bạn đăng tách ra tầm 10 câu mỗi lần đăng nha, chứ dài ntnay ngại làm lắm~

chu vi hình vuông là
50x4=200cm
độ dài 1 cạnh hình vuông là
50:4=12,5cm
diện tích hình vuông là
12,5x12,5=156,25cm²

a: \(P=\dfrac{8}{x\left(x+4\right)}+\dfrac{5x}{x\left(x+4\right)}-\dfrac{2x+8}{x\left(x+4\right)}=\dfrac{8+5x-2x-8}{x\left(x+4\right)}=\dfrac{3}{x+4}\)

b: Thay x=1/2 vào P, ta được:

P=3:9/2=3x2/9=6/9=2/3

Với khác 0 ; x khác 4 

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

Thay x = 1/2 vào P ta được \(\dfrac{3}{\dfrac{1}{2}+4}=\dfrac{3}{\dfrac{9}{2}}=3:\dfrac{9}{2}=\dfrac{2}{3}\)

a, \(n_{CaCO_3}=\dfrac{41,2}{100}=0,412\left(mol\right)\)

PTHH: CaO + H₂O → Ca(OH)₂

Mol:      0,412                 0,412                        

PTHH: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Mol:       0,412        0,412          0,412

\(m_{CaO}=0,412.56=23,072\left(g\right)\)

b, \(V_{CO_2}=0,412.22,4=9,2288\left(l\right)\)

\(m_{Na_2CO_3}=100.16,96\%=16,96\left(g\right)\Rightarrow n_{Na_2CO_3}=\dfrac{16,96}{106}=0,16\left(mol\right)\)

\(m_{BaCl_2}=200.10,4\%=20,8\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

PTHH: Na₂CO₃ + BaCl₂ → BaCO₃ + 2NaCl

Mol:      0,1             0,1                           0,2

Ta có: \(\dfrac{0,16}{1}>\dfrac{0,1}{1}\) ⇒ Na₂CO₃ dư, BaCl₂ hết

m_{dd sau pứ} = 100 + 200 = 300 (g)

\(C\%_{ddNaCl}=\dfrac{0,1.58,5.100\%}{300}=1,95\%\)

\(C\%_{ddNa_2CO_3}=\dfrac{\left(0,16-0,1\right).106.100\%}{300}=2,12\%\)

a) Ta có:

\(VT=\dfrac{\left(x-2\right)^3}{x^2-2x}\)

\(=\dfrac{\left(x-2\right)^3}{x\left(x-2\right)}\)

\(=\dfrac{\left(x-2\right)^2}{x}=VP\left(dpcm\right)\)

b) Ta có:

\(VT=\dfrac{x^2-4x+3}{x-3}\)

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

\(=\dfrac{\left(x-3\right)\left(x-1\right)}{x-3}=x-1\)

\(VP=\dfrac{x^2-5x+4}{x-4}\)

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

\(=\dfrac{\left(x-1\right)\left(x-4\right)}{x-4}=x-1\)

\(\Rightarrow VT=VP=x-1\left(dpcm\right)\)

c) Ta có:
\(VT=\dfrac{1-x}{-5x+1}\)

\(=\dfrac{-x+1}{-\left(5x-1\right)}\)

\(=\dfrac{-\left(x-1\right)}{-\left(5x-1\right)}\)

\(=\dfrac{x-1}{5x-1}=VP\left(dpcm\right)\)

d) Ta có: 

\(VT=\dfrac{x^3+27}{x^2-3x+9}\)

\(=\dfrac{x^3+3^3}{x^2-3x+9}\)

\(=\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}\)

\(=x+3=VP\left(dpcm\right)\)