Lấy mẫu thử, đánh số thứ tự:

- Cho quỳ tìm vào các mẫu thử:

+) HCl: quỳ hóa đỏ

+) NaOH: quỳ hóa xanh

+) MgSO₄; MgCl₂ : quỳ không chuyển màu  (1)

- Nhỏ vài giọt dd BaCl₂ vào (1):

+) MgSO₄: xuất hiện kết tủa trắng

+) MgCl₂: không hiện tượng

\(MgSO_4+BaCl_2\rightarrow BaSO_4\downarrow+MgCl_2\)

Câu 4.

\(\left\{{}\begin{matrix}x^2+y^2-2x-2y=6\\x+y-xy=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-2\left(x+y\right)=6\\x+y=5+xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-2\left(5+xy\right)=6\\x+y=5+xy\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-10-2xy=6\\x+y=5+xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=16\left(1\right)\\x+y=5+xy\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\left[{}\begin{matrix}x-y=4\\x-y=-4\end{matrix}\right.\)

`@` Với \(x-y=4\) \(\Leftrightarrow x=4+y\)

\(\left(2\right)\Rightarrow\left(4+y\right)+y=5+\left(4+y\right).y\)

\(\Leftrightarrow4+2y=5+4y+y^2\)

\(\Leftrightarrow y^2+2y+1=0\)

\(\Leftrightarrow\left(y+1\right)^2=0\)

\(\Leftrightarrow y=-1\Rightarrow x=4-1=3\)

`@`\(x-y=-4\Rightarrow x=-4+y\)

\(\left(2\right)\Rightarrow\left(-4+y\right)+y=5+\left(-4+y\right).y\)

\(\Leftrightarrow-4+2y=5-4y+y^2\)

\(\Leftrightarrow y^2-6y+9=0\)

\(\Leftrightarrow\left(y-3\right)^2=0\)

\(\Leftrightarrow y=3\Rightarrow x=-4+3=-1\)

Vậy \(\left(x;y\right)=\left(-1;3\right);\left(3;-1\right)\)

Câu 1.

\(\sqrt{2x^2-5x-9}=x-1\)

\(ĐK:x\ge1\)

\(\Leftrightarrow\left(\sqrt{2x^2-5x-9}\right)^2=\left(x-1\right)^2\)

\(\Leftrightarrow2x^2-5x-9=x^2-2x+1\)

\(\Leftrightarrow x^2-3x-10=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)

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

Vậy \(S=\left\{5\right\}\)

Bài 2.

\(\left(x-1\right)\left(x-3\right)+3\sqrt{x^2-4x+5}-2=0\)

\(ĐK:x\in R\)

\(\Leftrightarrow x^2-4x+3+3\sqrt{x^2-4x+5}-2=0\)

Đặt \(\sqrt{x^2-4x+5}=t;t\ge0\)

Ptr trở thành:

\(t^2+3t-4=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(tm\right)\\t=-4\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{x^2-4x+5}=1\)

\(\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy \(S=\left\{2\right\}\)

Câu 3.

\(\sqrt{3x-2}+\sqrt{x-1}=4x-9+2\sqrt{3x^2-5x+2}\)

`<=>`\(\sqrt{3x-2}+\sqrt{x-1}=4x-9+2\sqrt{\left(3x-2\right)\left(x-1\right)}\)

\(ĐK:x\ge1\) ; \(x\le\dfrac{2}{3}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{3x-2}=t\\\sqrt{x-1}=p\end{matrix}\right.\) \(\left(t,p\ge0\right)\)

\(\Rightarrow t^2+p^2=4x-3\)

Ptr trở thành:

\(t+p=t^2+p^2-6+2tp\)

\(\Leftrightarrow t+p=\left(t+p\right)^2-6\)

\(\Leftrightarrow\left(t+p\right)^2-\left(t+p\right)-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t+p=3\left(tm\right)\\t+p=-2\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{3x-2}+\sqrt{x-1}=3\)

\(\Leftrightarrow3x-2+x-1+2\sqrt{\left(3x-2\right)\left(x-1\right)}=9\)

\(\Leftrightarrow\sqrt{\left(3x-2\right)\left(x-1\right)}=6-2x\) ; \(x\le3\)

\(\Leftrightarrow3x^2-3x-2x+2=36-24x+4x^2\)

\(\Leftrightarrow x^2-19x+34=0\)

\(\Leftrightarrow\left(x-17\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=17\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{2\right\}\)

CTV rồi kìa:")

Đặt \(\left\{{}\begin{matrix}n_{FeO}=x\\n_{ZnO}=y\end{matrix}\right.\) ( mol )

\(\Rightarrow m_{hh}=72x+80Y=15,3\left(g\right)\left(1\right)\)

\(n_{H_2SO_4}=0,4.0,5=0,2\left(mol\right)\)

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

 x            x                                 ( mol )

\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

  y              y                                   ( mol )

\(\Rightarrow n_{H_2SO_4}=x+y=0,2\left(mol\right)\left(2\right)\)

\(\left(1\right);\left(2\right)\rightarrow\left\{{}\begin{matrix}x=0,0875\\y=0,1125\end{matrix}\right.\)

\(\left\{{}\begin{matrix}m_{FeO}=0,0875.72=6,3\left(g\right)\\m_{ZnO}=15,3-6,3=9\left(g\right)\end{matrix}\right.\)

\(y=\left(1-2m\right)x-m+2\) ; \(m\ne\dfrac{1}{2}\)

`a.` y đi qua gốc tọa độ khi \(-m+2=0\)

                                           \(\Leftrightarrow m=2\)

`b.` y đi qua A(2;3) 

\(\left(d\right)\Leftrightarrow3=\left(1-2m\right).2-m+2\)

\(\Leftrightarrow3=2-4m-m+2\)

\(\Leftrightarrow-5m=-1\)

\(\Leftrightarrow m=\dfrac{1}{5}\)

`c.` Cắt trục hoành tại điểm có hoành độ là 3

\(\Rightarrow x=3;y=0\)

\(\left(d\right)\Leftrightarrow0=\left(1-2m\right).3-m+2\)

\(\Leftrightarrow3-6m-m+2=0\)

\(\Leftrightarrow-7m+5=0\)

\(\Leftrightarrow m=\dfrac{5}{7}\)