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Bài 1 : Thực hiện phép tính :
a, \(\frac{4}{5}+1\frac{1}{6}\cdot\frac{3}{4}\)
= \(\frac{4}{5}+\frac{7}{6}\cdot\frac{3}{4}\)
= \(\frac{4}{5}+\frac{7}{8}\)
= \(\frac{32+35}{40}=\frac{67}{40}\)
b, \(\frac{2}{3}:\left(\frac{3}{4}\cdot\frac{4}{3}\right)+2\)
\(=\frac{2}{3}:1+2\)
\(=\frac{2}{3}+2=\frac{2+6}{3}=\frac{8}{3}\)
c, \(\frac{1}{2}\times\left(\frac{2}{3}+\frac{3}{5}\cdot\frac{5}{7}\right)+1\frac{1}{3}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{3}+\frac{9}{35}\right)+\frac{4}{3}\)
\(=\frac{1}{2}\cdot\frac{97}{105}+\frac{4}{3}\)
\(=\frac{97}{210}+\frac{4}{3}=\frac{377}{210}\)
Bài 2 : Tìm \(x\inℤ\), biết :
a, \(\frac{2}{3}< \frac{x}{6}\le\frac{10}{3}\)
\(\Leftrightarrow\frac{4}{6}< \frac{x}{6}\le\frac{20}{6}\)
mà \(x\inℤ\Rightarrow\text{x}\in\) {\(5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20\)}
b, \(\frac{1}{3}+x=1\frac{1}{2}\)
\(\frac{1}{3}+x=\frac{3}{2}\)
\(x=\frac{3}{2}+\frac{\left(-1\right)}{3}\)
\(x=\frac{7}{6}\) (loại vì \(x\notinℤ\))
\(\Rightarrow x\in\varnothing\)
c, \(\frac{1}{7}+x=\frac{25}{14}+\frac{5}{14}\)
\(\frac{1}{7}+x=\frac{15}{7}\)
\(x=\frac{15}{7}+\frac{(-1)}{7}\)
\(x=\frac{14}{7}=2\).
Bài 1
\(a,\left|x\right|=-\left|-\frac{5}{7}\right|=>x\in\varnothing\)
\(b,\left|x+4,3\right|-\left|-2,8\right|=0\)
\(=>\left|x+4,3\right|-2,8=0\)
\(=>\left|x+4,3\right|=0+2,8=2,8\)
\(=>x+4,3=\pm2,8\)
\(=>\hept{\begin{cases}x+4,3=2,8\\x+4,3=-2,8\end{cases}=>\hept{\begin{cases}x=-1,5\\x=-7,1\end{cases}}}\)
\(c,\left|x\right|+x=\frac{2}{3}\)
\(=>\hept{\begin{cases}x+x=\frac{2}{3}\\-x+x=\frac{2}{3}\end{cases}}=>\hept{\begin{cases}x=\frac{1}{3}\\x=-\frac{1}{3}\end{cases}}\)
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
\(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}=\left|-13\right|\)
\(=-8+\frac{1}{2}.8-5+13\)
\(=4\)
\(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(-\frac{2012}{2013}\right)^0\)
\(=\frac{1}{2}.10-\frac{1}{4}+1\)
\(=5-\frac{5}{4}\)
\(=\frac{15}{4}\)
\(\left(-2\right)^3+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+|-13|\)
\(=-12+\frac{1}{2}.8-5+13\)
\(=-12+4-5+13\)
\(=4\)
a) \(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
nên x + 1 = 0 => x = -1
Vậy x = -1
b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(1+\frac{x+4}{2000}+1+\frac{x+3}{2001}=1+\frac{x+2}{2002}+1+\frac{x+1}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}=\frac{2004+x}{2002}+\frac{2004+x}{2003}\)
\(\frac{2004+x}{2000}+\frac{2004+x}{2001}-\frac{2004+x}{2002}-\frac{2004+x}{2003}=0\)
\(\left(2004+x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\ne0\)
nên 2004 + x = 0 => x = -2004
Vậy x = -2004
=))
\(6.\left(-\frac{1}{3}\right)^2-\frac{5}{4}:0,5+3\frac{1}{2}\)
\(=6.\frac{1}{9}-\frac{5}{4}.2+\frac{7}{2}\)
\(=\frac{2}{3}-\frac{5}{2}+\frac{7}{2}\)
\(=-\frac{11}{6}+\frac{7}{2}\)
\(=\frac{5}{3}\)
\(\frac{2017}{2018}.\frac{15}{17}-\frac{32}{17}.\frac{2017}{2018}=\frac{2017}{2018}.\left(\frac{15}{17}-\frac{32}{17}\right)\)
\(=\frac{2017}{2108}.\left(-1\right)=-\frac{2017}{2018}\)