Giải phương trình sau:

\(\dfrac{x}{50}\) +\(\dfrac{x_{ }-1}{49}\)+\(\dfrac{x-2}{48}\)+\(\dfrac{x-3}{47}\)+\(\dfrac{x-150}{25}\)= 0

⇔ \(\dfrac{\left(x-50\right)+50}{50}\)+\(\dfrac{\left(x-50\right)+49}{49}\)+\(\dfrac{\left(x-50\right)+48}{48}\)+\(\dfrac{\left(x-50\right)-100}{25}\)= 0

⇔\(\dfrac{x-50}{50}\)+ 1 + \(\dfrac{x-50}{49}\)+1+\(\dfrac{x-50}{48}\)+1+\(\dfrac{x-50}{47}\)+1+\(\dfrac{x-50}{25}\)-4 = 0

⇔\(\dfrac{x-50}{50}\)+\(\dfrac{x-50}{49}\)+\(\dfrac{x-50}{48}\)+\(\dfrac{x-50}{47}\)+\(\dfrac{x-50}{25}\)= 0

⇔ (x - 50 ) ( \(\dfrac{1}{50}\)+ \(\dfrac{1}{49}\)+\(\dfrac{1}{48}\)+\(\dfrac{1}{47}\)+\(\dfrac{1}{25}\)) = 0

⇔ x-50 =\(\dfrac{0}{\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}}\)

⇔ x- 50 = 0

⇔ x = 50

vậy S = \(\left\{50\right\}\)

Ta có : \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{49}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)

\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{49}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)

\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{49}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)

<=> x - 100 = 0

<=> x = 100

Vậy ..

Ta có: \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{48}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)

\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{48}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)

\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{48}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)

mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}>0\)

nên x-100=0

hay x=100

Vậy: S={100}

          \(\frac{x}{50}+\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-150}{25}=0\)

\(\Leftrightarrow\)\(\frac{x}{50}-1+\frac{x-1}{49}-1+\frac{x-2}{48}-1+\frac{x-3}{47}-1+\frac{x-150}{25}+4=0\)

\(\Leftrightarrow\)\(\frac{x-50}{50}+\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{25}=0\)

\(\Leftrightarrow\)\(\left(x-50\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{25}\right)=0\)

\(\Leftrightarrow\)\(x-50=0\)    (vì  1/50 + 1/49 + 1/48 + 1/47 + 1/25 > 0  )

\(\Leftrightarrow\)\(x=50\)

Vậy....

1) 1996 + 3992 + 5988 + 7984 
= 1996 x 1 + 1996 x 2 + 1996 x 3 + 1996 x 4 
= 1996 x ( 1 + 2 + 3 + 4) 
= 1996 x 10 = 19960 
2) 2 x 3 x 4 x 8 x 50 x 25 x 125 
= 3 x (2 x 50) x (4 x 25) x (8 x 125) 
= 3 x 100 x 100 x 1000 
= 30.000.000 
3(45 x 46 + 47 x 48) x (51 x 52 - 49 x 48) x (45 x 128 - 90 x 64) 
x (1995 x 1996 + 1997 x 1998) 
Ta thÊy : 45 x 128 - 90 x 64 
= 45 x 2 x 64 - 90 x 64 
= 90 x 64 - 90 x 64 
= 0 

1) 1996 + 3992 + 5988 + 7984 
= 1996 x 1 + 1996 x 2 + 1996 x 3 + 1996 x 4 
= 1996 x ( 1 + 2 + 3 + 4) 
= 1996 x 10 = 19960 
2) 2 x 3 x 4 x 8 x 50 x 25 x 125 
= 3 x (2 x 50) x (4 x 25) x (8 x 125) 
= 3 x 100 x 100 x 1000 
= 30.000.000 
3(45 x 46 + 47 x 48) x (51 x 52 - 49 x 48) x (45 x 128 - 90 x 64) 
x (1995 x 1996 + 1997 x 1998) 
Ta thÊy : 45 x 128 - 90 x 64 
= 45 x 2 x 64 - 90 x 64 
= 90 x 64 - 90 x 64 
= 0 

L I K E NHÁ BẠN^^

1  1996 + 3992 + 5988 +7984

= [ 1996 + 7984 ] + [ 5988 + 3392 ]

=           9980      +            9380

=                         19360

x1+x2+x3+x4+...x49+x50+x51=.

\(\Rightarrow\)(x1+x2)+(x3+x4)+...+(x49+x50)+x51=0

Theo de bai ta co:\(\Rightarrow\)1+1+...+1+x51=0

                         \(\Rightarrow\)1*25+x51=0

                          \(\Rightarrow\)x51=-25

Ma x60+x51=0\(\Rightarrow\)-50+(-25)=1\(\Rightarrow\)x50=26

Nho tich nha

x₁+x₂+x₃+...+x₄₉+x₅₀+x₅₁=0

(x1+x2)+(X3+x4)+ ...+ (x49+x50) + x51=0

1.25+x51=0

=> x51= -25 ( vì -25+25=0 nhá ^^)

mà x₅₀+ (-25)(tức là  x_{51 ý ) =1}

_=>  x50=26

x₅₀=26 

đây là toán lớp 6 sao ???