Em thấy bài 2 còn hack hơn bài 1

a) \(\sqrt{x-3}\) xác định

\(\Leftrightarrow x-3\ge0\)

\(\Leftrightarrow x\ge3\)

Vậy..

b) \(\sqrt{3-2x}\) xác định

\(\Leftrightarrow3-2x\ge0\)

\(\Leftrightarrow x\le-\dfrac{3}{2}\)

Vậy..

c) \(\sqrt{4x^2-1}\) xác định

\(\Leftrightarrow4x^2-1\ge0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}2x-1\ge0\\2x+1\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ge\dfrac{-1}{2}\end{matrix}\right.\)\(\Rightarrow x\ge\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}2x-1\le0\\2x+1\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\le\dfrac{-1}{2}\end{matrix}\right.\) \(\Rightarrow x\le\dfrac{-1}{2}\)

Vậy ...

d) \(\sqrt{3x^2+2}\) xác định

\(\Leftrightarrow3x^2+2\ge0\)

mà \(3x^2\ge0\)

\(\Rightarrow3x^2+2>0\)

Vậy...

e) \(\sqrt{2x^2+4x+5}\) xác định

\(\Leftrightarrow2x^2+4x+5\ge0\)

mà \(2x^2+4x\ge0\)

\(2x\left(x+2\right)\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}2x\ge0\\x+2\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-2\end{matrix}\right.\)\(\Rightarrow x\ge0\)

\(\Rightarrow\left\{{}\begin{matrix}2x\le0\\x+2\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le0\\x\le-2\end{matrix}\right.\)\(\Rightarrow x\le-2\)

\(\Rightarrow2x^2+4x+5>0\)

Vậy...

( Câu này không chắc lắm nha )

Bài 2: Tách sẵn ra cho bạn luôn nhé, không thì bạn nhấn máy tính ra cũng được :v

a) \(-\dfrac{7}{9}\sqrt{\left(-27\right)^2+6\sqrt{1}}\)

\(=-\dfrac{7}{9}\sqrt{\left(-3\right)^2.\left(-9\right)^2+6}\)

\(=\dfrac{-7}{9}\sqrt{735}\)

\(=\dfrac{-7}{9}\sqrt{49.15}\)

\(=\dfrac{-49\sqrt{15}}{9}\)

b) \(\sqrt{49}\sqrt{12^2}+\sqrt{256}:\sqrt{8^2}\)

\(=84+2=86\)

c)\(\sqrt{\left(\sqrt{3-1}\right)^2-\sqrt{\left(\sqrt{3+1}\right)^2}}\)

\(=\sqrt{2-2}\)

= 0

Axit pemanganic

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

Gọi x, y , z lần lượt là số mol của Mg, Zn, Fe

\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\) (1)

x -------> 2x--------->x------> x

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) (2)

y ----> 2y -------->y --------> y

\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\) (3)

z -----> 2z --------> z ----------> z

(1)(2)(3) \(\Rightarrow\left\{{}\begin{matrix}24x+65y+56z=21\\x+y+z=0,4\end{matrix}\right.\) (I)

\(2KOH+MgCl_2\rightarrow Mg\left(OH\right)_2\downarrow+2KCl\)

x ----------> x

\(2KOH+ZnCl_2\rightarrow Zn\left(OH\right)_2\downarrow+2KCl\)

\(2KOH+Zn\left(OH\right)_2\rightarrow K_2ZnO_2+2H_2O\) ( tan hết )

\(2KOH+FeCl_2\rightarrow Fe\left(OH\right)_2\downarrow+2KCl\)

z -----------> z

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\) (4)

x ----------------> x

\(2Fe\left(OH\right)_2+\dfrac{1}{2}O_2+H_2O\rightarrow2Fe\left(OH\right)_3\downarrow\)

z -------------------------------------> z

\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\) (5)

z -----------------> 0,5z

(4)(5)\(\Rightarrow40x+160z=12\) (II)

(I)(II) \(\Rightarrow\left\{{}\begin{matrix}24x+65y+56z=21\\x+y+z=0,4\\40x+160.\dfrac{1}{2}z=12\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\\z=0,1\end{matrix}\right.\)

\(m_{Mg}=0,1.24=2,4\left(g\right)\)

\(m_{Zn}=0,2.65=13\left(g\right)\)

\(m_{Fe}=0,1.56=5,6\left(g\right)\)

p và p + 4 là hợp số

=> p lẻ (thõa mãn)

=> p + 7 chẵn nên p + 7 là hợp số (dpcm)      

Tên: Rain

Lớp: 4

Link: Click ava.