Câu 16: A

\(n_{CuSO_4}=0,2x\left(mol\right)\)

\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

0,2x    0,2x                         0,2x

\(m_{tăng}=m_{Cu}-m_{Fe}=64\cdot0,2x-56\cdot0,2x=1,6\)

\(\Rightarrow x=1M\)

Chọn C.

\(n_{Fe}=n_{FeSO_4}=n_{Cu}=n_{CuSO_4}=0,2.x\left(mol\right)\\ Fe+CuSO_4\rightarrow FeSO_4+Cu\\ m_{t\text{ăn}g}=m_{Cu.b\text{á}m.v\text{ào}}-m_{Fe.tan.ra}\\ \Leftrightarrow1,6=64.0,2x-56.0,2x\\ \Leftrightarrow x=1\\ \Rightarrow C\)

ĐK: \(-1\le x\le1\)

Đặt \(\sqrt{1-x}=a;\sqrt{x+1}=b\Rightarrow3-x=2a^2+b^2\)

\(pt\Leftrightarrow2a-b+3ab=2a^2+b^2\)

\(\Leftrightarrow2a^2+b^2-2a+b-3ab=0\)

\(\Leftrightarrow2a^2-a\left(3b+2\right)+b^2+b=0\)

\(\Delta=\left(3b+2\right)^2-4.2.\left(b^2+b\right)=9b^2+12b+4-8b^2-8b\)

\(=b^2+4b+4=\left(b+2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{3b+2-\left(b+2\right)}{4}=\dfrac{2b}{4}=\dfrac{b}{2}\Leftrightarrow2a=b\left(1\right)\\a=\dfrac{3b+2+b+2}{4}=\dfrac{4b+4}{4}=b+1\left(2\right)\end{matrix}\right.\)

pt (1) \(\Leftrightarrow2\sqrt{1-x}=\sqrt{x+1}\)

\(\Leftrightarrow4\left(1-x\right)=x+1\)

\(\Leftrightarrow5x=3\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)

\(pt\left(2\right)\Leftrightarrow\sqrt{1-x}=1+\sqrt{x+1}\)

\(\Leftrightarrow1-x=1+x+1+2\sqrt{x+1}\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2+4x+1=4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\left(l\right)\\x=-\dfrac{\sqrt{3}}{2}\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy, pt có tập nghiệm là: \(S=\left\{-\dfrac{\sqrt{3}}{2};\dfrac{5}{3}\right\}\)

???

`a,` ĐKXĐ: `x>=0;x\ne1`

`A=...=(sqrtx(1+sqrtx)+sqrtx(1-sqrtx)+sqrtx-3)/((1-sqrtx)(1+sqrtx))`

`=(sqrtx+x+sqrtx-x+sqrtx-3)/((1-sqrtx)(1+sqrtx))`

`=(3sqrtx-3)/((1-sqrtx)(1+sqrtx))`

`=-3/(1+sqrtx)`

`b,A=-3/(1+sqrtx)` 

Vì `x>=0` nên `1+sqrtx>=1` nên `3/(1+sqrtx)<=3` suy ra `A>=-3`

Dấu "=" xảy ra `<=>x=0`

Vậy `A_(min)=-3<=>x=0`

`a)->` ĐKXĐ : `x>=0;x\ne1`

`b)` Ta có :

`A=((\sqrtx)/(1-\sqrtx)+(\sqrtx)/(1+\sqrtx))+(3-\sqrtx)/(x-1)`

`A=[(\sqrtx(1+\sqrtx)+\sqrtx(1-\sqrtx))/(1-x)]+(3-\sqrtx)/(x-1)`

`A=[(\sqrtx+x+\sqrtx-x)/(1-x)]+(3-\sqrtx)/(x-1)`

`A=(2\sqrtx)/(1-x)+(3-\sqrtx)/(x-1)`

`A=(3-\sqrtx)/(x-1)-(2\sqrtx)/(x-1)`

`A=(3+\sqrtx)/(x-1)`

Vậy `A=(3+\sqrtx)/(x-1)` khi `x>=0;x\ne1`

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19 He has studied English for four years