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a, Nếu x là số nguyên dương
\(\Rightarrow2010.x>0\)
Nếu x là số nguyên âm;
\(\Rightarrow2010.x< 0\)
Nếu x = 0
\(\Rightarrow2010.x=0\)
b, Nếu y là số nguyên dương
\(\Rightarrow-2011.y< 0\)
Nếu y là số nguyên âm
\(\Rightarrow-2011.y>0\)
Nếu y = 0
\(\Rightarrow-2011.y=0\)
hok tốt!!!
a) Nếu x\(\in\)Z ; x > 0 \(\Rightarrow\)2010.x > 0 b ) Nếu y \(\in\)Z ;y > 0 \(\Rightarrow\)-2011.y < 0
Nếu x\(\in\)Z ; x < 0 \(\Rightarrow\)2010.x < 0 Nếu y\(\in\)Z ; y < 0 \(\Rightarrow\)-2010.y > 0
Nếu x=0 \(\Rightarrow\)2010.x=0 Nếu y=0 \(\Rightarrow\)-2011.y=0
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b: Ta có: \(\dfrac{x+2}{5}=\dfrac{3-2x}{11}\)
\(\Leftrightarrow11x+22=15-10x\)
hay \(x=-\dfrac{1}{3}\)
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a: x+1/4=1/5
=>x=1/5-1/4
=>x=-1/20
b: x-1/5=3/20
=>x=3/20+1/5
=>x=3/20+4/20=7/20
c: \(\dfrac{5}{6}-x=1\)
=>x=5/6-1=-1/6
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Bài 2:
c: \(\left(\dfrac{3}{2}+x\right):1\dfrac{2}{5}=\dfrac{1}{2}\cdot\dfrac{3}{5}+0,2\)
=>\(\left(x+\dfrac{3}{2}\right):\dfrac{7}{5}=\dfrac{3}{10}+\dfrac{1}{5}=\dfrac{5}{10}=\dfrac{1}{2}\)
=>\(x+\dfrac{3}{2}=\dfrac{1}{2}\cdot\dfrac{7}{5}=\dfrac{7}{10}\)
=>\(x=\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{7}{10}-\dfrac{15}{10}=-\dfrac{8}{10}=-\dfrac{4}{5}\)
f: \(-\dfrac{7}{5}-\left(\dfrac{2}{3}+x\right)=\dfrac{3}{10}\)
=>\(-\dfrac{7}{5}-\dfrac{2}{3}-x=\dfrac{3}{10}\)
=>\(x=-\dfrac{7}{5}-\dfrac{2}{3}-\dfrac{3}{10}=\dfrac{-42-20-9}{30}=\dfrac{-71}{30}\)
g: \(\left|x-\dfrac{1}{6}\right|-\dfrac{7}{12}=\dfrac{1}{4}\)
=>\(\left|x-\dfrac{1}{6}\right|=\dfrac{1}{4}+\dfrac{7}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{5}{6}\\x-\dfrac{1}{6}=-\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{6}=1\\x=-\dfrac{5}{6}+1=-\dfrac{4}{6}=-\dfrac{2}{3}\end{matrix}\right.\)
l: \(x:\left(\dfrac{1}{5}-\dfrac{7}{10}\right)=-2+\left(-1\dfrac{2}{3}\right)\)
=>\(x:\left(\dfrac{2}{10}-\dfrac{7}{10}\right)=-2+\dfrac{-5}{3}=\dfrac{-11}{3}\)
=>\(x:\dfrac{-1}{2}=-\dfrac{11}{3}\)
=>\(x=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)
k: \(x:\left(\dfrac{1}{7}-\dfrac{3}{14}\right)=-3+\left(-1\dfrac{2}{3}\right)\)
=>\(x:\left(\dfrac{2}{14}-\dfrac{3}{14}\right)=-3-\dfrac{5}{3}=\dfrac{-14}{3}\)
=>\(x:\dfrac{-1}{14}=\dfrac{-14}{3}\)
=>\(x=\dfrac{14}{3}\cdot\dfrac{1}{14}=\dfrac{1}{3}\)
Bài 3:
a: \(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
b: \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{4}-\dfrac{1}{10}=\dfrac{5-2}{20}=\dfrac{3}{20}\)
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b) \(\left(5^{22}.7-5^{21}.10\right):25^{11}\)
\(=5^{21}.\left(5.7-10\right):5^{22}\)
\(=5^{21}.25:5^{22}=5^{21}.5^2:5^{22}=5^{23}:5^{22}=5\)
c) \(\dfrac{3^6.15^5+9^3.15^6}{3^{10}.5^2.2^3}=\dfrac{3^6.3^5.5^5+3^6.3^6.5^6}{3^{10}.5^2.2^3}\)
\(=\dfrac{3^{11}.5^5+3^{12}.5^6}{3^{10}.5^2.2^3}=\dfrac{3^{11}.5^5\left(1+3.5\right)}{3^{10}.5^2.2^3}\)
\(=\dfrac{3^1.5^3.16}{2^3}=\dfrac{3^1.5^3.2^4}{2^3}=3.125.2=750\)
d) \(\dfrac{11.27^7.3^8-9^{15}}{\left(2.3^{14}\right)^2}=\dfrac{11.3^{21}.3^8-3^{30}}{2^2.3^{28}}=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\dfrac{3^1.8}{2^2}=\dfrac{3.2^3}{2^2}=3.2=6\)
ktr?
hình như là đề ôn thì phải