cái chỗ trên cạnh AB và AC là AB và BC đk

minh cung chua hoc bai nay

tt

liuliu

\(\frac{x}{3}-x+1=\frac{x}{4}-\frac{x+1}{3}\)

\(\Leftrightarrow12.\left(\frac{x}{3}-x+1\right)=12.\left(\frac{x}{4}-\frac{x+1}{3}\right)\)

\(\Leftrightarrow4x-12x+12=3x-4.\left(x+1\right)\)

\(\Leftrightarrow4x-12x+12=3x-4x-4\)

\(\Leftrightarrow4x-12x-3x+4x=\left(-12\right)-4\)

\(\Leftrightarrow\left(-7\right)x=-16\)

\(\Leftrightarrow x=\frac{16}{7}\)

Vậy \(x=\frac{16}{7}\)

\(4x-12x+12=3x-4\left(x+1\right)\Leftrightarrow-8x+12=-x-4\)

\(\Leftrightarrow-7x=-16\Leftrightarrow x=\dfrac{16}{7}\)

\(\left\{{}\begin{matrix}n_{CH_4}+n_{H_2}=\dfrac{6,72}{22,4}=0,3\\\dfrac{2.n_{H_2}}{16.n_{CH_4}+2.n_{H_2}}.100\%=30\%\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{CH_4}=\dfrac{21}{310}\left(mol\right)\\n_{H_2}=\dfrac{36}{155}\left(mol\right)\end{matrix}\right.\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

          \(\dfrac{21}{310}\)---------------------->\(\dfrac{21}{155}\)

            2H₂ + O₂ --t^o--> 2H₂O

           \(\dfrac{36}{155}\)------------>\(\dfrac{36}{155}\)

=> \(m_{H_2O}=\left(\dfrac{21}{155}+\dfrac{36}{155}\right).18=\dfrac{1026}{155}\left(g\right)\)

Trả lời:

 Vì \(\Delta ABC~\Delta MNP\Rightarrow\frac{AB}{MN}=\frac{AC}{PM}=\frac{BC}{NP}=\frac{AB}{18}=\frac{AC}{18}=\frac{BC}{15}\)

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

\(\frac{AB}{18}=\frac{AC}{18}=\frac{BC}{15}=\frac{AB+AC}{18+18}=\frac{12}{36}=\frac{1}{3}\)

\(\Rightarrow BC=\frac{1}{3}.15=5\left(cm\right)\)

Chu vi tam giác ABC là: AB + AC + BC = 6 + 6 + 5 = 17 (cm)