X sẽ bằng 2007 vì:

2032-x/25+2053-x/21+2070-x/21+2038-x/19 = 10 ( vì đỏi vế số 10 nên = 0+10=10)

10= 1+2+3+4 (Có 4 phân số thì mỗi phân số tương ứng lần lượt la 1 ,2 ,3 ,4)

Vậy x =2007

Chúc bạn học giỏi 

=>\(\left(\frac{2032-x}{25}-1\right)+\left(\frac{2053-x}{23}-2\right)+\left(\frac{2070-x}{21}-3\right)+\left(\frac{2083-x}{19}-4\right)=0\)

=>\(\frac{2007-x}{25}+\frac{2007-x}{23}+\frac{2007-x}{21}+\frac{2007-x}{19}=0\)

=>\(\left(2007-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)

Vì \(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\)

=> 2007 - x = 0 => x = 2007

\(\frac{x-2}{71}+\frac{x-4}{69}=\frac{x-6}{67}+\frac{x-8}{65}\)

\(\Leftrightarrow\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1\)

\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}=\frac{x-73}{67}+\frac{x-73}{65}\)

\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}-\frac{x-73}{67}-\frac{x-73}{65}=0\)

\(\Leftrightarrow\left(x-73\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)

Mà \(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\ne0\)

\(x-73=0\Leftrightarrow x=73\)

\(\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1\)

\(\Leftrightarrow\frac{x-73}{71}+\frac{x-73}{69}=\frac{x-73}{67}+\frac{x-73}{65}\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)

\(\Rightarrow...\)

Ta có: \(\frac{x-2}{71}+\frac{x-4}{69}=\frac{x-6}{67}+\frac{x-8}{65}\)

\(\frac{x-2}{71}+\frac{x-4}{69}-2=\frac{x-6}{67}+\frac{x-8}{65}\)

\(\frac{x-2}{71}-1+\frac{x-4}{69}-1=\frac{x-6}{67}-1+\frac{x-8}{65}-1=0\)

\(\frac{x-73}{71}+\frac{x-73}{69}-\frac{x-6}{67}-\frac{x-8}{65}=0\)

\(\left(x-73\right)\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)=0\)

\(\left(\frac{1}{71}+\frac{1}{69}-\frac{1}{67}-\frac{1}{65}\right)\ne0\)

=> x - 73 = 0 <=> x = 73

Vậy x = 73.

Ta có ; \(\frac{x+1}{65}+\frac{x+2}{66}=\frac{x+3}{67}+\frac{x+4}{68}\)

\(\Leftrightarrow\left(\frac{x+1}{65}-1\right)+\left(\frac{x+2}{66}-1\right)=\left(\frac{x+3}{67}-1\right)+\left(\frac{x+4}{68}-1\right)\)

\(\Leftrightarrow\frac{x-64}{65}+\frac{x-64}{66}=\frac{x-64}{67}+\frac{x-64}{68}\)

\(\Leftrightarrow\left(x-64\right)\left(\frac{1}{65}+\frac{1}{66}-\frac{1}{67}-\frac{1}{68}\right)=0\)

Vì \(\left(\frac{1}{65}+\frac{1}{66}-\frac{1}{67}-\frac{1}{68}\right)\ne0\)nên \(x-64=0\Leftrightarrow x=64\)

Vậy nghiệm của phương trình ; \(S=\left\{64\right\}\)

\(\text{Ta có ; }\)\(\frac{x+1}{65}+\frac{x+2}{66}=\frac{x+3}{67}+\frac{x+4}{68}\)

\(\Leftrightarrow\left(\frac{x+1}{65}-1\right)+\left(\frac{x+2}{66}-1\right)=\)\(\left(\frac{x+3}{67}-1\right)+\left(\frac{x+4}{68}-1\right)\)

\(\Leftrightarrow\frac{x-64}{65}+\frac{x-64}{66}=\frac{x-64}{67}+\frac{x-64}{68}\)

\(\Leftrightarrow\left(x-64\right)\left(\frac{1}{65}+\frac{1}{66}-\frac{1}{67}-\frac{1}{68}\right)=0\)

\(\text{Vì}\)\(\left(\frac{1}{65}+\frac{1}{66}-\frac{1}{67}-\frac{1}{68}\right)\ne0\)\(\text{nên}\)\(x-64=0\Leftrightarrow x=64\)

\(\text{Vậy nghiệm của phương trình ; }S=\left\{64\right\}\)

tham khảo nhé 

https://olm.vn/hoi-dap/detail/103171879928.html

\(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)

\(\Leftrightarrow\left(\frac{45-x}{1963}+1\right)+\left(\frac{40-x}{1968}+1\right)+\left(\frac{35-x}{1973}+1\right)+\left(\frac{30-x}{1978}+1\right)=0\)

\(\Leftrightarrow\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1973}=0\)

\(\Leftrightarrow\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

\(\Leftrightarrow x=2008\)

\(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)

=> \(\left(\frac{45-x}{1983}+1\right)+\left(\frac{40-x}{1968}+1\right)+\left(\frac{35-x}{1973}+1\right)+\left(\frac{30-x}{1978}+1\right)=0\)

=> \(\frac{2008-x}{1983}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=0\)

=> \(\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

Ta có \(\frac{1}{1963}>0,\frac{1}{1968}>0,\frac{1}{1973}>0,\frac{1}{1978}>0\)

=> \(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}>0\)

=> \(2008-x=0\)

=> \(x=2008\)

\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)

\(-18x^3+51x^2+9x-60=0\)

\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)

\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)

196345−x​+196840−x​+197335−x​+197830−x​=−4
\left(\frac{45-x}{1963}+1\right)+\left(\frac{40-x}{1968}+1\right)+\left(\frac{35-x}{1973}+1\right)+\left(\frac{30-x}{1978}+1\right)=0(196345−x​+1)+(196840−x​+1)+(197335−x​+1)+(197830−x​+1)=0
\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=019632008−x​+19682008−x​+19732008−x​+19782008−x​=0
\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0(2008−x)(19631​+19681​+19731​+19781​)=0
=> 2008 - x = 0 ( vì 1/ 1963 + ... khác 0 )
=> x = 2008

Ta có : \(\frac{45-x}{1963}+\frac{40-x}{1968}+\frac{35-x}{1973}+\frac{30-x}{1978}+4=0\)

\(\Leftrightarrow\frac{45-x}{1963}+1+\frac{40-x}{1968}+1+\frac{35-x}{1973}+1+\frac{30-x}{1978}=0\)

\(\Leftrightarrow\frac{2008-x}{1963}+\frac{2008-x}{1968}+\frac{2008-x}{1973}+\frac{2008-x}{1978}=0\)

\(\Leftrightarrow\left(2008-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

Vì \(\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)\ne0\)

Nên : 2008 - x = 0 

<=> x = 2008

Vậy x = 2008

\(\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0\)

\(\Leftrightarrow\left(\frac{55-x}{1963}+1\right)+\left(\frac{50-x}{1968}+1\right)+\left(\frac{45-x}{1973}+1\right)+\left(\frac{40-x}{1978}+1\right)=0\)

\(\Leftrightarrow\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0\)

\(\Leftrightarrow\left(2018-x\right).\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

\(\Leftrightarrow2018-x=0\)

\(\Leftrightarrow x=2018\)

Vậy \(x=2018\)

Dễ dàng :v

Có \(\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0\)

\(\Rightarrow\left(\frac{55-x}{1963}+1\right)+\left(\frac{50-x}{1968}+1\right)+\left(\frac{45-x}{1973}+1\right)+\left(\frac{40-x}{1978}+1\right)=0\)

\(\Rightarrow\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0\)

\(\Rightarrow\left(2018-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

Mà \(\Rightarrow\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)>0\Rightarrow2018-x=0\)

\(\Rightarrow x=2018-8=2018\)

Vậy x = 2018

a, \(\frac{x+16}{49}+\frac{x+18}{47}=\frac{x+20}{45}-1\)

\(\Leftrightarrow1+\frac{x+16}{49}+1+\frac{x+18}{47}=\frac{x+20}{45}-1+2\)

\(\Leftrightarrow\frac{x+16+49}{49}+\frac{x+18+47}{47}=\frac{x+20+45}{45}\)

\(\Leftrightarrow\frac{x+65}{49}+\frac{x+65}{47}-\frac{x+65}{45}=0\)

\(\Leftrightarrow\left(x+65\right)\left(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\right)=0\)

Ta có: \(\frac{1}{49}+\frac{1}{47}-\frac{1}{45}\)>0

\(\Rightarrow x+65=0\)

\(\Leftrightarrow x=-65\)

Vậy x = -65

b, \(\frac{x-69}{30}+\frac{x-67}{32}+\frac{x-65}{34}=\frac{x-63}{36}+\frac{x-61}{38}+\frac{x-59}{40}\)

\(\Leftrightarrow\frac{x-69}{30}-1+\frac{x-67}{32}-1+\frac{x-65}{34}-1+\frac{x-63}{36}-1+\frac{x-61}{38}-1+\frac{x-59}{40}-1\)

\(\Leftrightarrow\frac{x-99}{30}+\frac{x-99}{32}+\frac{x-99}{34}-\frac{x-99}{36}-\frac{x-99}{38}-\frac{x-99}{40}=0\)

\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\right)=0\)

Vì \(\frac{1}{30}+\frac{1}{32}+\frac{1}{34}-\frac{1}{36}-\frac{1}{38}-\frac{1}{40}\)>0

\(\Rightarrow x-99=0\)

\(\Leftrightarrow x=99\)

Vậy x =99