\(n_{CuO}=\dfrac{8}{80}=0,1mol\)

\(PTHH:CuO+HCl\rightarrow CuCl_2+H_2O\)

\(\Rightarrow n_{HCl}=0,1mol\Rightarrow m_{HCl}=0,1.36,5=3,65g\)

\(\Rightarrow C\%_{dd\:HCl}=\dfrac{3,65}{100}.100\%=3,65\%\)

\(ta\:có:\:x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\:\forall x\in R\)

\(do\:đó\:\sqrt{x^2-x+1}\:có\:nghĩa\:\forall x\in R\)

mình giải cách này ko bt đúng hay sai nha :))

\(\left|x-2015\right|+\left|x-2016\right|+\left|x-2017\right|\ge\left|x-2015\right|+\left|x-2017\right|\ge\left|2015-x+x-2017\right|\ge2\)

đẳng thức xảy ra khi \(2015\le x\le2017\)

a)

\(P=\dfrac{a+2\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{a-1}{\sqrt{a}-1}-\sqrt{a}\\ P=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}+\sqrt{a}+1-\sqrt{a}\\ P=\dfrac{a+2\sqrt{a}+1}{\sqrt{a}-1}+1\\P=\dfrac{a+2\sqrt{a}+1+\sqrt{a}-1}{\sqrt{a}-1}\\ P=\dfrac{a+3\sqrt{a}}{\sqrt{a}-1}\)

ai nhanh nhất **** cho 

\(A=\sqrt{x-2\sqrt{x-3}}\\ A=\sqrt{x-3-2\sqrt{x-3}+1+2}\\ A=\sqrt{\left(\sqrt{x-3}-1\right)^2+2}\)

vì:

\(\sqrt{x-3}\ge0\\ \Rightarrow\sqrt{x-3}+1\ge1\\ do\:đó\:\left(\sqrt{x-3}+1\right)^2+2\ge2\Rightarrow A\ge\sqrt{2}\)

đẳng thức xảy ra khi x=3

vậy \(MIN_A=\sqrt{2}\) tại x=3

\(n_{Zn}=\dfrac{19,5}{65}=0,3mol\)

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

a)\(\Rightarrow n_{H_2}=0,3mol\)

\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b)\(\Rightarrow n_{HCl}=0,3.2=0,6mol\)

\(\Rightarrow V_{HCl}=0,6.1=0,6\left(l\right)\)

cho hỗn hợp khí CO và CO2 vào dd Ca(OH)2 dư

khí CO2 phản ứng hết với Ca(OH)2 còn khí CO thì không phản ứng

do đó ta thu được khí CO tinh khiết

\(PTHH:\:CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

a) A có nghĩa khi \(x>0;x\ne1\)

b)

\(A=\dfrac{1}{x^2-\sqrt{x}}:\dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}\\ A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}^3-1\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\\ A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{1}{x-1}\)