xét tam giác ABH và tam gia ACH có

AB=AC[tm gic ABC cân tại A]

góc AHC=góc AHB= ̣90 độ

góc B =góc C[ABC cân tại A]

suy ra tam giác ABH = tam giác ACH[cạnh huyền góc nhọn]

x bằng bao nhiêu vậy bạn

\(\Leftrightarrow\frac{x-5}{325}-1+\frac{x-6}{324}-1=\frac{x-7}{323}-1+\frac{x-8}{322}-1\)

\(\Leftrightarrow\frac{x-330}{325}+\frac{x-330}{324}-\frac{x-330}{323}-\frac{x-330}{322}=0\)

\(\Leftrightarrow\left(x-330\right)\left(\frac{1}{325}+\frac{1}{324}-\frac{1}{323}-\frac{1}{322}\right)=0\)

Ma \(\frac{1}{325}+\frac{1}{324}-\frac{1}{323}-\frac{1}{322}\ne0\)

\(\Rightarrow x-330=0\)

\(\Rightarrow x=330\)

\(\frac{x-5}{325}+\frac{x-6}{324}=\frac{x-7}{323}+\frac{x-8}{322}\)

\(\Leftrightarrow\left[\frac{x-5}{325}-1\right]+\left[\frac{x-6}{324}-1\right]=\left[\frac{x-7}{323}-1\right]+\left[\frac{x-8}{322}-1\right]\)

\(\Leftrightarrow\frac{x-5-325}{325}+\frac{x-6-324}{324}=\frac{x-7-323}{323}+\frac{x-8-322}{322}\)

\(\Leftrightarrow\frac{x-330}{325}+\frac{x-330}{324}=\frac{x-330}{324}+\frac{x-330}{332}\)

\(\Leftrightarrow\frac{x-330}{325}+\frac{x-330}{324}=\frac{x-330}{324}+\frac{x-330}{332}=0\)

Auto làm nốt :v