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\(1,=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+0,5=1-1+0,5=0,5\\ 2,=-12:\left(-\dfrac{1}{12}\right)^2=12\cdot\dfrac{1}{144}=\dfrac{1}{12}\\ 3,=\dfrac{7}{23}\left(-\dfrac{23}{6}\right)=-\dfrac{7}{6}\\ 4,=\dfrac{7}{5}\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right)=\dfrac{7}{5}\cdot10=14\\ 5,=\dfrac{17}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2=\dfrac{17}{12}\cdot\left(\dfrac{1}{20}\right)^2=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\\ 6,=\dfrac{5}{3}\left(-16\dfrac{2}{7}+28\dfrac{2}{7}\right)=\dfrac{5}{3}\cdot12=20\\ 7,=\left(3-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=-14\\ 8,=\left[\dfrac{1}{9}\cdot\left(-9\right)\right]^{25}-\dfrac{2}{3}\cdot\dfrac{1}{4}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)
\(9,=\dfrac{3}{5}:\left(-\dfrac{7}{30}\right)+\dfrac{3}{5}:\left(-\dfrac{7}{5}\right)=\dfrac{3}{5}\left(-\dfrac{30}{7}-\dfrac{5}{7}\right)=\dfrac{3}{5}\left(-5\right)=-3\\ 10,=5,7\cdot\left(-10\right)=-57\\ 11,=10\cdot\dfrac{1}{10}\cdot\dfrac{4}{3}+21-\dfrac{1}{3}=\dfrac{4}{3}-\dfrac{1}{3}+21=1+21=22\\ 12,=\dfrac{2^{10}}{2^{10}}-\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
a: Xét ΔIMC vuông tại I và ΔINC vuông tại I có
IM=IN
CI chung
Do đó: ΔIMC=ΔINC
b: Xét ΔCKB có
M là trung điểm của BC
MN//KB
Do đó: N là trung điểm của CK
\(a,\Leftrightarrow2x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(b,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3+x>0\\2x-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\2x-5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{5}{2}\\x< -3\end{matrix}\right.\)
\(c,\Leftrightarrow x\left(x+3\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-3< x< 0\)
\(d,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\x+5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-3\\x< -5\end{matrix}\right.\)
\(e,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-2x\ge0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-2x\le0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x< 1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< x\le\dfrac{3}{2}\)
Bài 1:
a, \(\dfrac{2}{3}\) + \(\dfrac{1}{5}\). \(\dfrac{10}{7}\)
= \(\dfrac{2}{3}\) + \(\dfrac{2}{7}\)
= \(\dfrac{20}{21}\)
b, \(\dfrac{7}{12}\) - \(\dfrac{27}{7}\). \(\dfrac{1}{18}\)
= \(\dfrac{7}{12}\) - \(\dfrac{3}{14}\)
= \(\dfrac{31}{84}\)
c, \(\dfrac{3}{10}\). \(\dfrac{-5}{6}\) - \(\dfrac{1}{8}\)
= - \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\)
= - \(\dfrac{3}{8}\)
d, - \(\dfrac{4}{9}\): \(\dfrac{8}{3}\) + \(\dfrac{1}{18}\)
= - \(\dfrac{1}{6}\) + \(\dfrac{1}{18}\)
= - \(\dfrac{1}{9}\)
e, {[(\(\dfrac{1}{2}\) - \(\dfrac{2}{3}\))2 : 2 ] - 1}. \(\dfrac{4}{5}\)
= {[ (-\(\dfrac{1}{6}\))2 : 2] - 1}. \(\dfrac{4}{5}\)
= { [\(\dfrac{1}{36}\) : 2] - 1}. \(\dfrac{4}{5}\)
= { \(\dfrac{1}{72}\) - 1}. \(\dfrac{4}{5}\)
=- \(\dfrac{71}{72}\).\(\dfrac{4}{5}\)
= -\(\dfrac{71}{90}\)
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:
\(\dfrac{x}{20}=\dfrac{y}{10}=\dfrac{4z}{60}=\dfrac{x+4z}{20+60}=\dfrac{240}{80}=3\)
\(\Rightarrow x=3\cdot20=60\)
\(y=3\cdot10=30\)
\(z=3\cdot15=45\)
Câu 5:
\(\dfrac{13}{6}+x=-2,4\)
\(\Rightarrow\dfrac{13}{6}+x=-\dfrac{12}{5}\)
\(\Rightarrow x=-\dfrac{12}{5}-\dfrac{13}{6}\)
\(\Rightarrow x=-\dfrac{137}{30}\)
Câu 6:
\(3,7-x=\dfrac{7}{10}\)
\(\Rightarrow\dfrac{37}{10}-x=\dfrac{7}{10}\)
\(\Rightarrow x=\dfrac{37}{10}-\dfrac{7}{10}\)
\(\Rightarrow x=3\)
Câu 7:
\(\dfrac{3}{7}+x=\dfrac{2}{14}\)
\(\Rightarrow\dfrac{3}{7}+x=\dfrac{1}{7}\)
\(\Rightarrow x=\dfrac{1}{7}-\dfrac{3}{7}\)
\(\Rightarrow x=-\dfrac{2}{7}\)
Câu 8:
\(\dfrac{3}{7}\cdot y=\dfrac{-2}{5}\)
\(\Rightarrow y=\dfrac{-2}{5}:\dfrac{3}{7}\)
\(\Rightarrow y=\dfrac{-2}{5}\cdot\dfrac{7}{3}\)
\(\Rightarrow y=-\dfrac{14}{15}\)
Câu 2:
a, Vì m⊥MN và n⊥MN nên m//n
b, Vì m//n nên \(\widehat{D_1}=\widehat{C}=45^0\) (so le trong)
c, Vì m//n nên \(\widehat{D_1}=\widehat{C_1}\) (đồng vị)