<=>\(\left(x^3-4x^2\right)+\left(x^2-4x\right)+\left(5x-20\right)=0\)

<=>\(x^2\left(x-4\right)+x\left(x-4\right)+5\left(x-4\right)=0\)

<=>\(\left(x^2+x+5\right)\left(x-4\right)=0\)

Vì \(x^2+x+5>0\)=>x-4=0

<=>x=4

Áp dụng tính chất của dãy tỉ số bằng nhau, có:

    \(\frac{x}{2}=\frac{y}{3}=\frac{x+y}{2+3}=\frac{20}{5}=4\)

Suy ra: \(\frac{x}{2}=4\Rightarrow x=4\cdot2=8\)

              \(\frac{y}{3}=4\Rightarrow y=3\cdot4=12\)

\(\dfrac{x}{18}=\dfrac{4}{3}\Rightarrow x=\dfrac{18.4}{3}=24\\ \dfrac{20}{y}=\dfrac{4}{3}\Rightarrow y=\dfrac{20.3}{4}=15\\ \dfrac{z}{21}=\dfrac{4}{3}\Rightarrow z=\dfrac{21.4}{3}=28\)

Ta có:

\(\dfrac{x}{18}\) = \(\dfrac{4}{3}\)

⇒ x = \(\dfrac{4}{3}\) . 18

⇒ x = 24

\(\dfrac{20}{y}\) = \(\dfrac{4}{3}\)

⇒ y = 20 : \(\dfrac{4}{3}\)

⇒ y = 15

\(\dfrac{z}{21}\) = \(\dfrac{4}{3}\)

⇒ z = \(\dfrac{4}{3}\) . 21

⇒ z = 28

⇒ x + y + z = 24 + 15 + 28 = 67

Vậy x + y + z = 67

x = 24, y = 15, z = 28;

Ta có: \(\dfrac{x}{y}=\dfrac{7}{20}\)

nên \(\dfrac{x}{7}=\dfrac{y}{20}\)(1)

Ta có: \(\dfrac{y}{z}=\dfrac{5}{8}\)

nên \(\dfrac{y}{5}=\dfrac{z}{8}\)

hay \(\dfrac{y}{20}=\dfrac{z}{32}\)(2)

Từ (1) và (2) suy ra \(\dfrac{x}{7}=\dfrac{y}{20}=\dfrac{z}{32}\)

hay \(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}\)

mà 2x-5y+2z=100

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x}{14}=\dfrac{5y}{100}=\dfrac{2z}{64}=\dfrac{2x-5y+2z}{14-100+64}=\dfrac{100}{-22}=\dfrac{-50}{11}\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{7}=\dfrac{-50}{11}\\\dfrac{y}{20}=\dfrac{-50}{11}\\\dfrac{z}{32}=-\dfrac{50}{11}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{350}{11}\\y=\dfrac{-1000}{11}\\z=\dfrac{-1600}{11}\end{matrix}\right.\)

Ta có:  \(\dfrac{x}{y}=\dfrac{7}{20}\Rightarrow\dfrac{x}{7}=\dfrac{y}{20}\Rightarrow\dfrac{x}{14}=\dfrac{y}{40}\Rightarrow\dfrac{2x}{28}=\dfrac{5y}{200}\) \(\left(1\right)\)

Lại có:  \(\dfrac{y}{z}=\dfrac{5}{8}\Rightarrow\dfrac{y}{5}=\dfrac{z}{8}\Rightarrow\dfrac{y}{40}=\dfrac{z}{64}\Rightarrow\dfrac{5y}{200}=\dfrac{2z}{128}\)   \(\left(2\right)\)

Kết hợp ( 1 ) và ( 2 ) ta có:     \(\dfrac{2x+5y-2z}{28+200-128}=\dfrac{100}{100}=1\)

⇒  \(\dfrac{2x}{28}=1\Rightarrow x=\dfrac{1.28}{2}=14\)

⇒  \(\dfrac{5y}{200}=1\Rightarrow y=\dfrac{1.200}{5}=40\)

⇒  \(\dfrac{2z}{128}=1\Rightarrow z=\dfrac{1.128}{2}=64\)

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\(d_{\dfrac{X}{kk}}=1,034\\ M_{kk}=29\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_X=d_{\dfrac{X}{kk}}.M_{kk}=1,034.29=30\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow m_C=\%C.M_X=80\%.30=24\left(g\right)\\ m_H=m_X-m_C=30-24=6\left(g\right)\)

\(\Rightarrow n_C=\dfrac{m}{M}=\dfrac{24}{12}=2\left(mol\right)\\ n_H=\dfrac{m}{M}=\dfrac{6}{1}=6\left(mol\right)\\ CTHH.của.X:C_2H_6\)

\(M_X=1,034.29=30\left(\dfrac{g}{mol}\right)\)

\(m_C=\dfrac{30.80\%}{100\%}=24\left(g\right)\)

\(n_C=\dfrac{23}{12}=2\left(mol\right)\)

\(m_H=\dfrac{30.20\%}{100\%}=6\left(g\right)\)

\(n_H=\dfrac{6}{1}=6\left(mol\right)\)

\(=>CTHH:C_2H_6\)

    \(x+x+x+81=-3-x\)

⇔\(4x=-84\)

⇔\(x=-21\)

`x xx 95 + x + x : 0,25 = 7963`

`x xx (95 + 1 + 4) = 7963`

`x xx 100 = 7963`

`=> x = 79,63`.

\(x\times\left(95+1+4\right)=7963\)

\(100\times x=7963\)

\(x=79,63\)

\(=>x\left(95+1+4\right)=7963\)

\(x=\dfrac{7963}{95+1+4}=\dfrac{7963}{100}=79,63\)

`x xx95+x+x:0,25=102,34`

`x xx95+x+x xx4=102,34`

`x xx(95+1+4)=102,34`

`x xx100=102,34`

`x=102,34:100=1,0234`

`x xx 95 + x + x :0,25 = 102,34`

`x xx 95 + x + x xx 4 = 102,34`

`x xx ( 95+1+4)=102,34`

`x xx 100 = 102,34`

`x = 102,34 : 100`

`x = 1,0234`

`x xx95+x+x:0,25=102,34`

`x xx(95+1+4)=102,34`

`x xx100=102,34`

`x=102,34:100`

`x=1,0234`