91:x; 26:xvà 10<x<30
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PT
1
NQ
0
GL
2
TT
8 tháng 2 2021
\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
=> x = - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
8 tháng 2 2021
Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
HG
12
19 tháng 7 2018
\(\left(x-27\right)\cdot91-72=215\cdot91-72\)
\(\left(x-27\right)\cdot91=215\cdot91-72+72\)
\(\left(x-27\right)\cdot91=215\cdot91\)
\(\left(x-27\right)=215\cdot91:91\)
\(x-27=215\)
\(x=242\)
19 tháng 7 2018
\(\left(x-27\right).91-72=215.91-72\)
Do 2 đẳng thức trên bằng nhau nên rút gọn 2 vế với - 7 ta đc
\(\Rightarrow\left(x-27\right).91=215.91\)
\(\Rightarrow\left(x-27\right)=215\)
\(\Rightarrow x=215+27=242\)
PB
4
TN
4
24 tháng 1 2018
Ta có\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
<=> \(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+9}{91}+1\right)=\left(\frac{x+91}{9}+1\right)+\left(\frac{x+92}{8}+1\right)+\left(\frac{x+61}{39}+1\right)\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}=\frac{x+100}{9}+\frac{x+100}{8}+\frac{x+100}{39}\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}-\frac{x+100}{9}-\frac{x+100}{8}-\frac{x+100}{39}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\right)=0\)
Do \(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\ne0\)
Nên x+100=0 => x=-100
1 tháng 7 2018
\(\dfrac{1909-x}{91}+\dfrac{1907-x}{93}+\dfrac{1905-x}{95}+\dfrac{1903}{97}+4=0\) ( Sửa đề)
⇔ \(\dfrac{1909-x}{91}+1+\dfrac{1907-x}{93}+1+\dfrac{1905-x}{95}+1+\dfrac{1903}{97}+1=0\) ⇔ \(\dfrac{2000-x}{91}+\dfrac{2000-x}{93}+\dfrac{2000-x}{95}+\dfrac{2000-x}{97}=0\)
⇔ \(\left(2000-x\right)\left(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}\right)=0\)
Do : \(\dfrac{1}{91}+\dfrac{1}{93}+\dfrac{1}{95}+\dfrac{1}{97}>0\)
\(\text{⇔}2000-x=0\)
\(\text{⇔}x=2000\)
Vậy ,....
QA
11 tháng 8 2016
91 x 157 - 29 x 91 - 91 x 28
=9100
127 x 63 + 19 x 127 - 82 x 26
=8282
2 x 9 x 41 + 36 + 3 x 57 x 6
=1800
29 x 17 + 31 x 29 + 48 x 93 - 48 x 22
=4800
1125 - 25 x 30 + 0,06 : 0,2
=375.3
Bn k cho mk nha!
CW
11 tháng 8 2016
91 x 157 - 29 x 91 - 91 x 28
= 19 (157 - 29 - 28)
=19 * 100
=1900
127 x 63 + 19 x 127 - 82 x 26
= 127 (63 + 19) - 82 * 26
= 127 * 82 - 82 * 26
= 82 (127 - 26)
= 82 * 101
= 8282
2 x 9 x 41 + 36 + 3 x 57 x 6
= 738 + 36 + 1026
= 1800
29 x 17 + 31 x 29 + 48 x 93 - 48 x 22
= 29 (17 + 31) + 48 (93 - 22)
= 29 * 48 + 48 * 71
= 48 (29 + 71)
= 48 * 100
= 4800
1125 - 25 x 30 + 0,06 : 0,2
= 1125 - 750 + 0,3
= 375 + 0,3
= 375,3