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Ta có:\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{x}{3}=\dfrac{2y}{2.4}=\dfrac{3z}{3.6}\)
Áp dung tcdtsbn , ta có:
\(\dfrac{x}{3}=\dfrac{2y}{2.4}=\dfrac{3z}{3.6}=\dfrac{x+2y-3z}{3+8-18}=\dfrac{-14}{-7}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=8\\z=12\end{matrix}\right.\)
\(-\dfrac{4}{3}\cdot x=\dfrac{2}{3}:\dfrac{7}{12}:\dfrac{4}{18}\)
\(\Rightarrow-\dfrac{4}{3}\cdot x=\dfrac{36}{7}\)
\(\Rightarrow x=\dfrac{\dfrac{36}{7}}{-\dfrac{4}{3}}=-\dfrac{27}{7}\)
-\(\dfrac{4}{3}\).\(x\) = \(\dfrac{2}{3}\): \(\dfrac{7}{12}\):\(\dfrac{4}{18}\)
-\(\dfrac{4}{3}.x\) = \(\dfrac{2}{3}\times\)\(\dfrac{12}{7}\)\(\times\)\(\dfrac{18}{4}\)
-\(\dfrac{4}{3}.\)\(x\)= \(\dfrac{36}{7}\)
\(x\) = \(\dfrac{36}{7}\):(-\(\dfrac{4}{3}\))
\(x\) = - \(\dfrac{27}{7}\)
đường thẳng xx' và yy' vuông góc với nhau nên \(\widehat{xOy}=90^o\)
Ta có \(\widehat{xOy}=90^o\)
xOt + tOy = 90o
1/2 xOt + 1/2 tOy = 45o
1/7 tOy + 1/2 tOy = 45o
tOy = 70o
xOt = 20o
tOy' = 110o
cho hai đường thẳng xx' và yy' vuông góc với nhau và cắt nhau tại điểm O. Vẽ tia Ot nằm giữa hai tia Ox và Oy sao cho 1/2 xOt = 1/7 tOy. Số đo tOy' bằng
giải nhanh hộ mình với ạ
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
\(a,x=\dfrac{18}{z};y=10\Leftrightarrow x:y=\dfrac{18}{z}:10=\dfrac{9}{5z}=9:5z\)
\(b,y=x:\dfrac{9}{5z}=\dfrac{9}{5xz}\)
\(c,x=-2\Leftrightarrow z=-9\Leftrightarrow y=\dfrac{9}{5\cdot\left(-9\right)\cdot\left(-2\right)}=\dfrac{1}{10}\\ x=\dfrac{1}{5}\Leftrightarrow z=90\Leftrightarrow y=\dfrac{9}{5\cdot\dfrac{1}{5}\cdot90}=\dfrac{1}{10}\)
\(\dfrac{x}{7}=\dfrac{18}{14}\)
<=> 14 . x = 18 . 7
<=> 14x = 126
<=> x = \(\dfrac{126}{14}=9\)