\(xy\left(x-y\right)-xz\left(x+z\right)+yz\left(2x-y+z\right)\)

\(=xy\left(x-y\right)-xz\left(x+z\right)+yz\left[\left(x-y\right)+\left(x+z\right)\right]\)

\(=xy\left(x-y\right)-xz\left(x+z\right)+yz\left(x-y\right)+yz\left(x+z\right)\)

\(=\left(x-y\right)\left(xy+yz\right)+\left(x+z\right)\left(yz-xz\right)\)

\(=y\left(x-y\right)\left(x+z\right)-z\left(x+z\right)\left(x-y\right)\)

\(=\left(x-y\right)\left(x+z\right)\left(y-z\right)\)

\(\frac{1}{ab}+\frac{1}{a^2+b^2}=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}\ge\frac{1}{2ab}+\frac{4}{a^2+2ab+b^2}\)

\(\ge\frac{1}{\frac{\left(a+b\right)^2}{2}}+\frac{4}{\left(a+b\right)^2}=\frac{2}{1}+\frac{4}{1}=6\)

A/ 

Số mol NH3 có trong 6,72l NH3 ở đkt :

\(n_{NH_3}=\frac{V_{NH_3}}{24}=\frac{6,72}{24}=0,28\left(mol\right)\)

Khối lượng của 6,72l NH3 ở đkt :

\(m_{NH_3}=n_{NH_3}.M_{NH_3}=0,28.17=4,76\left(gam\right)\)

B/Số nguyên tử có trong 1,5 mol Nhôm :

\(n_{Al}.6.10^{23}=1,5.6.10^{23}=9.10^{23}\) ( nguyên tử )

C/ Số mol có trong 8 gam khí Oxi :

\(n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{8}{16}=0,25\left(mol\right)\)

Thể tích đktc của 8 gam khí Oxi :

\(V_{O_2}=n_{O_2}.22,4=0,25.22,4=5,6\left(l\right)\)

Còn Câu D/ XD

Số mol CO2 có trong 0,336l CO2 ở đktc :

\(n_{CO_2}=\frac{V_{CO_2}}{22,4}=\frac{0,336}{22,4}=0,015\left(mol\right)\)

Khối lượng của 0,25 mol O2 :

\(m_{O_2}=n_{O_2}.M_{O_2}=0,25.32=8\left(gam\right)\)

Khối lượng của 0,015mol CO2 :

\(m_{CO_2}=n_{CO_2}.M_{CO_2}=0,015.44=0,66\left(gam\right)\)

Khối lượng hỗn hợp CO2 và O2 :

\(0,66+8=8,66\left(gam\right)\)

Thể tích hỗn hợp CO2 và O2 ở đktc :

\(V_{hỗnhợp}=\left(n_{CO_2}+n_{O_2}\right).22,4=\left(0,25+0,015\right).22,4=5,936\left(l\right)\)

Xog r =.= Fine

a) ĐKXĐ : \(x\ne\left\{\pm1\right\}\)

\(B=\frac{x-1}{x+1}-\frac{x+1}{x-1}-\frac{4}{1-x^2}\)

\(B=\frac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{4}{\left(x-1\right)\left(x+1\right)}\)

\(B=\frac{\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+4}{\left(x+1\right)\left(x-1\right)}\)

\(B=\frac{x^2-2x+1-x^2-2x-1+4}{\left(x+1\right)\left(x-1\right)}\)

\(B=\frac{4-4x}{\left(x+1\right)\left(x-1\right)}\)

\(B=\frac{-4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

\(B=\frac{-4}{x+1}\)

b) \(x^2-x=0\Leftrightarrow x\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\left(Chon\right)\\x=1\left(Loai\right)\end{cases}}\)

Thay x = 0 vào B ta có :

\(B=\frac{-4}{0+1}=-4\)

Chọn C.

ĐKXĐ : \(x\ne2\)

\(\frac{3x^2+6x+12}{x^3-8}\)

\(=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\frac{3}{x-2}\)

3x²+6x+12/x³-8

=3*(x^2+2x+4)/x^3-2^3

=3*(x^2+2x+4)/(x-2)*(x^2+2x+4)

=3/x-2

\(x^{11}+x^7+1\)

\(=\left(x^{11}-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x^9-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x^3-1\right)\left(x^6+x^3+1\right)+x\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x-1\right)\left(x^2+x+1\right)\left(x^6+x^3+1\right)+x\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)\left(x^6+x^3+1\right)+x\left(x-1\right)\left(x^3+1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left(x^9+x^6+x^3-x^8-x^5-x^2+x^5+x^2-x^4-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^4+x^3-x+1\right)\)

\(x^7+x^2+1\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)

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

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

\(=\left(x^2+x+1\right)\left(x^5+x^2-x^4-x+1\right)\)