\(\dfrac{x+1}{98}+\dfrac{x+2}{97}+\dfrac{x+90}{9}+\dfrac{x+84}{15}>-4\\ \Leftrightarrow\left(\dfrac{x+1}{98}+1\right)+\left(\dfrac{x+2}{97}+1\right)+\left(\dfrac{x+90}{9}+1\right)+\left(\dfrac{x+84}{15}+1\right)>0\\ \Leftrightarrow\dfrac{x+99}{98}+\dfrac{x+99}{97}+\dfrac{x+99}{9}+\dfrac{x+99}{15}>0\\ \Leftrightarrow\left(x+99\right)\left(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}\right)>0\)

Vì \(\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{9}+\dfrac{1}{15}>0\Rightarrow x+99>0\Rightarrow x>-99\)

cho tớ hỏi ông  

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

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

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