\(Cu\left(II\right),SO_4\left(II\right)\)

Ta có : \(Cu_x\left(SO_4\right)_y\)

\(\dfrac{x}{y}=\dfrac{II}{II}=\dfrac{2}{2}\)

\(\Rightarrow CTHH:CuSO_4\)

Bài 1 .

a. \(\sqrt{25a}+\sqrt{81a}-\sqrt{64a}\\ =\sqrt{25}\sqrt{a}+\sqrt{81}\sqrt{a}-\sqrt{64}\sqrt{a}\\ =5\sqrt{a}+9\sqrt{a}-8\sqrt{a}\\ =6\sqrt{a}\)

b. \(\sqrt{27}\left(\sqrt{3}-\sqrt{5}\right)^2\\ =\sqrt{27}\left(3-2\sqrt{3}.\sqrt{5}+5\right)\\ =\sqrt{9.3}.\left(3-2\sqrt{3}.\sqrt{5}+5\right)\\ =3\sqrt{3}.8-2\sqrt{3}.\sqrt{5}\\ =8\sqrt{15}\)

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

d. \(\dfrac{x-2}{\sqrt{x^2-4x+4}}\\ =\dfrac{x-2}{\sqrt{\left(x-2\right)^2}}\\ =\dfrac{x-2}{\left|x-2\right|}\\ =\dfrac{x-2}{x-2}=1\)

Bài 1 :

a. Ta có : Ba(NO₃)_y 

\(\Rightarrow137+\left(14+48\right)y=261\\ \Rightarrow y=2\)

NO3 có hóa trị I

b. Ta có : Mn₂O_x

\(\Rightarrow55.2+16x=222\\ \Rightarrow x=7\)

Hoa trị của Mn là 7

\(PTK_{Mg\left(HCO_3\right)_2}=24+\left(1+12+16.3\right).2=24+122=146\)

\(\left(x^2+1\right)^2-6\left(x^2+1\right)^2+5\\ \Leftrightarrow\left(x^2+1\right)\left(-6+5\right)\\ \Leftrightarrow\left(x^2+1\right).-1\\ \Leftrightarrow-x^2-1\)

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

Đề thiếu % á

a. \(n_{H_2SO_4}=\dfrac{200.14,7\%}{98}=0,3\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2         0,3                    0,1        0,3

\(m_{Al}=0,2.27=5,4\left(g\right)\)

\(m_{H_2}=0,3.2=0,6\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{5,4+200-0,6}=16,69\%\)

b. \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

\(\left[\left(x-\dfrac{1}{2}\right):6+4\right]\times\dfrac{2}{3}=\dfrac{2}{3}\times\dfrac{40}{6}\\ \Leftrightarrow\left[\left(x-\dfrac{1}{2}\right):6+4\right]\times\dfrac{2}{3}=\dfrac{40}{9}\\ \Leftrightarrow\left[\left(x-\dfrac{1}{2}\right):6+4\right]=\dfrac{20}{3}\\ \left(x-\dfrac{1}{2}\right):6=\dfrac{8}{3}\\ x-\dfrac{1}{2}=16\\ x=\dfrac{33}{2}\)