9652% × 6320% × 63207 ÷ 963 × 6542=
giải hộ tớ nhé
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12:
\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2004}\right)\)
\(B=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{2004}{2004}-\dfrac{1}{2004}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}...\cdot\dfrac{2003}{2004}\)
\(B=\dfrac{1\cdot2\cdot3\cdot...\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2003\cdot2004}\)
\(B=\dfrac{1}{2004}\)
`12) B=(1-1/2)(1-1/3)(1-1/4)(1-1/5)....(1-1/2003)(1-1/2004)`
`B=1/2 . 2/3 . 3/4 . 4/5 .... 2002/2003 . 2003/2004`
`B= [1.2.3.4......2002.2003]/[2.3.4.5.....2003.2004]`
`B=1/2004`
`7)`
`a)[254xx399-145]/[254+399xx253]`
`=[(253+1)xx399-145]/[254+399xx253]`
`=[253xx399+399-145]/[254+399xx253]`
`=[254+253xx399]/[254+253xx399]=1`
`b)[5932+6001xx5931]/[5932xx6001-69]`
`=[5932+6001xx5931]/[(5931+1)xx6001-69]`
`=[5932+6001xx5931]/[5931xx6001+6001-69]`
`=[5932+6001xx5931]/[5932+6001xx5931]=1`
Câu 17:
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b+c) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)=n_{MgCl_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\m_{ddHCl}=\dfrac{0,3\cdot36,5}{14,6\%}=75\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\\m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\end{matrix}\right.\)
d) Ta có: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=78,3\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{14,25}{78,3}\cdot100\%\approx18,2\%\)
a) $Mg + 2HCl \to MgCl_2 + H_2$
b) n H2 = n Mg = 3,6/24 = 0,15(mol)
V H2 = 0,15.22,4 = 3,36 lít
c) n HCl = 2n Mg = 0,3(mol)
=> m dd HCl = 0,3.36,5/14,6% = 75(gam)
d)
n MgCl2 = n Mg = 0,15(mol)
Sau phản ứng :
m dd = m Mg + mdd HCl - m H2 = 3,6 + 75 - 0,15.2 = 78,3(gam)
C% MgCl2 = 0,15.95/78,3 .100% = 18,2%
Ta có: \(\left\{{}\begin{matrix}8x-7y=5\\12x+13y=-8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}24x-21y=15\\24x+26y=-16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-47y=31\\8x-7y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-31}{47}\\8x=5+7y=5+7\cdot\dfrac{-31}{47}=\dfrac{18}{47}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{188}\\y=\dfrac{-31}{47}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left(x,y\right)=\left(\dfrac{9}{188};\dfrac{-31}{47}\right)\)
\(\frac{18181818}{81818181}+\frac{7}{9}=\frac{18\cdot\left(101010\right)}{81\cdot\left(101010\right)}+\frac{7}{9}=\frac{18}{81}+\frac{7}{9}=\frac{2}{9}+\frac{7}{9}=1\)
HOK TỐT~
a ) 13/20
B)
C..........................................................
minh dang tính
= 2619291430 nhé
TL
=2619291430
bn nhé
HT