a)\(\frac{x^2+xy}{x^2-y^2}=\frac{x\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}=\frac{x}{x-y}\)

b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{-5x-2}{x^2-4}\)

\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x-8+3x+6-5x-2}{\left(x-2\right)\left(x+2\right)}\)

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

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

a, x(x-y)+2(x-y)=(x-y)(x+2)

b, \(x^2-6xy+9y^2=\left(x-3y\right)^2\)Thay x=16, y=2 có

                \(x^2-6xy+9y^2=\left(x-3y\right)^2=\left(16-2\cdot3\right)^2=10^2=100\)

Learning English is very important.

=> It is very important to learn English.

It is very important for learning English.

Đúng thì nha

Bài 1.

a) \(m_{CO_2}=n_{CO_2}\times M_{CO_2}=2,5\times44=110\left(g\right)\)

b) \(n_{SO_2}=\frac{V_{SO_2}}{22,4}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

\(m_{SO_2}=n_{SO_2}\times M_{SO_2}=0,1\times64=6,4\left(g\right)\)

Bài 2.

a) \(n_{CO_2}=\frac{m_{CO_2}}{M_{CO_2}}=\frac{4,4}{44}=0,1\left(mol\right)\)

\(V_{CO_2}=n_{CO_2}\times22,4=0,1\times22,4=2,24\left(l\right)\)

b) Ta có 1 mol NH₃ = 6.10²³ phân tử NH₃

=> 3.10²³ phân tử NH₃ = 0, 5 mol NH₃

\(V_{NH_3}=n_{NH_3}\times22,4=0,5\times22,4=11,2\left(l\right)\)

Bài 1:

a) \(m_{CO_2}=n.M=2,5.44=110\left(g\right)\)

b) \(n_{SO_2}=\frac{V}{22,4}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Leftrightarrow m_{SO_2}=n.M=0,1.64=6,4\left(g\right)\)

Bài 2:

a) \(n_{CO_2}=\frac{m}{M}=\frac{4,4}{44}=0,1\left(mol\right)\)

\(\Leftrightarrow V_{CO_2}=n.22,4=0,1.22,4=2,24\)(lít)

b) \(n_{NH_3}=\frac{a}{N}=\frac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

\(\Leftrightarrow V_{NH_3}=n.22,4=0,5.22,4=11,2\)(lít)

Lưu ý: a là số phân tử.

a³ + b³ + c³ = 3abc

⇔ ( a³ + b³ ) + c³ - 3abc = 0

⇔ ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

⇔ [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

⇔ ( a + b + c )[ ( a + b )² - ( a + b ).c + c² ] - 3ab( a + b + c ) = 0

⇔ ( a + b + c )( a² + 2ab + b² - ac - bc + c² - 3ab ) = 0

⇔ ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

⇔ \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

Từ đây tự làm tiếp nhé :))

Ta có : \(a^3+b^3+c^3=3abc\)

\(\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Rightarrow\left(a+b+c\right)^3-3\left(a+b\right)c\left(a+b+c\right)-3ab\left(a+b+c\right)=0\)

\(\Rightarrow\left(a+b+c\right)[\left(a+b+c\right)^2-3ac-3bc-3ab]=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac\right)=0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

​Để \(N\)có nghĩa thì \(\left(a+b+c\right)^2\ne0\)

Hay \(a+b+c\ne0\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(c-a\right)^2\ge0\forall c,a\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)\(\Rightarrow a=b=c\)

Thay \(a=b=c\)vào \(N\), ta có : \(N=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

Vậy \(N=\frac{1}{3}\)