Câu 2:

 \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

  2a           3a              a                 3a

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

 b         b                 b           b

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25mol\)

Ta có: \(\left\{{}\begin{matrix}54a+65b=9.2\\3a+b=0.25\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0.05\\b=0.1\end{matrix}\right.\)

a.\(\%m_{Al}=\dfrac{0.05\times54\times100}{9.2}=29.3\%\)

\(\%m_{Zn}=100-29.3=70.7\%\)

Vdd sau phản ứng = 9.2 + 600 - 0.0056 = 609.2ml

\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0.05}{0.6092}=0.08M\)

\(CM_{ZnSO_4}=\dfrac{0.1}{0.6092}=0.16M\)

Câu 3: 

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

 0.2       0.2            0.2          0.2

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

\(n_{H_2}=\dfrac{4.48}{22.4}=0.2mol\)

a. \(\%m_{Mg}=\dfrac{0.2\times24\times100}{12}=40g\)

\(\%m_{FeO}=100-40=60\%\)

b. \(n_{FeO}=\dfrac{12-0.2\times24}{72}=0.1mol\)

m muối khan \(=m_{MgSO_4}+m_{FeSO_4}=0.2\times120+0.1\times152=39.2g\)

\(C2:y=ax^2+bx+c\) \(\left(P\right)\) \(đi\) \(qua\)\(M\left(1;5\right)vàN\left(-2;8\right)\)

\(\Rightarrow\left\{{}\begin{matrix}a+b+2=5\\4a-2b+2=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) \(\Rightarrow S=3a+4b+c=3.2+4+2=12\)

\(C3a,\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}\left(2;-4\right)\\\overrightarrow{AC}\left(2;-1\right)\end{matrix}\right.\)

\(b,D\left(xo;yo\right)\Rightarrow ABCD\) \(là\) \(hbh\Leftrightarrow\overrightarrow{AB}=\overrightarrow{DC}\Leftrightarrow\left\{{}\begin{matrix}3-xo=2\\1-yo=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xo=1\\yo=5\end{matrix}\right.\) \(\Rightarrow D\left(1;5\right)\)

Bài 3: 

a: Thay x=4 vào A, ta được:

\(A=\dfrac{2\cdot4}{4-9}=\dfrac{8}{-5}=-\dfrac{8}{5}\)

b: Ta có: \(B=\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{5}{\sqrt{x}+3}+\dfrac{2x+12}{9-x}\)

\(=\dfrac{2x+6\sqrt{x}-5\sqrt{x}+15-2x-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{1}{\sqrt{x}-3}\)

Luôn cả câu 4 nữa ạ

Bài 2: 

a: \(\Leftrightarrow x^2+2x-x^2=0\)

=>2x=0

hay x=0

b: \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

c: \(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)

hay \(x\in\left\{-3;5\right\}\)

Wlk=(Z*mp-(A-Z)*mn-mCo)*c^2

=(27*1,0073u +31*1,0087u-56,9540u)*c^2=1,5128u*c^2=1,5128*931,5

=1409,1732MeV

--->Wr=Wlk/A=1409,1732/58=24,296MeV

Bài 3:

a: Thay x=1,44 vào A, ta được:

\(A=\frac{1,44+7}{\sqrt{1,44}}=\frac{8.44}{1,2}=\frac{844}{120}=\frac{211}{30}\)

b: \(B=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-1}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\)

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

\(=\frac{x-3\sqrt{x}+2x+6\sqrt{x}-\sqrt{x}-3-2x+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}\)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{3x}-5\sqrt{12x}+7\cdot\sqrt{27x}=12\)

=>\(\sqrt{3x}-5\cdot2\sqrt{3x}+7\cdot3\sqrt{3x}=12\)

=>\(12\sqrt{3x}=12\)

=>\(\sqrt{3x}=1\)

=>3x=1

=>x=1/3(nhận)

b: ĐKXĐ: x>=-1

\(\frac{x}{1+\sqrt{1+x}}=\sqrt{1+x}-1\)

=>\(x=\left(\sqrt{1+x}-1\right)\left(\sqrt{1+x}+1\right)=1+x-1\)

=>x=x

=>0x=0(luôn đúng)

Vậy: x>=-1

bài 1: 

\(\left\{{}\begin{matrix}x+y=57\\4x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x+4y=228\\4x-2y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6y=234\\x+y=57\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=39\\x=18\end{matrix}\right.\)