\(a)\)\(\left(x+1\right)\left(x+3\right)-x\left(x-1\right)=8\)

\(\Leftrightarrow x^2+4x+3-x^2+x=8\)

\(\Leftrightarrow5x=5\)

\(\Leftrightarrow x=1\)

Vậy x = 1.

\(b)\)\(9x^2=1-\left(3x+1\right)\left(2x-9\right)\)

\(\Leftrightarrow\left(1-9x^2\right)-\left(3x+1\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(1-3x\right)-\left(3x+1\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(1-3x+9-2x\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+1=0\\10-5x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=-1\\5x=10\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=2\end{cases}}\)

Vậy\(x=-\frac{1}{3}\)hoặc\(x=2\)

Dumflinz

Ta có : 24x³ + 18x² + 36x + 27

= ( 24x³ + 18x² ) + ( 36x + 27 )

= 2x²( 12x + 9 ) + 3( 12x + 9 )

= ( 12x + 9 )( 2x² + 3 )

Vậy mỗi học sinh trồng được  ( 24x³ + 18x² + 36x + 27 ) : ( 12x + 9 ) = ( 12x + 9 )( 2x² + 3 ) : ( 12x + 9 ) = 2x² + 3 cây [ x nguyên dương ]

Cần ko gấp

Travelers to vietnam expect to experience the country's rich culture and vibrant traditions

-> Traverlers to Vietnam look_____________for experiencing the country's rich culture and vibrant traditions____________________________________________

cạnh huyền là 5^2 + 7^2=9^2

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6,5 cm nha nb

1.\(4FeS_2+11O_2\rightarrow8SO_2+2Fe_2O_3\)

2.\(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\)

3. \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+NaSO_4\)

\(1.FeS_2+O_2---->SO_2+Fe_2O_3\)

 \(4FeS_2+11O_2\rightarrow8SO_2+2Fe_2O_3\)

\(2.Cu+HNO_3---->CU\left(NO_3\right)_2+NO+H_2O\)

  \(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\)

\(3.CuSO_4+NaOH---->Cu\left(OH\right)_2+Na_2SO_4\)

    \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

\(n_S=\frac{m}{M}=\frac{16}{32}=0,5\left(mol\right)\)

a,PTHH: \(S+O_2\rightarrow SO_2\)(có nhiệt độ nữa nhé)

(mol)    1         1          1

(mol)    0,5      0,5      0,5

b) Theo pt, ta có: \(n_S=n_{SO_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{SO_2}=n.M=0,5.64=32\left(gam\right)\)

c)Theo pt, ta có: \(n_S=n_{O_2}=n_{SO_2}=0,5\left(mol\right)\)

\(\Rightarrow V_{O_2}\left(đktc\right)=n.22,4=0,5.22,4=11,2\left(lít\right)\)

\(\Rightarrow V_{SO_2}\left(đktc \right)=n.22,4=0,5.22,4=11,2\left(lit\right)\)

TL

a)\(S+O_2\underrightarrow{t^0}SO_2\)

b)\(n_S=\frac{16}{32}=0,5\left(mol\right)\)

\(S+O_2\underrightarrow{t^0}SO_2\)\(\)

0,5->0,5        0,5 (mol)

\(m_{SO_2}=0,5.64=32\left(g\right)\)

c)

\(V_{SO_2}=0,5.22,4=11,2\left(l\right)\)