\(25-x+29-x+33-x+...+101-x=128\)

\(\left(25+29+33+...+101\right)-\left(x+x+x+...+x\right)=128\)

Tính tổng như thường:

\(1260+20x=128\)

\(20x=-1132\)

\(x=-56,6\)

Ta có:

B. ( x - 21.13 ) : 11 = 30

=> x = 30 x 11 + 21 x 13

=> x = 330 + 273

=> x = 603

C. ( 25 - x ) + ( 29 - x ) + ( 33 - x ) + ... + ( 101 - x ) = 128

=> ( 25 + 29 + 33 + ... + 101 ) - ( x + x + x +...+ x ) = 128

=> 1260 - 20x = 128

=> 20x = 1260 - 128

=> x = 1132 : 20

=> x = 283/5

B. ( x - 21.13 ) : 11 = 30

=> x = 30 x 11 + 21 x 13

=> x = 330 + 273

=> x = 603

C. ( 25 - x ) + ( 29 - x ) + ( 33 - x ) + ... + ( 101 - x ) = 128

=> ( 25 + 29 + 33 + ... + 101 ) - ( x + x + x +...+ x ) = 128

=> 1260 - 20x = 128

=> 20x = 1260 - 128

=> x = 1132 : 20

=> x = 283/5

283/5

a: =>7x=42

hay x=6

b: =>5x=35

hay x=7

c: =>x-14=16

hay x=30

d: =>36-4x=4

=>4x=32

hay x=8

e: =>x-12=144

hay x=156

f: =>3x-16=14

hay x=10

g: =>x+33=45

hay x=12

h: =>(x+9):2=39

=>x+9=78

hay x=69

a: =>7x=42

hay x=6

b: =>5x=35

hay x=7

c: =>x-14=16

hay x=30

d: =>36-4x=4

=>4x=32

5 ( năm )

a, <=> (x-5/100) -1 +(x-4/101) -1 +(x-3/102) -1= (x-100/5) -1+(x-101/4) -1 +(x-102/3) -1
<=> (x-105)(1/100 +1/101 +1/102)= (x-105)(1/5+1/4+1/3)
<=> (x-105)(1/100+1/101+1/102-1/5-1/4-1/3)=0
vì 1/100+1/101+1/102-1/5-1/4-1/3 khác 0 <=> x-105=0
<=> x=105

b, 29-x/21 +1+27-x/23 +1+25-x/25 +1+23-x/27 +1+21-x/29 +1=0
<=> 50-x/21 +50-x/23 +50-x/25 +50-x/27 +50-x/29=0
<=> (50-x)(1/21 +1/23 +1/25 +1/27 +1/29)=0
vì 1/21+1/23+1/25+1/27+1/29 lớn hơn 0
nên 50-x=0
<=> x=50

ai làm ơn trả lời hộ mình câu này với

a) \(\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(\Leftrightarrow\left(\frac{x-5}{100}-1\right)+\left(\frac{x-4}{101}-1\right)+\left(\frac{x-3}{102}-1\right)=\left(\frac{x-100}{5}-1\right)+\left(\frac{x-101}{4}-1\right)+\left(\frac{x-102}{3}-1\right)\)
\(\Leftrightarrow\frac{x-105}{100}+\frac{x-105}{101}+\frac{x-105}{102}=\frac{x-105}{5}+\frac{x-105}{4}+\frac{x-105}{3}\)
\(\Leftrightarrow\left(x-105\right)\left(\frac{1}{100}+\frac{1}{101}+\frac{1}{102}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x=105\)
b) \(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5\)
\(\Leftrightarrow\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)=0\)
\(\Leftrightarrow\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
\(\Leftrightarrow x=50\)

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+2542)=1933×(842+2542)

=1933×3342=1933×3342

=19×3333×42=19×3333×42

=1942