`a.`Ta có: \(AM+5=MB\)

ABCD là hình thoi \(\Rightarrow AB=\dfrac{C_{ABCD}}{4}=\dfrac{60}{4}=15\left(cm\right)\)

Mà \(AB=AM+MB\)

\(\Rightarrow15=AM+AM+5\)

\(\Rightarrow AM=5\left(cm\right);BM=10\left(cm\right)\)

Tứ giác MBCN là hình bình hành \(\Rightarrow MN=BC=15\left(cm\right)\)

                                                         \(MB=NC=10\left(cm\right)\)

\(\Rightarrow C_{MBCN}=MN+NC+BC+MB\)

                 \(=15+10+15+10=50\left(cm\right)\)

`b.` Đường cao hình thoi ABCD: \(216:15=14,4\left(cm\right)\)

Kẻ đường cao DA

`=>` DA bằng đường cao hình thoi

\(\Rightarrow S_{AMND}=DA.AM=14,4.5=72\left(cm^2\right)\)

\(\left(\dfrac{1}{2}\right)\times\left(-\dfrac{1}{2}\right)^6=\left(\dfrac{1}{2}\right)^7=\dfrac{1}{128}\)

\(\left(\dfrac{2}{3}\right)^3\times\left(-\dfrac{2}{3}\right)^4=\left(\dfrac{2}{3}\right)^7=\dfrac{128}{2187}\)

\(\left(\dfrac{4}{5}\right)^5\times\left(-\dfrac{4}{5}\right)^4=\left(\dfrac{4}{5}\right)^9\)

9.\(\sqrt{x}>1\) ; \(\left(x\ge0\right)\)

\(\Leftrightarrow\left(\sqrt{x}\right)^2>1^2\)

\(\Leftrightarrow x>1\)

Vậy \(S=\left\{x|x>1\right\}\)

10.\(-\sqrt{2x}>-3\) ; \(x\ge0\)

\(\Leftrightarrow\sqrt{2x}< 3\)

\(\Leftrightarrow2x< 9\)

\(\Leftrightarrow x< \dfrac{9}{2}\)

Vậy \(S=\left\{x|0\le x< \dfrac{9}{2}\right\}\)

11.\(\sqrt{2-x}< 4\) ; \(x\le2\)

\(\Leftrightarrow2-x< 16\)

\(\Leftrightarrow-x< 14\)

\(\Leftrightarrow x>-14\)

Vậy \(S=\left\{x|-14< x\le2\right\}\)

12.\(-\sqrt{2+x}< 4\)  

\(\Rightarrow x\ge-2\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,25      0,5         0,25      0,25     ( mol )

\(m_{Zn}=0,25.65=16,25\left(g\right)\)

\(m_{dd\left(HCl\right)}=\dfrac{0,5.36,5}{20\%}=91,25\left(g\right)\)

\(m_{ddspứ}=16,25+91,25-0,25.2=107\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{0,25.136}{107}.100=31,77\%\)

4.

a.\(A=8.32^4=2^3.\left(2^5\right)^4=2^3.2^{20}=2^{23}\)

b.\(B=27^3.9^4.243^2=\left(3^3\right)^3.\left(3^2\right)^4.\left(3^5\right)^2=3^9.3^8.3^{10}=3^{27}\)

c.\(C=625:5^3=5^4:5^3=5\)

d.\(D=11^5:121=11^5:11^2=11^3\)

5.

\(25< 5^n< 625\)

\(\Rightarrow5^2< 5^n< 5^4\)

\(\Rightarrow n=3\)

\(-4\dfrac{5}{7}.33\dfrac{1}{3}+4\dfrac{5}{7}.47\dfrac{1}{3}\)

\(=4\dfrac{5}{7}.\left(-33\dfrac{1}{3}+47\dfrac{1}{3}\right)\)

\(=4\dfrac{5}{7}.\left(-\dfrac{100}{3}+\dfrac{142}{3}\right)\)

\(=4\dfrac{5}{7}.14\)

\(=\dfrac{33}{7}.14\)

\(=66\)

ũa r Sun nam hay nữ?:"))

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,15     0,3                      0,15  ( mol )

\(m_{Zn}=0,15.65=9,75\left(g\right)\)

\(m_{Cu}=16,25-9,75=6,5\left(g\right)\)

\(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{9,75}{16,25}.100=60\%\\\%m_{Cu}=100\%-60\%=40\%\end{matrix}\right.\)

\(C_{M\left(HCl\right)}=\dfrac{0,3}{0,15}=2\left(M\right)\)

Vụ gì?:")) hãaa

\(420-\left\{121:\left[56-\left(5+2^3\times5^1\right)\right]\right\}\)

\(=420-\left\{121:\left[56-45\right]\right\}\)

\(=420-\left(121:11\right)\)

\(=420-11\)

\(=409\)