(x-5)²=36

<=>\(\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\)<=> \(\orbr{\begin{cases}x=11\\x=-1\end{cases}}\)

Vậy:...

\(\left(x-5\right)^2-36=0\)

\(\left(x-5\right)^2=0+36\)

\(\left(x-5\right)^2=36\)

\(\left(x-5\right)^2=\pm6^2\)

\(\Rightarrow x-5=6\text{ hoặc }x-5=-6\)

     \(x=6+5\)        \(x=-6+5\)

     \(x=11\)              \(x=-1\)

Vậy x = 11; x = -1 

ủa anh nghĩ vợ anh lớp 7 mà :( 

Bài 1 : em tự kết luận nghiệm nha, anh lười quáaaaa

a, \(8+2x=5x-12\Leftrightarrow-3x=-20\Leftrightarrow x=\frac{20}{3}\)

b, \(4x\left(x-3\right)+3\left(x-3\right)=0\Leftrightarrow\left(4x+3\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{3}{4};x=3\)

c, \(\frac{x+2}{4}+\frac{x-3}{3}=\frac{x-1}{6}\Leftrightarrow\frac{3x+6+4x-12}{12}=\frac{2x-2}{12}\)

\(\Rightarrow7x-6=2x-2\Leftrightarrow5x=4\Leftrightarrow x=\frac{4}{5}\)

d, \(\frac{3}{x-2}-\frac{1}{x+2}=\frac{x-1}{x^2-4}\)ĐK : \(x\ne\pm2\)

\(\Leftrightarrow\frac{3x+6-x+2}{\left(x-2\right)\left(x+2\right)}=\frac{x-1}{\left(x-2\right)\left(x+2\right)}\Rightarrow2x+8=x-1\Leftrightarrow x=-9\)( tmđk )

Làm bài 3

\(\frac{1}{9+x}-\frac{1}{x}=\frac{1}{5}+\frac{1}{4}\)

\(\frac{1}{9+x}-\frac{1}{x}=\frac{9}{20}\)

\(\frac{1}{x}+\frac{1}{9}-\frac{1}{x}=\frac{9}{20}\)

\(0=\frac{9}{20}-\frac{1}{9}\)

Pt vô nghiệm :)

đạo nhái ảnh đại diện

\(\frac{x+2}{2019}+\frac{x+3}{2018}=\frac{x+4}{2017}+\frac{x}{2021}\)

\(\Leftrightarrow\frac{x+2}{2019}+1+\frac{x+3}{2018}+1=\frac{x+4}{2017}+1+\frac{x}{2021}+1\)

\(\Leftrightarrow\frac{x+2021}{2019}+\frac{x+2021}{2018}=\frac{x+2021}{2017}+\frac{x+2021}{2021}\)

\(\Leftrightarrow x+2021=0\)

\(\Leftrightarrow x=-2021\)