`(x^2-10x-29)/1971+(x^2-10x-27)/1973=(x^2-10x-1971)/1929+(x^2-10x-1973)/1927`

`<=>(x^2-10x-29)/1971-1+(x^2-10x-27)/1973-1=(x^2-10x-1971)/1929-1+(x^2-10x-1973)/1927-1`

`<=>(x^2-10x-200)/1971+(x^2-10x-200)/1973=(x^2-10x-200)/1971+(x^2-10x-200)/1927`

`<=>(x^2-10x-200)(1/1971+1/1973-1/1929-1/1927)=0`

`<=>x^2-10x-200=0` do `1/1971+1/1973-1/1929-1/1927<0`

`<=>x^2-20x+10x-200=0`

`<=>x(x-20)+10(x-20)=0`

`<=>(x-20)(x+10)=0`

`<=>` \(\left[ \begin{array}{l}x=20\\x=-10\end{array} \right.\) 

Vậy `S={20,-10}`

Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\left(\dfrac{x^2-10x-27}{1973}-1\right)+\left(\dfrac{x^2-10x-29}{1971}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1973}+\dfrac{x^2-10x-2000}{1971}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)

\(\Leftrightarrow\left(x^2-10x-2000\right)=0\) vì \(\left(\dfrac{1}{1973}+\dfrac{1}{1971}-\dfrac{1}{29}-\dfrac{1}{27}\right)\ne0\)

\(\Leftrightarrow x^2-50x+40x-2000=0\)

\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)

\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-50=0\\x+40=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=50\\x=-40\end{matrix}\right.\)

Vậy: Giá trị x thỏa mãn là: \(x=-40;50\)

\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)

\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)

vì \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)

Nên \(x^2-10x-2000=0\)

<=> \(x^2-50x+40x-2000=0\)

<=> \(x\left(x-50\right)+40\left(x-50\right)=0\)

<=> \(\left(x-50\right)\left(x+40\right)=0\)

<=> \(x=50\) hoặc \(x=-40\)

Vậy tập nghiệm của phương trình là \(S=\left\{50;-40\right\}\)

\(\text{a) }\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\\ \Leftrightarrow\dfrac{2-x-2002}{2002}=\left(\dfrac{1-x}{2003}-1\right)+\left(1-\dfrac{x}{2004}\right)\\ \Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2003-x}{2003}-\dfrac{2004-x}{2004}=0\\ \Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\\ \Leftrightarrow2004-x=0\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\right)\\ \Leftrightarrow x=2004\)

Vậy phương trình có nghiệm \(x=2004\)

\(\text{b) }\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\left(\text{ Chữa đề }\right)\\ \Leftrightarrow\left(\dfrac{x^2-10x-29}{1971}-1\right)+\left(\dfrac{x^2-10x-27}{1973}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\\ \Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-2000}{29}-\dfrac{x^2-10x-2000}{27}=0\\ \Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\\ \Leftrightarrow x^2-10x-2000=0\left(\text{Vì }\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\right)\\ \Leftrightarrow x^2-20x+10x-2000=0\\ \Leftrightarrow x\left(x-20\right)+10\left(x-20\right)=0\\ \Leftrightarrow\left(x+10\right)\left(x-20\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+10=0\\x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{-10;20\right\}\)

em moi hoc lop 7 thoi a doi xong ki 2 nha

em mới học lớp 7 thôi

\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1973}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\frac{x^2-10x-29}{1971}-1+\frac{x^2-10x-27}{1973}-1=\frac{x^2-10x-1971}{29}-1+\frac{x^2-10x-1973}{27}-1\)

\(\Leftrightarrow\frac{x^2-10x-29-1971}{1971}+\frac{x^2-10x-27-1973}{1973}=\frac{x^2-10x-1971-29}{29}+\frac{x^2-10x-193-27}{27}\)

\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}-\frac{x^2-10x-2000}{29}-\frac{x^2-10x-2000}{27}=0\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)

\(\Leftrightarrow x^2-10x-2000=0\)

\(\Leftrightarrow x^2-50x+40x-2000=0\)

\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)

\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)

     Th1  \(x-50=0\Leftrightarrow x=50\)

    Th2  \(x+40=0\Leftrightarrow x=-40\)

Vậy tập nghiệm của phương trình là   \(S=\left\{50;-40\right\}\)

\(\frac{x^2-10x-29}{1971}+\frac{x^2-10x-27}{1972}=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}\)

\(\Leftrightarrow\frac{x^2-10x-29}{1971}-1+\frac{x^2-10x-27}{1973}-1=\frac{x^2-10x-1971}{29}+\frac{x^2-10x-1973}{27}-1\)

\(\Leftrightarrow\frac{x^2-10x-29-1971}{1971}+\frac{x^2-10x-27-1973}{1973}=\frac{x^2-10x-1971-29}{29}+\frac{x^2-10x-1973-27}{27}\)

\(\Leftrightarrow\frac{x^2-10x-2000}{1971}+\frac{x^2-10x-2000}{1973}-\frac{x^2-10x-2000}{29}-\frac{x^2-10x-2000}{27}=0\)

\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\frac{1}{1971}+\frac{1}{1973}-\frac{1}{29}-\frac{1}{27}\right)=0\)

\(\Leftrightarrow x^2-10x-2000=0\)

\(\Leftrightarrow x^2-50x+40x-2000=0\)

\(\Leftrightarrow x\left(x-50\right)+40\left(x-50\right)=0\)

\(\Leftrightarrow\left(x-50\right)\left(x+40\right)=0\)

         Th1:   \(x-50=0\Leftrightarrow x=50\)

        Th2:  \(x+40=0\Leftrightarrow x=-40\)

Vậy tập nghiệm của phương trình là    \(S=\left\{50;-40\right\}\)

hơi mất thời gian để chiều tôi làm cho

x= - 40 hoặc x=50

Bài khá dài nên mình chỉ có thể ghi được kết quả

a) \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)

\(\Leftrightarrow\left(\frac{x-45}{55}-1\right)+\left(\frac{x-47}{53}-1\right)=\left(\frac{x-55}{45}-1\right)+\left(\frac{x-53}{47}-1\right)\)

\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)

\(\Leftrightarrow\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)

Vì \(\hept{\begin{cases}\frac{1}{55}< \frac{1}{45}\\\frac{1}{53}< \frac{1}{47}\end{cases}}\Rightarrow\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}< 0\)

\(\Rightarrow x-100=0\Rightarrow x=100\)

Vậy x = 100

Các phần sau tương tự nhé bạn

\(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)=> \(\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)=>\(\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-2000}{29}+\dfrac{x^2-10x-2000}{27}\) => \(\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)\)

=> \(x^2-10x-2000=0\)

Tự giải ra nhé hi hi