\(Xx\left(9-2\right)=63\)
\(Xx7=63\)
\(X=63:7=9\)

x(9-2)=63

x . 7= 63

x=63:7

x=9

X x (9 - 2) = 63

X x 7 = 63

x = 63 : 7

x = 9

X + 6 = 33

X = 33 − 6

X = 27

9 + x = 22

X = 22 − 9

X = 13

X − 41 = 41

X = 41 + 41

X = 82

A = (91,5 X 26,3 - 17,4 x 52,6)x(41x11  -4100 x0,1 - 41)

A = (91,5 x 26,3 - 17,4 x 52,6) x 41 x (11 - 100X0,1 - 1)

A = (91,5 x26,,3 - 17,4 x 52,6) x 41 ( 11 - 10 - 1)

A = (91,5 x26,,3 - 17,4 x 52,6) x 41 x 0 

A = 0

0 nhé chị

\(\dfrac{x+1}{x-3}-\dfrac{41}{x+3}+\dfrac{x^2+22}{9-x^2}=0\left(ĐKXĐ:x\ne3;x\ne-3\right)\\ \Leftrightarrow\dfrac{x+1}{x-3}-\dfrac{41}{x+3}-\dfrac{x^2+22}{x^2-9}=0\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)-41\left(x-3\right)-x^2-22}{x^2-9}=0\\ \Leftrightarrow x^2+4x+3-41x+123-x^2-22=0\\ \Leftrightarrow-37x+104=0\\ \Leftrightarrow-37x=-104\\ \Leftrightarrow x=\dfrac{104}{37}\left(tmđk\right)\)

Vậy \(x=\dfrac{104}{37}\) là nghiệm của pt.

Bài 3:

\(a,\dfrac{x-1}{10}+\dfrac{x-1}{11}=\dfrac{x-1}{12}+\dfrac{x-1}{13}\)

\(\Rightarrow\dfrac{x-1}{10}+\dfrac{x-1}{11}-\dfrac{x-1}{12}-\dfrac{x-1}{13}=0\)

\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)=0\)

Mà \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\ne0\)

\(\Rightarrow x-1=0\Rightarrow x=1\)

Vậy x = 1

b, \(\dfrac{x-2000}{10}+\dfrac{x-1999}{9}=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}\)

\(\Rightarrow\dfrac{x-2000}{10}+1+\dfrac{x-1999}{9}+1=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}+1\)

\(\Rightarrow\dfrac{x-1990}{10}+\dfrac{x-1990}{9}-\dfrac{x-1990}{8}-\dfrac{x-1990}{7}=0\)

\(\Rightarrow\left(x-1990\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\right)=0\)

Mà \(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\ne0\)

\(\Rightarrow x-1990=0\Rightarrow x=1990\)

1) \(x^2+\frac{8}{9}=\frac{41}{36}\)\(\Leftrightarrow x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)

2) \(\left(x-3\right)^2+-\frac{9}{25}=\frac{2}{5}.\frac{8}{5}\)

\(\Leftrightarrow\left(x-3\right)^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=1\\x-3=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=2\end{cases}}\)

3) \(\frac{3}{11}.\frac{22}{6}-\left(x-1\right)^2=\frac{7}{16}\)

\(\Leftrightarrow\left(x-1\right)^2=\frac{9}{16}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{3}{4}\\x-1=-\frac{3}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=\frac{1}{4}\end{cases}}\)

4) \(1+\left(x+1\right)^3=\frac{37}{64}\)

\(\Leftrightarrow\left(x+1\right)^3=-\frac{27}{64}\)

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

\(\Leftrightarrow x=-\frac{7}{4}\)

5) \(\left(x-\frac{1}{2}\right)^2-\frac{9}{16}=1\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{25}{16}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{5}{4}\\x-\frac{1}{2}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{4}\\x=-\frac{3}{4}\end{cases}}\)

6) Bn ghi rõ đề nha mk ko hiểu 

6 ) \(\frac{3-x}{5-x}=\left(\frac{-3}{5}\right)^2\)

\(\frac{3-x}{5-x}=\frac{9}{25}\Leftrightarrow\frac{3-x}{5-x}-\frac{9}{25}\Leftrightarrow\frac{75-25x}{125-25x}-\frac{45-9x}{125-5x}=0\)

\(\Rightarrow\frac{75-25-45+9x}{125-25x}=0\Leftrightarrow5+9x=0\Leftrightarrow x=\frac{-5}{9}\)