Quy hỗn hợp thành \(\left\{{}\begin{matrix}FeO\\Fe_2O_3\end{matrix}\right.\)

Ta có: \(\dfrac{n_{FeO}}{n_{Fe_2O_3}}=\dfrac{1}{1}\) \(\Rightarrow n_{FeO}=n_{Fe_2O_3}=x\left(mol\right)\)

\(\Rightarrow m_{hh}=72x+160x=23,2\)

\(\Leftrightarrow x=0,1\left(mol\right)\)

\(FeO+H_2SO_4\rightarrow FeSO_4+H_2O\)

0,1          0,1                           ( mol )

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

   0,1             0,3                                  ( mol )

\(\Rightarrow a=n_{H_2SO_4}=0,1+0,3=0,4\left(mol\right)\)

`1.`\(f\left(x\right)=7x^2+5x\)

\(f\left(x\right)=7\left(x^2+\dfrac{5}{7}x\right)\)

\(f\left(x\right)=7\left(x^2+\dfrac{5}{7}x+\dfrac{25}{196}-\dfrac{25}{196}\right)\)

\(f\left(x\right)=7\left(x+\dfrac{5}{14}\right)^2-\dfrac{25}{28}\ge-\dfrac{25}{28}\)

Dấu "=" xảy ra `<=>x+5/14=0`

                           `<=>x=-5/14`

Vậy \(Min_{f\left(x\right)}=-\dfrac{25}{28}\) khi `x=-5/14`

`2.`\(f\left(x\right)=-3x^2+5x\)

\(f\left(x\right)=-3\left(x^2-\dfrac{5}{3}x\right)\)

\(f\left(x\right)=-3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{25}{36}\right)\)

\(f\left(x\right)=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{25}{12}\le\dfrac{25}{12}\)

Dấu "=" xảy ra `<=>x-5/6=0`

                          `<=>x=5/6`

Vậy \(Max_{f\left(x\right)}=\dfrac{25}{12}\) khi `x=5/6`

`a.`\(f\left(x\right)=4x^2+9x+1\)

\(f\left(x\right)=\left(4x^2+9x+\dfrac{81}{16}\right)-\dfrac{81}{16}+1\)

\(f\left(x\right)=\left(2x+\dfrac{9}{4}\right)^2-\dfrac{65}{16}\ge-\dfrac{65}{16}\)

Dấu "=" xảy ra `<=>2x+9/4=0`

                          `<=>x=-9/8`

Vậy \(Min_{f\left(x\right)}=-\dfrac{65}{16}\) khi `x=-9/8`

`b.`\(f\left(x\right)=-5x^2-12x+3\)

\(f\left(x\right)=-\left(5x^2+12x-3\right)\)

\(f\left(x\right)=-5\left(x^2+\dfrac{12}{5}x-\dfrac{3}{5}\right)\)

\(f\left(x\right)=-5\left(x^2+\dfrac{12}{5}x+\dfrac{36}{25}-\dfrac{36}{25}-\dfrac{3}{5}\right)\)

\(f\left(x\right)=-5\left(x+\dfrac{6}{5}\right)^2+\dfrac{51}{5}\le\dfrac{51}{5}\)

Dấu "=" xảy ra `<=>x+6/5=0`

                           `<=>x=-6/5`

Vậy \(Max_{f\left(x\right)}=\dfrac{51}{5}\) khi \(x=-\dfrac{6}{5}\)

`n_{Fe}={5,6}/{56}=0,1(mol)`

`Fe+2HCl->FeCl_2+H_2`

0,1     0,2                     0,1  ( mol )

`V_{H_2}=0,1.22,4=2,24(l)`

`V_{dd(HCl)}={0,2}/{2}=0,1(l)`

\(f\left(x\right)=7x^2+5x\)

\(f\left(x\right)=5\left(x^2+x\right)+2x^2\)

\(f\left(x\right)=5\left(x^2+x+\dfrac{1}{4}\right)+2x^2-\dfrac{5}{4}\)

\(f\left(x\right)=5\left(x+\dfrac{1}{2}\right)^2+2x^2-\dfrac{5}{4}^2\ge-\dfrac{5}{4}\)

Dấu "=" xảy ra `<=>x+1/2=0`

                            `<=>x=-1/2`

Vậy \(Min_{f\left(x\right)}=-\dfrac{5}{4}\) khi \(x=-\dfrac{1}{2}\)

\(f\left(x\right)=4x^2-9x+1\)

\(f\left(x\right)=\left(4x^2-9x+\dfrac{81}{16}\right)-\dfrac{81}{16}+1\)

\(f\left(x\right)=\left(2x-\dfrac{9}{4}\right)^2-\dfrac{65}{16}\ge-\dfrac{65}{16}\)

Dấu "=" xảy ra `<=>2x-9/4=0`

                            `<=>x=9/8`

Vậy \(Min_{f\left(x\right)}=-\dfrac{65}{16}\) khi `x=9/8`

\(C=x^2+2x+1\dfrac{1}{2}\)

\(C=x^2+2x+\dfrac{3}{2}\)

\(C=\left(x^2+2x+1\right)+\dfrac{1}{2}\)

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

Dấu "=" xảy ra `<=>x+1=0`

                        `<=>x=-1`

Vậy `Min_C=1/2` khi `x=-1`

`a.23+x=48`

`=>x=48-23`

`=>x=25`

`b.(x+3).9=108`

`=>x+3=12`

`=>x=9`

`c.`\(5^{x+1}:5=125\)

`=>`\(5^{x+1}=625\)

`=>5^x.5=625`

`=>5^x=125`

`=>x=3`

`(x+1)^2-(x-1)^2-4x`

`=x^2+2x+1-x^2+2x-1-4x`

`=0`