a,=3/13.17/29.29/34

=3/13.1/2

=3/26

b,=(4/21+25/42).19/33

=11/14.19/33

=19/42

`a, = 3/13 xx (17/29 xx 29/34)`

`= 3/13 xx 1/2`

`= 3/26`.

`b, = 19/33(4/21 + 25/42)`

`= 19/33 xx ( 33/42 )`

`= 19/42`

`a)(3/13xx17/29)xx29/34=3/13xx17/29xx29/34=[3xx17xx29]/[13xx19xx17xx2]=3/26`

`b)4/21xx19/33+25/42xx19/33=19/33xx(4/21+25/42)=19/33xx(8/42+25/42)=19/33xx33/42=19/42`

 

 

313×1729)×2934(313×1729)×2934

=313×1729×2934=313×1729×2934

=3×17×2913×29×34=3×17×2913×29×34

=3×17×2913×29×2×17=3×17×2913×29×2×17

=313×2=313×2

=326=326

     

421×1933+2542×1933421×1933+2542×1933

=1933×(421+2542)=1933×(421+2542)

=1933×(842+25

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}-4=0\)

\(\dfrac{\left(x-17\right)\times725}{33\times725}+\dfrac{\left(x-21\right)\times825}{29\times825}+\dfrac{x\times957}{25\times957}-\dfrac{4\times23925}{23925}=0\)

\(725x-12325+825x-17325+957x-95700=0\)

\(2507x-125350=0\)

\(2507x=125350\)

\(x=50\)

Nếu mà theo cách x - 50 = 0 thì bạn theo cách này nha:

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}-4=0\)

\(\dfrac{x-17}{33}-1+\dfrac{x-21}{29}-1+\dfrac{x}{25}-2=0\)

\(\dfrac{x-50}{33}+\dfrac{x-50}{29}+\dfrac{x-50}{25}=0\)

\(\left(x-50\right)\left(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{5}\right)=0\)

Vì  \(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}>0\)

=> \(x-50=0\)
=> \(x=50\)

 

 

da co cach nao de chung minh x-50=0 roi tinh k ak

tai mik dg lm theo cach do ak

a) Ta có: \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Leftrightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)

\(\Leftrightarrow\frac{x-17-33}{33}+\frac{x-21-29}{29}+\frac{x-2\cdot25}{25}=0\)

\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\)

nên x-50=0

hay x=50

Vậy: x=50

b) Ta có: \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)=2\)

\(\Leftrightarrow-15x^2+46x-35+15x^2-4x-4-2=0\)

\(\Leftrightarrow42x-41=0\)

\(\Leftrightarrow42x=41\)

hay \(x=\frac{41}{42}\)

a, \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=4-4\)

\(\Leftrightarrow\left(\frac{x-17-33}{33}\right)+\left(\frac{x-21-29}{29}\right)+\left(\frac{x-2.25}{25}\right)=0\)

\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\) (*)

Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\Rightarrow\) Phương trình (*) xảy ra khi: \(x-50=0\Leftrightarrow x=50\)

Vậy phương trình có nghiệm duy nhất là x = 50.

a) (x-17)/33 + (x-21)/29 + x/25 = 4 <=> [(x-17)/33-1] + [(x-21)/29-1] + x/25-2] = 0 <=> (x-50)/33 + (x-50)/29 + (x-50)/25 = 0 <=> (x-50)(1/33+1/29+1/25) = 0 Mà 1/33+1/29+1/25 khác 0. <=> x- 50 = 0 <=> x=50 b) (3x−5)(7−5x)+(5x+2)(3x−2)=2 <=> 21x−15x2−35+25x+15x2−10x+6x−4−2=0 <=> (15x2−15x2)+(25x+21x−10x+6x)−(35+4+2)=0 <=> 42x=41 <=> x=41/42

PT<=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

<=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

<=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Mà \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\)

<=> x -50 = 0

<=> x = 50

\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Rightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)

\(\Rightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Dễ  thấy\(\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)>0\Rightarrow x-50=0\Rightarrow x=50\)

Vậy x = 50

Ta có 

\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)

\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Mà : \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\)

\(\Rightarrow x-50=0\)

\(\Rightarrow x=50\)

Vậy : \(x=50\)

\(\Leftrightarrow\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}-4=0\)

\(\Rightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

\(\Rightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

Mà \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\)nên 

\(\Rightarrow x-50=0\Rightarrow x=50\)

Vậy x=50

theo đề trên thì tử và mẫu cùng trừ một số thì được phân số 7/11

hiệu tử và mẫu lúc đầu là :

29 - 21 = 8 

hiệu tử và mẫu lúc sau là :

11 - 7 = 4 

mak cùng bớt 1 số nên hiệu k thay đổi

để hiệu k thay đổi ta nhân hiệu lên 2 lần , vậy tử và mẫu cũng phải nhân lên 2 lần

7/11 = 14/22

vậy x là : 21 - 14 = 7 

\(\frac{21-x}{29-x}=\frac{7}{11}\)

\(\Leftrightarrow11\left(21-x\right)=7\left(29-x\right)\)

\(\Leftrightarrow231-11x=203-7x\)

\(\Leftrightarrow231-203=-7x+11x\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

Nhớ k với nhé