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Bài 1
\(=-\frac{21}{60}=-\frac{7}{20}\)
\(b,\left(2-\frac{1}{3}\right)^2+|-\frac{5}{6}|+\frac{-7}{12}-\frac{25}{9}\)
\(=\frac{25}{9}+\frac{5}{6}-\frac{7}{12}-\frac{25}{9}\)
\(=\left(\frac{25}{9}-\frac{25}{9}\right)+\left(\frac{5}{6}-\frac{7}{12}\right)\)
\(=0+\frac{1}{4}=\frac{1}{4}\)
Bài 2
\(a,x+\frac{2}{5}=-\frac{3}{10}\)
\(x=-\frac{3}{10}-\frac{2}{5}\)
\(x=-\frac{3}{10}-\frac{4}{10}\)
\(x=-\frac{7}{10}\)
\(b,|\frac{2}{3}+x|=\frac{5}{7}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}+x=\frac{5}{7}\\\frac{2}{3}+x=-\frac{5}{7}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{7}-\frac{2}{3}\\x=-\frac{5}{7}-\frac{2}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{21}\\x=-\frac{29}{21}\end{cases}}}\)
== chắc trog quá trình lm lỡ xóa đó
\(a,-\frac{3}{4}.\frac{7}{15}\)
\(=-\frac{21}{60}=-\frac{7}{20}\)
với lại bài trên mk tính nhẩm ko bấm máy sai == sửa giúp
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( 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) =
\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)
Ta xét 2 trường hợp
\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)
tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian
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, \(\frac{1}{7}+x=\frac{25}{14}+\frac{5}{14}\)
\(\frac{1}{7}+x=\frac{15}{7}\)
\(x=\frac{15}{7}+\frac{(-1)}{7}\)
\(x=\frac{14}{7}=2\).
a)\(=\frac{-7}{10}+\frac{-1}{5}=\frac{-7}{10}+\frac{-2}{10}=\frac{-9}{10}\)
b)5x=125
5x=53=>x=3