a) A=x²+2xy-3x³+2y³+3x³-y³

      = x²+2xy+y³

thay x=5, y=4 ta có:

A=5²+2.5.4+4³

    =129

b) B= xy-x²y²+x⁴y⁴-x⁶y⁶+x⁸y⁸

thay x=-1, y=-1

B=(-1)(-1)-(-1)².(-1)²+(-1)⁴.(-1)⁴-(-1)⁶.(-1)⁶+(-1)⁸.(-1)⁸

    =1-1+1-1+1

    = 1

mik trả lời nhầm, mình xin lỗi

Bài 8:

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

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

0,1--->0,15------>0,05---------->0,15

a. \(V_{H_2SO_4}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\)

b. \(V'=0,15.22,4=3,36\left(l\right)\)

c. \(CM_{Al_2\left(SO_4\right)_3}=\dfrac{0,05}{0,15}=\dfrac{1}{3}M\)

Bài 9:

\(n_{Fe}=\dfrac{2,8}{56}=0,2\left(mol\right)\)

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

0,2-->0,4---->0,2------>0,2

a. \(V_{HCl}=\dfrac{0,4}{1}=0,4\left(l\right)=400\left(ml\right)\)

b.\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c. \(CM_{FeCl_2}=\dfrac{0,2}{0,4}=0,5M\)

Bài 10:

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

0,1<--0,2<----0,1<----0,1

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

b. \(V_{HCl}=\dfrac{0,2}{0,1}=2\left(l\right)=200\left(ml\right)\)

c.\(CM_{FeCl_2}=\dfrac{0,1}{2}=0,05M\)

32.24.25=768.25

              =19200

Chúc Bạn Học Giỏi

32.24.25=(32.6).(25.4)

=192.100

=19200

Đáp án là 19200 nha.

A

A : mang tính giáo dục 

A