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}

(1 + \(\dfrac{1}{49}\))\(\times\)(1 + \(\dfrac{1}{50}\))\(\times\)(1 + \(\dfrac{1}{51}\))\(\times\)(1 + \(\dfrac{1}{52}\))\(\times\)...\(\times\)(1 + \(\dfrac{1}{60}\))

= \(\dfrac{49+1}{49}\) \(\times\) \(\dfrac{50+1}{50}\)\(\times\) \(\dfrac{51+1}{51}\)\(\times\)\(\dfrac{52+1}{52}\)\(\times\)...\(\times\)\(\dfrac{61}{60}\)

= \(\dfrac{50}{49}\)\(\times\)\(\dfrac{51}{50}\)\(\times\)\(\dfrac{52}{51}\)\(\times\)...\(\times\)\(\dfrac{61}{60}\)

= \(\dfrac{50\times51\times52\times53\times...\times60}{50\times51\times52\times53\times...\times60}\)\(\times\)\(\dfrac{61}{49}\)

= \(\dfrac{61}{49}\)

cảm ơn

a/50

a/50 

đáp án:d/49

Bài làm

\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)

\(\Leftrightarrow\frac{x+52}{21}+\frac{x+51}{22}+2=\frac{x+50}{23}+\frac{x+49}{24}+2\)

\(\Leftrightarrow\left(\frac{x+52}{21}+1\right)+\left(\frac{x+51}{22}+1\right)=\left(\frac{x+50}{23}+1\right)+\left(\frac{x+49}{24}+1\right)\)

\(\Leftrightarrow\left(\frac{x+52}{21}+\frac{21}{21}\right)+\left(\frac{x+51}{22}+\frac{22}{22}\right)=\left(\frac{x+50}{23}+\frac{23}{23}\right)+\left(\frac{x+49}{24}+\frac{24}{24}\right)\)

\(\Leftrightarrow\frac{x+52+21}{21}+\frac{x+51+22}{22}=\frac{x+50+23}{23}+\frac{x+49+24}{24}\)

\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}=\frac{x+73}{23}+\frac{x+73}{24}\)

\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}-\frac{x+73}{23}-\frac{x+73}{24}=0\)

\(\Leftrightarrow\left(x+73\right)\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\right)=0\)

\(\Leftrightarrow x+73=0\)

\(\Leftrightarrow x=-73\)

Vậy x = -73

 # Học tốt #

\(\frac{x+52}{21}+\frac{x+51}{22}=\frac{x+50}{23}+\frac{x+49}{24}\)

\(\Leftrightarrow\frac{x+52}{21}+\frac{x+51}{22}-\frac{x+50}{23}-\frac{x+49}{24}=0\)

\(\Leftrightarrow\left(\frac{x+52}{21}+1\right)+\left(\frac{x+51}{22}+1\right)-\left(\frac{x+50}{23}+1\right)-\left(\frac{x+49}{24}+1\right)=0\)

\(\Leftrightarrow\frac{x+73}{21}+\frac{x+73}{22}-\frac{x+73}{23}-\frac{x+73}{24}=0\)

\(\Leftrightarrow\left(x+73\right)\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\right)=0\)

=> \(x+73=0\left(\frac{1}{21}+\frac{1}{22}-\frac{1}{23}-\frac{1}{24}\ne0\right)\)

<=> x=-73

Thanks các bn nha

1.

a. 455

b. 600

2.

x = - 5 nhưng nếu đây thực sự là toán 4 thì ko tồn tại x

kb vs mk nhe

Bài 1:

a,Ta thấy:QLC là 1

SSH là:(50-41):1+1=10(số)

Tổng là:(41+50):10+1=455

b,45.6+55.6=(45+55)6=100.6=600

Bài 2:

50+x-5=3+37

50+x-5=40

  50+x=45

        x=45-50=-5

1 tính nhanh

a, 41 +42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 +50

= ( 41 + 50 ) x 10 : 2

= 455

b, 45 x 6 + 55 x 6

= 6 x ( 45 + 55 )

= 6 x 100

=600

2 tìm x

50 + x - 5 = 3 + 37

50 + x - 5 = 40

50 + x      = 45

x              =-5

ĐKXĐ : x khác 49 , x khác 50

Ta có :

\(\frac{x-49}{50}+\frac{x-50}{49}=\frac{49}{x-50}+\frac{50}{x-49}\)

\(\Leftrightarrow\frac{x-49}{50}-1+\frac{x-50}{49}-1=\frac{49}{x-50}-1+\frac{50}{x-49}-1\)

\(\Leftrightarrow\frac{x-99}{50}+\frac{x-99}{49}=\frac{99-x}{x-50}+\frac{99-x}{x-49}\)

\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{x-50}+\frac{1}{x-49}\right)=0\) 

Mà 1/50 + 1/49 + 1/x-50 + 1/x-49 khác 0

\(\Leftrightarrow x-99=0\)

\(\Leftrightarrow x=99\)

@Kyo-kun

\(\Leftrightarrow\left(x-99\right)\left(\frac{1}{50}+\frac{1}{x-50}+\frac{1}{49}+\frac{1}{x-49}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-99=0\\\frac{x}{50\left(x-50\right)}+\frac{x}{49\left(x-49\right)}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=99\\x\left(\frac{1}{50\left(x-50\right)}+\frac{1}{49\left(x-49\right)}\right)=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=99\\x=0\end{cases}\left(t.m\right)}}\)

Vậy x = 99 hoặc x = 0