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#)Giải :
\(\frac{-5}{12}< \frac{a}{5}< \frac{1}{4}\Leftrightarrow\frac{-25}{60}< \frac{12a}{60}< \frac{15}{60}\Leftrightarrow-25< 12a< 15\)
\(\Leftrightarrow12a\in\left\{\pm12;-24\right\}\)
\(\Leftrightarrow a\in\left\{\pm1;2\right\}\)
Bài giải
Ta có :
\(-\frac{5}{12}< \frac{a}{5}< \frac{1}{4}\)
\(\Leftrightarrow\text{ }-\frac{25}{60}< \frac{12a}{60}< \frac{15}{60}\) \(\Rightarrow\text{ }-25< 12a< 15\)
\(\Rightarrow\text{ }-1,25< a< 1,25\)
\(\text{Do }a\in Z\text{ }\Rightarrow\text{ }x\in\left\{-1\text{ ; }0\text{ ; }1\right\}\)
A=24x27-23/24+23x27
A=(23+1)x27-23/24+23x27
A=23x27+27-23/24+23x27
A=23x27+4/24+23x27
A=4/24
A=1/6
A=\(\frac{23x27+27-23}{23x27+24}\)
A=\(\frac{23x27+4}{23x27+24}\)
A=\(\frac{625}{645}\)
\(\frac{3}{1}+\frac{4}{5}=\frac{15}{5}+\frac{4}{5}=\frac{19}{5}\)
Trước hết :
7,5,1,6,2,4/3,2,4,8,2,5 = 136,8/38,4
=> : 3/2,2
<=> =3/4 .
mk làm rồi !
\(\frac{7,5\cdot1,6\cdot2,4}{3,2\cdot4,8\cdot2,5}=\frac{136,8}{38,4}\)
a. Ta có : \(x-8.2018⋮4\) mà \(8.2018⋮4\Rightarrow x⋮4\Rightarrow x\inƯ\left(4\right)=\left\{1;2;4\right\}\)
b.Ta có : \(75.2015-x⋮5\)mà \(75.2015⋮5\Rightarrow x⋮5\Rightarrow x\inƯ\left(5\right)=\left\{1;5\right\}\)
\(P=\frac{1}{5^2}+\frac{2}{5^3}+\frac{3}{5^4}+\frac{4}{5^5}+...+\frac{11}{5^{12}}\)
\(\Rightarrow\)\(5P=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+\frac{4}{5^4}+...+\frac{11}{5^{11}}\)
\(\Rightarrow\)\(4P=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+\frac{1}{5^4}+...+\frac{1}{5^{11}}-\frac{1}{5^{12}}\)
\(\Rightarrow\)\(20P=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{10}}-\frac{1}{5^{11}}\)
\(\Rightarrow\)\(16P=1-\frac{1}{5^{11}}+\frac{1}{5^{12}}-\frac{1}{5^{11}}\)\(< 1\)
\(\Rightarrow\)\(P< \frac{1}{16}\)
P/s: nguyên tác: https://olm.vn/thanhvien/nhatphuonghocgiot