(25-x)+(29-x)+(33-x)+....+(101-x) =12
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\(25-x+29-x+33-x+...+101-x=128\)
\(\left(25+29+33+...+101\right)-\left(x+x+x+...+x\right)=128\)
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
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
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
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`
=> 25 - x + 29 - x + 33 - x + ....+ 101 - x = 12
=> ( 25 + 29 + 33 +....+ 101 ) - ( x + x + .....+ x ) = 12
=> \(\frac{\left(25+101\right).20}{2}-20x=12\)
=> 1260 - 20x = 12
=> x = 62,4