\(tan\alpha=\dfrac{3}{4}\\ \Rightarrow cot\alpha=1:\dfrac{3}{4}=\dfrac{4}{3}\)

Có:

\(1+cot^2\alpha=\dfrac{1}{sin^2\alpha}\\ \Rightarrow sin\alpha=\sqrt{1:\left(1+\left(\dfrac{4}{3}\right)^2\right)}=\dfrac{3}{5}\)

\(\Rightarrow cos\alpha=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\)

a)

MSC: \(60x^4y^5\)

\(\dfrac{4}{15x^3y^5}=\dfrac{4.4.x}{15.4.x^3.x.y^5}=\dfrac{16x}{60x^4y^5}\)

\(\dfrac{11}{12x^4y^2}=\dfrac{11.5.y^3}{12.5.x^4.y^2.y^3}=\dfrac{55y^3}{60x^4y^5}\)

b)

MSC: \(4x^2-16\)

\(\dfrac{x}{2x-4}=\dfrac{2\left(2x+4\right)}{4x^2-16}=\dfrac{4x+8}{4x^2-16}\\ \dfrac{1}{2x+4}=\dfrac{2x-4}{4x^2-16}\)

c)

MSC: \(6x+6y\)

\(\dfrac{1}{3x+3y}=\dfrac{2}{6x+6y}\\ \dfrac{1}{2y+2x}=\dfrac{3}{6x+6y}\)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

0,2<---0,4<--------------0,2

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

b) \(CM_{HCl.đã.dùng}=\dfrac{0,4}{0,1}=4M\)

Bài 21:

\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\\ PTHH:MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

               0,1---->0,1--------->0,1

a. \(V_{dd.H_2SO_4}=\dfrac{0,1}{1}=0,1\left(l\right)=100\left(ml\right)\)

b. \(CM_{MgSO_4}=\dfrac{0,1}{0,1}=1M\)

Bài 22:

\(n_{CuO}=\dfrac{16}{160}=0,1\left(mol\right)\\ n_{H_2SO_4}=\dfrac{100.19,6\%}{100\%}:98=0,2\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,1----->0,1------>0,1

Xét \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow n_{H_2SO_4.dư}=0,2-0,1=0,1\left(mol\right)\)

a. Tìm V cái gì vậy?

b. \(C\%_{CuSO_4}=\dfrac{0,1.160.100\%}{16+100}=13,79\%\)

\(C\%_{H_2SO_4}=\dfrac{0,1.98.100\%}{16+100}=8,45\%\)

Bài 32:

\(n_{H_2SO_4}=\dfrac{180.19,6\%}{100\%}:98=0,36\left(mol\right)\\ n_{BaCl_2}=\dfrac{320.20,8\%}{100\%}:208=0,32\left(mol\right)\\ H_2SO_4+BaCl_2\rightarrow BaSO_4+2HCl\)

0,32<----0,32----->0,32-------->0,64

Xét \(\dfrac{0,36}{1}>\dfrac{0,32}{1}\Rightarrow n_{H_2SO_4.dư}=0,36-0,32=0,04\left(mol\right)\)

a. \(m_{BaSO_4}=0,32.233=74,56\left(g\right)\)

b. \(C\%_{HCl}=\dfrac{0,64.36,5.100\%}{180+320-74,56}=5,49\%\)

\(C\%_{H_2SO_4}=\dfrac{0,04.98.100\%}{180+320-74,56}=0,92\%\)

Bài 33:

\(n_{CuO}=\dfrac{1,6}{160}=0,01\left(mol\right)\\ n_{H_2SO_4}=\dfrac{30.9,8\%}{100\%}:98=0,03\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

0,01--->0,01----->0,01

Xét \(\dfrac{0,03}{1}>\dfrac{0,01}{1}\Rightarrow n_{H_2SO_4.dư}=0,03-0,01=0,02\left(mol\right)\)

\(C\%_{CuSO_4}=\dfrac{0,01.160.100\%}{1,6+30}=5,06\%\\ C\%_{H_2SO_4}=\dfrac{0,02.98.100\%}{1,6+30}=6,2\%\)

\(8x^3+36x^2y+54xy^2+27y^3\\ =\left(2x\right)^3+3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2+\left(3y\right)^3\\ =\left(2x+3y\right)^3\\ =\left(2x+3y\right)\left(2x+3y\right)\left(2x+3y\right)\)

\(\left(x-y\right)^3-\left(x+y\right)^3\\ =\left(x-y-x-y\right)\left(x^2-2xy+y^2+x^2-y^2+x^2+2xy+y^2\right)\\ =-2y\left(3x^2+y^2\right)\)

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

\(\left(x-1\right)^2-\left(x+1\right)^2\\ =\left(x-1-x-1\right)\left(x-1+x+1\right)\\ =-2.2x=-4x\)

\(m_{NaOH\left(20\%\right)}=\dfrac{400.20\%}{100\%}=80\left(g\right)\)

Gọi `x` (g) là khối lượng nước cần thêm:

Có: \(C\%_{NaOH}=\dfrac{80.100\%}{x+400}=15\%\)

=> `x=133,33` gam

\(m_{HCl}=\dfrac{200.20\%}{100\%}+\dfrac{400.10\%}{100\%}=80\left(g\right)\)

Nồng độ % HCl mới:

\(C\%_{HCl}=\dfrac{80.100\%}{200+400}=13,33\%\)

Nồng độ % NaOH là:

\(C\%=\dfrac{m_{NaOH}.100\%}{m_{dd}}=\dfrac{2.40.100\%}{2.40+420}=16\%\)