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Ta có:
x-y+x+z+y-x=8+10+12=30
=>2x = 30
=>x = 15
=> y = 15-8=7
z = 15-12=3
Vậy x=15;y=7;z=3
\(2\cdot3^x=10\cdot3^{12}\)
\(2\cdot3^x=2\cdot5\cdot3^{12}\)
\(\Rightarrow5\cdot3^{12}=3^x\)
\(3^{x+2}+3^x=10\)
\(3^x\cdot\left(3^2+1\right)=10\)
\(3^x\cdot\left(9+1\right)=10\)
\(3^x\cdot10=10\)
\(3^x=10:10\)
\(3^x=1\)
\(3^x=3^0\)
\(\Rightarrow x=0\)
\(a,210-5x=200\)
\(5x=10\)
\(x=2\)
Vậy \(x=2\)
\(b,210-5\left(x-10\right)=200\)
\(5\left(x-10\right)=10\)
\(x-10=2\)
\(x=12\)
Vậy \(x=12\)
\(c,450\div\left[41-\left(2x-5\right)\right]=3^2.5\)
\(450\div\left[41-\left(2x-5\right)\right]=45\)
\(41-\left(2x-5\right)=10\)
\(2x-5=31\)
\(2x=36\)
\(x=18\)
Vậy \(x=18\)
\(d,350\div x+10=20\)
\(350\div x=10\)
\(x=35\)
Vậy \(x=35\)
\(e,36\div\left(x-5\right)=2^2\)
\(36\div\left(x-5\right)=4\)
\(x-5=9\)
\(x=14\)
Vậy \(x=14\)
\(f,\left[3\left(70-x\right)+5\right]\div2=46\)
\(3\left(70-x\right)+5=92\)
\(3\left(70-x\right)=87\)
\(70-x=29\)
\(x=41\)
Vậy \(x=41\)
a)210-5x=200 b)210-5(x-10)=200
5x=210-200=10 5(x-10)=210-200=10
x=10:5=2 x-10 =10:5=2 c)45:[41-(2x-5)]=32.5 x =2+10=12
45:[41-(2x-5)] =9.5=45 d)350:x+10=20
41-(2x-5)=45:45=1 350:x =20-10=10
2x-5 =41-1=40 x =350:10=35
2x =40+5=45
x =45:2=22,5
e)36:(x-5)=22 f)[3.(70-x)+5]:2=46
36:(x-5)=4 [3.(70-x)+5] =46.2=92
x-5 =36:4=9 3.(70-x) =92-5=87
x =9+5=14 70-x =87:3=29
x =70-29=41
\(10-x=-3\)
\(x=10--3\)
\(x=13\)
10 - x = -3
x = 10 + 3
x = 13
HT và $$$