothing\left(\left|x-\dfrac{1}{9}\right|\ge0>-\dfrac{4}{5}\right)\\ d,\Rightarrow\left\{{}\begin{matrix}3x-2=0\\4y-7=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{7}{4}\end{matrix}\right.\\ e,\Rightarrow\left\{{}\begin{matrix}2x+1=0\\x-y=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=y=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow x=y=-\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Dạ là
a)góc BCB+g ABC+ g ADC
b)g BCD=90°
c)góc DAC=2 góc ACD
d)góc BCD=60°
Câu nào sai ạ?
![](https://rs.olm.vn/images/avt/0.png?1311)
tam giác KGI vuông tại K có
góc GIK = 90 độ - góc KGI
góc GIK = 90 độ - 70 độ = 20 độ
=> góc IFJ = 90 độ - góc GIK
góc IFJ = 90 độ - 20 độ =70 độ
góc HFI = 180 độ - góc IFJ
góc HFI = 180 độ - 70 độ = 110 độ
G, 1/1x2 +1/2x3 +1/3x4 +...........+1/2003x2004
= (1 - 1/2 )+(1/2 - 1/3)+............+(1/2003 -1/2004)
=1 -1/2 + 1/2 -1/3 +...............+ 1/2003 -1/2004
=1+(-1/2 +1/2)+(-1/3 +1/3) +.............+(-1/2003 +1/2003) -1/2004
=1-1/2004
=2003/2004
f, Đặt A= 1/1x3 +1/3x5 +........+1/2003x2005
=>2A=2/1x3 +2/3x5+.....+2/2003x2005
<=>2A=(1-1/3)+(1/3-1/5)+............+(1/2003-1/2005)
<=>2A=1-1/3+1/3-1/5+.......+1/2003-1/2005
<=>2A=1+(-1/3+1/3)+(-1/5+1/5)+...........+(-1/2003+1/2003)-1/2005
<=>2A= 1-1/2005
<=>2A=2004/2005
=> A= 2A:2=2004/2005 : 2=1002/2005
Chúc em học tốt nha :3