\(-18< 24-3x< 39\)

\(\Rightarrow24-39< 3x< 24-\left(-18\right)\)

\(\Leftrightarrow-15< 3x< 42\)

\(\Leftrightarrow-5< x< 14\)

\(\Rightarrow\) có 14 giá trị nguyên không âm

a) \(x\in B\left(12\right);20< x\le50\)

\(x\in\left\{24;36;48\right\}\)

d) \(x⋮12;x\le60\)

\(x\in\left\{0;12;24;36;48;60\right\}\)

g) \(\left\{{}\begin{matrix}x⋮25=5^2\\x⋮90=2.3^2.5\end{matrix}\right.\)

\(\Rightarrow x=2.3^2.5^2=450\)

\(\Rightarrow x\in\left\{0;450\right\}\)

\(D=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+...+\dfrac{5^2}{26.31}\)

\(=5\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+...+\dfrac{5}{26.31}\right)\)

\(=5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{26}-\dfrac{1}{31}\right)\)

\(=5\left(1-\dfrac{1}{31}\right)\)

\(=5.\dfrac{30}{31}=\dfrac{150}{31}\)

Ta có:

Vì tứ giác ABCD là hình vuông

=> AC=BD

Mà AC=10cm

=> BC=10cm

\(\dfrac{1.4+2.6+3.8+4.10+5.12}{5.2+10.3+15.4+20.5+25.6}\)

\(=\dfrac{2\left(1.2+2.3+3.4+4.5+5.6\right)}{5\left(1.2+2.3+3.4+4.5+5.6\right)}\)

\(=\dfrac{2}{5}\)

\(C=\left(x-2\right)^2+\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(MinC=\dfrac{3}{4}\Leftrightarrow x=2;y=\dfrac{1}{2}\)

\(\dfrac{1}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{2001.2003}+\dfrac{1}{2003.2005}+...+\dfrac{1}{2011.2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2003}+\dfrac{1}{2003}-\dfrac{1}{2005}+...+\dfrac{1}{2011}-\dfrac{1}{2013}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2001}-\dfrac{1}{2013}\right)=\dfrac{2}{1342671}\)

Ta có: 

BM là phân giác góc ABC

=> \(\widehat{ABM}=\dfrac{1}{2}\widehat{ABC}\)

Mặt khác:

\(\widehat{ABC}=\widehat{DBE}\) (đối đỉnh)

=> \(\widehat{ABM}=\dfrac{1}{2}\widehat{DBE}\)

Ta có:

\(\left\{{}\begin{matrix}\text{O là trung điểm của AC và HD}\\\widehat{H}=90^o\end{matrix}\right.\)

=> ADHC là hình chữ nhật

=> AD=HC=BH; AD//BC

Xét tứ giác ADHB có: 

\(\left\{{}\begin{matrix}\text{AD//BH}\\AD=BH\end{matrix}\right.\)

=> ADHC là hình bình hành

a) \(3.4.5+21k\left(k\in N\right)\)

Vì \(21k⋮7;3.4.5⋮̸7\)

\(\Rightarrow3.4.5+21k⋮̸7\)

b) \(196.3-2.7^4=7.3.28-2.7^4=7\left(3.28-2.7^3\right)⋮7\)