\(x^4+2x^3+6x-9=0\)\(\)

\(\Rightarrow x^4+3x^3-x^3+9x-3x-9+3x^2-3x^2=0\)

\(\Rightarrow\left(x^4+3x^3+3x^2+9x\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow x\left(x^3+3x^2+3x+9\right)-\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x^3+3x^2+3x+9\right)=0\)

\(\Rightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]=0\)

\(\Rightarrow\left(x-1\right)\left(x^2+3\right)\left(x+3\right)=0\)

Mà \(x^2+3>0\) với mọi x\(\in\)R.

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

Chọn C.

\(3Ba\left(OH\right)_2+P_2O_5\rightarrow Ba_3\left(PO_4\right)_2+3H_2O\)

\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=1\\d=3\end{matrix}\right.\) \(\Rightarrow a+b+d=3+1+3=7\)

a) \(5\sqrt{72}-12\sqrt{18}+4\sqrt{8}+3\sqrt{\dfrac{2}{9}}\)

 \(=5\cdot6\sqrt{2}-12\cdot3\sqrt{2}+4\cdot2\sqrt{2}+\sqrt{2}\)

 \(=30\sqrt{2}-36\sqrt{2}+8\sqrt{2}+\sqrt{2}=3\sqrt{2}\)

e) \(sin^22x-6sin2x+5=0\Rightarrow\) \(\left[{}\begin{matrix}sin2x=5\left(loại\right)\\sin2x=1\end{matrix}\right.\)

    \(\Rightarrow sin2x=sin\left(\dfrac{\pi}{2}\right)\)

    \(\Rightarrow2x=\dfrac{\pi}{2}+k2\pi\Rightarrow x=\dfrac{\pi}{4}+k\pi\)

\(n_{NO}=\dfrac{6,72}{22,4}=0,3mol\)

BTe:  \(3n_{Fe}+2n_{Cu}=3n_{NO}=0,9\)

BTKL:  \(56n_{Fe}+64n_{Cu}=24,8\)

  \(\Rightarrow\left\{{}\begin{matrix}n_{Fe}=0,1mol\\n_{Cu}=0,3mol\end{matrix}\right.\)  \(\Rightarrow m_{Fe}=0,1\cdot56=5,6\left(g\right)\)

Cho \(Ba\left(OH\right)_2\) dư vào dung dịch X sản sinh hai kết tủa\(\left\{{}\begin{matrix}n_{BaSO_4}=n_{SO^{2-}_4}=0,01mol\\n_{BaCO_3}=n_{CO^{2-}_3}=0,01mol\end{matrix}\right.\)

\(\Rightarrow m_{\downarrow}=0,01\cdot233+0,01\cdot197=4,3\left(g\right)\)

\(n_{NO_2}=\dfrac{10,08}{22,4}=0,45mol\)

BTe:  \(3n_{Al}=n_{NO_2}\) \(\Rightarrow n_{Al}=0,15mol\) \(\Rightarrow m_{Al}=0,15\cdot27=4,05\left(g\right)\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,4     0,8                          0,4

\(m_{Fe}=0,4\cdot56=22,4g\)

\(C_M=\dfrac{0,8}{0,05}=16M\)

Câu 13. Chọn C.

Do x,y\(\ge\)0, x\(\ne\)y ta có:

\(A=\dfrac{x-\sqrt{xy}}{x-y}=\dfrac{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\cdot\left(\sqrt{x}+\sqrt{y}\right)}\)

    \(=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)

Câu 12.

   \(5\sqrt{a}+6\sqrt{\dfrac{a}{4}}-a\sqrt{\dfrac{4}{a}}+5\sqrt{\dfrac{4a}{25}}\)

\(=5\sqrt{a}+6\dfrac{\sqrt{a}}{2}-a\cdot\dfrac{2}{\sqrt{a}}+5\dfrac{2\sqrt{a}}{5}\)

\(=5\sqrt{a}+3\sqrt{a}-2\sqrt{a}+2\sqrt{a}\) (vì a>0)

\(=8\sqrt{a}\)