Question

bài 1:

a, (-\(2\frac{1}{5}\)).(\(\frac{-9}{11}\)).(-1\(\frac{1}{14}\)).\(\frac{2}{5}\)

b, [\(\frac{5}{3}\)- (-\(\frac{1}{4}\)): 1\(\frac{1}{5}\)]: (\(\frac{5}{8}\)+\(\frac{9}{4}\))

c,-66. (\(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{11}\))+124. -37+ 63. (-123)

d, C= \(\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}\). \(\frac{\frac{1}{3}-0,25+0,2}{1\frac{1}{6}-0,875+0,7}\)+ \(\frac{6}{7}\)

e, D= (\(\frac{1}{4}\)-1). (\(\frac{1}{9}\)-1).(\(\frac{1}{16}\)-1)....(\(\frac{1}{81}\)-1).(\(\frac{1}{100}\)-1)

bài 2:

a, x: (\(\frac{2}{9}\)-\(\frac{1}{5}\))=\(\frac{8}{16}\)

b,(2x-1).(2x+3)=0

c,\(\frac{4-3x}{2x+5}\)=0

d,(x-2). (x+\(\frac{2}{3}\))≥0

e, \(\frac{x-2}{3x+2}\)<0

MỌI NGƯỜI LÀM NHANH GIÚP MÌNH VỚI. MÌNH ĐANG CẦN GẤP TRONG 3H CHIỀU NAY

Answer 1

Bài 2:

a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)

\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)

b) \(\left(2x-1\right).\left(2x+3\right)=0\)

\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)

\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)

d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

Answer 2

Bài 2:

a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)

=> \(x:\frac{1}{45}=\frac{1}{2}\)

=> \(x=\frac{1}{2}.\frac{1}{45}\)

=> \(x=\frac{1}{90}\)

Vậy \(x=\frac{1}{90}.\)

b) \(\left(2x-1\right).\left(2x+3\right)=0\)

=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)

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