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\(1.\) \(\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{7+32-24}{36}=\dfrac{5}{12}.\)
\(2.\) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}=\dfrac{-9-2-7}{18}+\dfrac{4+3}{7}=\dfrac{-18}{18}+\dfrac{7}{7}=-1+1=0.\)
\(3.\) \(-\dfrac{10}{3}+\dfrac{13}{10}-\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{13+1}{10}+\dfrac{-20-1}{6}=\dfrac{14}{10}+\dfrac{-21}{6}=\dfrac{7}{5}-\dfrac{7}{2}=-\dfrac{21}{10}.\)
\(4.\) \(\dfrac{10}{17}-\dfrac{5}{13}-\left(-\dfrac{7}{17}\right)-\dfrac{8}{13}+\dfrac{11}{25}=\dfrac{10+7}{17}+\dfrac{-5-8}{13}+\dfrac{11}{25}=\dfrac{17}{17}-\dfrac{13}{13}+\dfrac{11}{25}=\dfrac{11}{25}.\)
1: \(=\dfrac{7}{36}+\dfrac{32}{36}+\dfrac{-24}{36}=\dfrac{15}{36}=\dfrac{5}{12}\)
2: \(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\dfrac{3}{7}+\dfrac{4}{7}\)
\(=\dfrac{-9-2-7}{18}+1=-1+1=0\)
3: \(=\left(-\dfrac{10}{3}-\dfrac{1}{6}\right)+\dfrac{14}{10}\)
\(=-\dfrac{21}{6}+\dfrac{14}{10}=\dfrac{-7}{2}+\dfrac{14}{10}=\dfrac{-35+14}{10}=-\dfrac{21}{10}\)
4: \(=\left(\dfrac{10}{17}+\dfrac{7}{17}\right)-\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\dfrac{11}{25}=\dfrac{11}{25}\)
a: =30-22=8
b: =10*(23+17)=10*40=400
c: =21*3-7*(-14)
=63+98=161
d: =-20-[10*10*5^2+16]
=-20-100*25-16
=-36-2500
=-2536
a) \(4^8\cdot4^4=\left(2^2\right)^8\cdot\left(2^2\right)^4=2^{16}\cdot2^8=2^{16+8}=2^{24}\)
b) \(5^{12}\cdot7-5^{11}\cdot10\)
\(=5^{11}\cdot\left(5\cdot7-10\right)=5^{11}\cdot\left(35-10\right)=5^{11}\cdot25\)
\(=5^{11}\cdot5^2=5^{11+2}=5^{13}\)
d) \(27^{16}:9^{10}\)
\(=\left(3^3\right)^{16}:\left(3^2\right)^{10}=3^{48}:3^{20}=3^{48-20}=3^{28}\)
e) \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5^{9-8}=5\)
f) \(24^4:3^4-32^{12}:16^{12}\)
\(=\left(24:4\right)^4-\left(32:16\right)^{12}\)
\(=6^4-2^{12}\)
\(=2^4\cdot\left(3^4-2^8\right)=2^4\cdot-175=-2800\)
\(a,x+\dfrac{5}{7}=\dfrac{10}{-35}\\ x=\dfrac{-10}{35}-\dfrac{5}{7}\\ x=-1\\ b,\dfrac{2}{x}=\dfrac{x}{8}\\ x.x=2.8\\ x^2=16\\ x=\pm4\)
a, \(x\) + \(\dfrac{5}{7}\) = \(\dfrac{10}{-35}\)
\(x\) = \(-\dfrac{10}{35}\) - \(\dfrac{5}{7}\)
\(x\) = -1
b, \(\dfrac{2}{x}\) = \(\dfrac{x}{8}\)
\(x^2\) = 16
\(\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -4; 4}
c, \(\dfrac{x-3}{2}\) = \(\dfrac{32}{x-3}\)
(\(x-3\))(\(x-3\)) = 32\(\times\) 2
(\(x-3\))(\(x\) - 3) = 64
(\(x-3\))2 = 82
\(\left[{}\begin{matrix}x-3=-8\\x-3=8\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-8+3\\x=8+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=11\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -5; 11}
cho mik hỏi bài này làm thế nào
\(10\cdot7^2-10\cdot5+10+32\)
\(=10\cdot\left(7^2-5+1\right)+32\)
\(=10\cdot45+32\)
\(=450+32\)
\(=482\)
hok tot