`c.`\(x^3-x=0\)

\(\Leftrightarrow x\left(x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

`d.`\(4x^2-\left(x-2\right)^2=0\)

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

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

\(\Leftrightarrow\left(x+2\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{2}{3}\end{matrix}\right.\)

`e.`\(x^3-25x=0\)

\(\Leftrightarrow x\left(x^2-25\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.\)  \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm5\end{matrix}\right.\)

`a.`\(\left(4+\sqrt{15}\right)\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\)

\(=\sqrt{\left(4+\sqrt{15}\right)\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}\left(\sqrt{10}-\sqrt{6}\right)\)

\(=\sqrt{\left(4+\sqrt{15}\right).1}\left(\sqrt{10}-\sqrt{6}\right)\)

\(=\sqrt{\left(4+\sqrt{15}\right)}.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left|\sqrt{5}+\sqrt{3}\right|.\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(\sqrt{5}+\sqrt{3}\right).\left(\sqrt{5}-\sqrt{3}\right)\)

\(=2\)

`b.` Đề sai

\(\dfrac{1}{\sqrt{\left(\sqrt{6}-1\right)^2+1}}-\dfrac{1}{\sqrt{\left(\sqrt{6}+1\right)^2-1}}=0\)

\(\Leftrightarrow\left(\sqrt{6}-1\right)^2+1=\left(\sqrt{6}+1\right)^2-1\)

\(\Leftrightarrow6-2\sqrt{6}+1+1=6+2\sqrt{6}+1-1\)

\(\Leftrightarrow4\sqrt{6}=2\) ( vô lý )

`c.`\(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\dfrac{1}{\sqrt{x}-\sqrt{y}}\)

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

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

\(=x-y\)

`d.`\(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}=\dfrac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1}{a-1}\)

\(=\dfrac{4\sqrt{a}}{a-1}\)

`e.`\(\left(\dfrac{1}{\sqrt{x}+4}+\dfrac{1}{\sqrt{x}-4}\right).\dfrac{\sqrt{x}+4}{\sqrt{x}}\)

\(=\left(\dfrac{\sqrt{x}-4+\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\right).\dfrac{\sqrt{x}+4}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}.\dfrac{\sqrt{x}+4}{\sqrt{x}}\)

\(=\dfrac{2}{\sqrt{x}-4}\)

lạc đề r ak

`.a` \(A=\left(x^2-3x+1\right)\left(x^2-3x-1\right)\)

\(A=\left(x^2-3x\right)^2-1^2\)

\(A=\left(x^2-3x\right)^2-1\ge-1\)

Dấu "=" xảy ra \(\Leftrightarrow x^2-3x=0\)

                         \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Vậy \(Min_A=-1\) khi \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

`b.`\(B=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)

\(B=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)

\(B=\left(x^2+4x\right)^2-5^2\)

\(B=\left(x^2+4x\right)^2-25\ge-25\)

Dấu "=" xảy ra \(\Leftrightarrow x^2+4x=0\)

                         \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy \(Min_B=-25\) khi \(\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

\(n_{HNO_3}=1,2.0,45=0,54\left(mol\right)\)

Đặt \(\left\{{}\begin{matrix}n_{FeO}=x\\n_{CuO}=y\\n_{MO}=z\end{matrix}\right.\) ( mol ) \(\Rightarrow m_{hh}=72x+80y+\left(M+16\right)z=11,52\left(g\right)\left(\cdot\right)\)

`@`TH₁: M bị H₂ khử

\(FeO+H_2\rightarrow\left(t^o\right)Fe+H_2O\) (1)

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\) (2)

\(MO+H_2\rightarrow\left(t^o\right)M+H_2O\)   (3)

\(Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\) (4)

\(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\) (5)

\(3M+8HNO_3\rightarrow3M\left(NO_3\right)_2+2NO+4H_2O\) (6)

\(MO+2HNO_3\rightarrow M\left(NO_3\right)_2+H_2O\) (7)

Theo pt \(\left(1\right);\left(2\right);\left(3\right);\left(4\right);\left(5\right);\left(6\right)\):

\(\Rightarrow n_{HNO_3}=4x+\dfrac{8}{3}y+\dfrac{8}{3}z=0,54\)

 Ta có: \(x:y:z=5:3:1\) 

`=>` không có giá trị thỏa mãn

`@`TH₂: M không bị khử bởi H₂

Theo pt: \(\left(1\right);\left(2\right);\left(4\right);\left(5\right);\left(7\right)\) :

\(n_{HNO_3}=4x+\dfrac{8}{3}y+2z=0,54\)

Ta có: \(x:y:z=5:3:1\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,09\\y=0,054\\z=0,018\end{matrix}\right.\)

\(\left(\cdot\right)\Leftrightarrow72.0,09+80.0,054+\left(M+16\right).0,018=11,52\)

\(\Rightarrow M=24\) \((g/mol)\)

`=>` \(M\) là Magie ( Mg )

Theo pt: \(\left(1\right);\left(2\right);\left(4\right);\left(5\right)\) :

\(n_{NO}=x+2y=0,09+2.0,054=0,198\left(mol\right)\)

\(\Rightarrow V_{NO}=0,198.22,4=4,4352\left(mol\right)\)

`a.b.`\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

1,5         1,5                          1,5     ( mol )

\(m_{Fe}=1,5.56=84\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{1,5}{0,5}=3\left(M\right)\)

`c.`\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

       1,5    0,75                     ( mol )

\(V_{O_2}=0,75.22,4=16,8\left(l\right)\)

A: Fe₃O₄     B: H₂SO₄          C: FeSO₄          D: Fe₂(SO₄)₃          E: H₂O   

 F: Fe(OH)₂            G: Fe(OH)₃     H: K₂SO₄             K: FeCl₃         I: HCl

PTHH:

\(Fe_3O_4+4H_2SO_4\rightarrow FeSO_4+Fe_2\left(SO_4\right)_3+4H_2O\)

\(FeSO_4+2NaOH\rightarrow Fe\left(OH\right)_2\downarrow+Na_2SO_4\)

\(10FeSO_4+2KMnO_4+8H_2SO_4\rightarrow5Fe_2\left(SO_4\right)_3+2MnSO_4+K_2SO_4+8H_2O\)

\(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

\(4Fe\left(OH\right)_2+O_2+2H_2O\rightarrow4Fe\left(OH\right)_3\)

\(Fe_2\left(SO_4\right)_3+2KI\rightarrow2FeSO_4+K_2SO_4+I_2\)

hiccc e hongg có giỏi Sinh:"((