\(\dfrac{2}{x^2+4x+3}+\dfrac{5}{x^2+11x+24}+\dfrac{2}{x^2+18x+80}=\dfrac{9}{52}\\ ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\\ \Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{52\left(x+10\right)}{52\left(x+1\right)\left(x+10\right)}-\dfrac{52\left(x+1\right)}{52\left(x+1\right)\left(x+10\right)}=\dfrac{9\left(x+1\right)\left(x+10\right)}{52\left(x+1\right)\left(x+10\right)}\\ \Leftrightarrow52\left(x+10\right)-52\left(x+1\right)=9\left(x+1\right)\left(x+10\right)\\ \Leftrightarrow9\left(x^2+10x+x+10\right)=52\left(x+10-x-1\right)\\ \Leftrightarrow9\left(x^2+11x+10\right)=52\cdot9\\ \Leftrightarrow x^2+11x+10=52\\ \Leftrightarrow x^2+14x-3x-42=0\\ \Leftrightarrow x\left(x+14\right)-3\left(x+14\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+14\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\left(T/m\right)\)

Vậy.............

ĐKXĐ: \(x\ne\left\{-10;-8;-3;-1\right\}\)

\(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=52\)

\(\Leftrightarrow x^2+11x-42=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\)

\(\Rightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+10x+8x+80}=\frac{9}{52}\)

\(\Rightarrow\frac{2}{x\left(x+1\right)+3\left(x+1\right)}+\frac{5}{x\left(x+3\right)+8\left(x+3\right)}+\frac{2}{x\left(x+10\right)+8\left(x+10\right)}=\frac{9}{52}\)

\(\Rightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\Rightarrow\frac{x+10-x-1}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\Rightarrow\frac{9}{x^2+11x+10}=\frac{9}{52}\)

\(\Rightarrow x^2+11x+10=52\Rightarrow x^2+2\cdot\frac{11}{2}x+\frac{121}{4}-\frac{81}{4}=52\)

\(\Rightarrow\left(x+\frac{11}{2}\right)^2=\frac{289}{4}\Rightarrow x+\frac{11}{2}=\frac{17}{2}\Rightarrow x=\frac{17}{2}-\frac{11}{2}=\frac{6}{2}=3\Rightarrow x=3\)

\(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{52}\)(ĐKXĐ: x khác -1;-3;-8;-10)

\(\Leftrightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+8x+10x+80}=\frac{9}{52}\)

\(\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\frac{2\left(x+8\right)\left(x+10\right)+5\left(x+1\right)\left(x+10\right)+2\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\frac{9x^2+99x+216}{x^4+22x^3+155x^2+374x+240}=\frac{9}{52}\)

\(\Rightarrow468x^2+5148x+11232=9x^4+198x^3+1395x^2+3366x+2160\)

\(\Leftrightarrow9x^4+198x^3+927x^2-1782x-9072=0\)

\(\Leftrightarrow x^4+22x^3+103x^2-198x-1008=0\)

\(\Leftrightarrow x^4-3x^3+25x^3-75x^2+178x^2-534x+336x-1008=0\)

\(\Leftrightarrow x^3\left(x-3\right)+25x^2\left(x-3\right)+178x\left(x-3\right)+336\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+25x^2+178x+336\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2+22x^2+66x+112x+336\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[x^2\left(x+3\right)+22x\left(x+3\right)+112\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+22x+112\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+8x+14x+112\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left[x\left(x+8\right)+14\left(x+8\right)\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left(x+8\right)\left(x+14\right)=0\)

\(\Leftrightarrow\frac{\orbr{\begin{cases}x+3=0\\x-3=0\end{cases}}}{\orbr{\begin{cases}x+8=0\\x+14=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-3\left(\times\right)\\x=3\end{cases}}}{\orbr{\begin{cases}x=-8\left(\times\right)\\x=-14\end{cases}}}\)(Vì x=-3 và x=-8 không t/m ĐKXĐ)

Vậy tập nghiệm của pt là \(S=\left\{3;-14\right\}.\)

Đáp án B

23 x23 x23 x23 x23 x23 = 23⁶ = 148035889

23 x 23 x 23 x 23 x 23 x 23 = 23⁶ nha bạn

=23^6= 148 035 889

a \(\dfrac{1}{x-y}+\dfrac{2}{x+y}+\dfrac{3x}{y^2-x^2}\)

\(=\dfrac{x+y+2x-2y-3x}{\left(x-y\right)\left(x+y\right)}=\dfrac{-y}{\left(x-y\right)\left(x+y\right)}\)

b: \(\dfrac{1}{x-2}+\dfrac{1}{x+2}-\dfrac{4x-4}{x^2-4}\)

\(=\dfrac{x+2+x-2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{-2x+4}{\left(x-2\right)\left(x+2\right)}\)

=-2/x+2

c: \(\dfrac{x+1}{x+3}-\dfrac{x-1}{3-x}+\dfrac{2x-2x^2}{x^2-9}\)

\(=\dfrac{\left(x+1\right)\left(x-3\right)+\left(x-1\right)\left(x+3\right)+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{x^2-2x-3+x^2+2x-3+2x-2x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{2x-6}{\left(x+3\right)\left(x-3\right)}=\dfrac{2}{x+3}\)

80 − (4 . 52 – 3 . 23)

= 80 – ( 4 . 25 – 3 . 8)

= 80 – (100 – 24 )

= 80 – 76 = 4

52 - | x - 3 | = 80

\(\Rightarrow\)| x - 3 | = 52 - 80

\(\Rightarrow\)| x - 3 | = -28

\(\Rightarrow\)x không có giá trị thỏa mãn

52 - | x - 3 | = 80

       | x - 3 | = 52 - 80

       | x - 3 | = - 28

         \(52-\left|x-3\right|=80\)

\(\Leftrightarrow\)\(\left|x-3\right|=52-80\)

\(\Leftrightarrow\)\(\left|x-3\right|=-28\)  (vô lí   vì  |a| luôn lớn hơn hoặc = 0 )

Vậy ko tìm đc x thỏa mãn

=80-(100-24)

=80-76

=4 

80-( 4.25-3.8)

=80-(100-24)

=80-100+24

=-20+24

=4