3/x = 39/91
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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}
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
13 tháng 2 2019
Theo đề ta có 3/x=-39/91 => 3.13/13x=39/91=>39/13x=39/91=> x= 91:13=7(*)
Ta lại có từ (*) x=7 => 3/7=y/28=> 3.4/7.4=y/28=>12/28=y/28=> y=12
TM
1
28 tháng 3 2021
Ta có: \(\dfrac{3}{x}=\dfrac{y}{8}=\dfrac{-39}{91}\)
\(\Rightarrow\) \(\dfrac{3}{x}=\dfrac{y}{8}=\dfrac{-3}{7}\)
\(\Rightarrow\) \(\dfrac{-3}{-x}=\dfrac{\dfrac{y}{4}}{7}=\dfrac{-3}{7}\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}-x=7\\\dfrac{y}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x=-7\\y=-12\end{matrix}\right.\) (TM)
Vậy x = -7; y = -12
Chúc bn học tốt!
LT
1
21 tháng 1 2024
\(\dfrac{3}{x-5}=\dfrac{-4}{x-2}\left(x\notin\left\{5;2\right\}\right)\)
\(\Rightarrow3\left(x-2\right)=-4\left(x-5\right)\)
\(\Rightarrow3x-6=-4x+20\)
\(\Rightarrow3x+4x=20+6\)
\(\Rightarrow7x=26\)
\(\Rightarrow x=\dfrac{26}{7}\) (thỏa)
_________
\(\dfrac{3}{x}=\dfrac{y}{28}=\dfrac{-39}{91}\left(x\ne0\right)\)
\(\Rightarrow\dfrac{3}{x}=\dfrac{y}{28}=\dfrac{-3}{7}\)
+) \(\dfrac{3}{x}=-\dfrac{3}{7}\)
\(\Rightarrow x=\dfrac{3\cdot-7}{3}=-7\) (thỏa)
+) \(\dfrac{y}{28}=-\dfrac{3}{7}\)
\(\Rightarrow y=\dfrac{28\cdot-3}{7}=-12\)
18 tháng 7 2017
a)\(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\dfrac{9}{49}}=\sqrt{\dfrac{3}{7}}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{7^2}+\sqrt{91^2}}=\dfrac{\sqrt{9}+\sqrt{1521}}{\sqrt{49}+\sqrt{8281}}=\dfrac{3+39}{7+91}=\dfrac{42}{98}\)
c)Tương tự câu b, ta đc:
\(\dfrac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{7^2}-\sqrt{91^2}}=\dfrac{3-39}{7-91}=\dfrac{-36}{86}=\dfrac{3}{7}\)
d)Tương tự câu a, ta đc:
\(\dfrac{\sqrt{39^2}}{\sqrt{91^2}}=\dfrac{39}{91}\)
Chúc Bạn Học Tốt!!!
18 tháng 7 2017
a) \(\sqrt{\dfrac{3^2}{7^2}}=\sqrt{\left(\dfrac{3}{7}\right)^2}=\left|\dfrac{3}{7}\right|=\dfrac{3}{7}\)
b) \(\dfrac{\sqrt{3}^2+\sqrt{39}^2}{\sqrt{7}^2+\sqrt{91}^2}=\dfrac{\left|3\right|+\left|39\right|}{\left|7\right|+\left|91\right|}=\dfrac{3+39}{7+91}=\dfrac{42}{98}=\dfrac{3}{7}\)
c) \(\dfrac{\sqrt{3}^2-\sqrt{39}^2}{\sqrt{7}^2-\sqrt{91}^2}=\dfrac{\left|3\right|- \left|39\right|}{\left|7\right|-\left|91\right|}=\dfrac{3-39}{7-91}=\dfrac{-36}{-84}=\dfrac{3}{7}\)
d) \(\sqrt{\dfrac{39^2}{91^2}}=\sqrt{\left(\dfrac{39}{91}\right)^2}=\left|\dfrac{39}{91}\right|=\dfrac{39}{91}=\dfrac{3}{7}\)
LD
2
26 tháng 7 2023
a,91-5(5+x)=61
=>25+5x=91-61=30
=>5x=30-25=5
=>x=1
b,\([\left(x+34\right)-50]\)x2=56
=>(x+34)-50=56:2=28
=>x+34=28+50=78
=>x=78-34=44.
c,1045-\([2015-\left(3x-24\right)]\)=5
=>2015-(3x-24)=1045-5=1040
=>3x-24=2015-1040=975
=>3x=975+24=999
=>x=999:3=333
d,\([195-\left(3x-27\right)]\)x39=4212
=>195-(3x-27)=4212:39=108
=>3x-27=195-108=87
=>3x=87+27=117
=>x=39
e,30-3(x-2)=18
=>30-3x+6=18
=>30-3x=18-6=12
=>3x=30-12=18
=>x=18:3=6
26 tháng 7 2023
a) \(...\Rightarrow5\left(5+x\right)=91-61=30\)
\(\Rightarrow\left(5+x\right)=30:5=6\Rightarrow x=6-5=1\)
b) \(...\Rightarrow\left(x+34\right)-50=56:2=28\)
\(\Rightarrow\left(x+34\right)=28+50=78\Rightarrow x=78-34=44\)
c) \(...\Rightarrow2015-\left(3x-24\right)=1045-5=1040\)
\(\Rightarrow\left(3x-24\right)=2015-1040=975\)
\(\Rightarrow3x=975+24=999\Rightarrow x=999:3=333\)
d) \(...\Rightarrow195-\left(3x-27\right)=4212:39=108\)
\(\Rightarrow\left(3x-27\right)=195-108=87\)
\(\Rightarrow3x=87+27=114\Rightarrow x=114:3=38\)
e) \(...\Rightarrow3\left(x-2\right)=30-18=12\Rightarrow x-2=12:3=4\)
\(\Rightarrow x=4+2=6\)
PT
1
NM
20 tháng 9 2023
\(a,91-5.\left(5+x\right)=61\\ \Rightarrow=5.\left(5+x\right)=30\\ \Rightarrow5+x=6\\ \Rightarrow x=1.\\ b,\left[195-\left(3x-27\right)\right].39=4212\\ \Rightarrow195-\left(3x-27\right)=108\\ \Rightarrow3x-27=87\\ \Rightarrow3x=114\\ \Rightarrow x=38.\)
20 tháng 9 2023
\(91-5.\left(5+x\right)=61\)
\(\Rightarrow5.\left(5+x\right)=91-61\)
\(\Rightarrow5.\left(5+x\right)=30\)
\(\Rightarrow5+x=\dfrac{30}{5}=6\)
\(\Rightarrow x=6-5=1\)
\(\left[195-\left(3x-27\right)\right].39=4212\)
\(\Rightarrow195-3x+27=\dfrac{4212}{39}=108\)
\(\Rightarrow222-3x=108\)
\(\Rightarrow3x=222-108=114\)
\(\Rightarrow x=\dfrac{114}{3}=38\)
NT
0
3/x=3/7
x=3*7/3
x=7
\(\frac{3}{x}\) = \(\frac{39}{91}\)
\(\frac{3}{x}=\frac37\)
\(x=7\)
Vậy \(x=7\)