Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(35=5\cdot7;105=3\cdot5\cdot7\)
=>\(ƯCLN\left(35;105\right)=35\)
\(35⋮x;105⋮x\)
=>\(x\inƯC\left(35;105\right)\)
mà x lớn nhất
nên x=ƯLCN(35;105)
=>x=35
b:
\(72=2^3\cdot3^2;54=3^3\cdot2\)
=>\(ƯCLN\left(72;54\right)=3^2\cdot2=18\)
\(72⋮x;54⋮x\)
=>\(x\inƯC\left(72;54\right)\)
=>\(x\inƯ\left(18\right)\)
=>\(x\in\left\{1;-1;2;-2;3;-3;6;-6;9;-9;18;-18\right\}\)
mà 10<x<20
nên x=18
c:
\(21=3\cdot7;35=5\cdot7;50=5^2\cdot2\)
=>\(BCNN\left(21;35;50\right)=5^2\cdot2\cdot3\cdot7=1050\)
\(x⋮21;x⋮35;x⋮50\)
=>\(x\in BC\left(21;35;50\right)\)
=>\(x\in B\left(1050\right)\)
mà x nhỏ nhất
nên x=1050
d:
\(39=3\cdot13;65=5\cdot13;26=2\cdot13\)
=>\(BCNN\left(39;65;26\right)=2\cdot3\cdot5\cdot13=390\)
\(x⋮39;x⋮65;x⋮26\)
=>\(x\in BC\left(39;65;26\right)\)
=>\(x\in B\left(390\right)\)
=>\(x\in\left\{390;780;1170;...\right\}\)
mà 100<=x<=999
nên \(x\in\left\{390;780\right\}\)
=>X.15+x +15=463.
=>X.(15+1)+15=463
=>X.16+15=463
=>X.16=448
=> X=28
\(x\cdot15+x+15=463\)
\(x\cdot\left(15+1\right)+15=463\)
\(x\cdot16+15=463\)
\(x\cdot6=463-15\)
\(x\cdot6=448\)
\(x=448:6\)
\(x=\frac{448}{6}\)
\(x=\frac{224}{3}\)
\(\Rightarrow16-3x=504:72=7\\ \Rightarrow3x=16-7=9\Rightarrow x=3\\ \left(2^2\cdot x-5^2\right)\cdot3^8=3^9\\ \Rightarrow4x-25=3^9:3^8=3\\ \Rightarrow4x=28\Rightarrow x=7\)
a) \(21^{15}=21^{3.5}=\left(21^3\right)^5=9261^5\)
Vì \(9261>27\Rightarrow9261^5>27^5\Rightarrow21^{15}>27^5\)
b) \(15^{12}=\left(3.5\right)^{12}=3^{12}.5^{12}\)
\(81^3.125^5=\left(3^4\right)^3.\left(5^3\right)^5=3^{4.3}.5^{3.5}=3^{12}.5^{15}\)
Vì \(3^{12}=3^{12}\)mà \(5^{12}< 5^{15}\Rightarrow3^{12}.5^{12}< 3^{12}.5^{15}\Rightarrow15^{12}< 81^3.125^5\)
a. 15-(15+x)=21
=> 15-15-x=21
=> 0-x=21
=> -x=21
=> x=-21
b. 39+(x-39)=50
=> 39+x-39=50
=> x=50
a)\(45+\left(x-6\right).3=60\)
\(\Rightarrow45+\left(x-6\right)=60:3=20\)
\(\Rightarrow x-6=20-45=-25\)
\(\Rightarrow x=-25+6=-19\)
Vậy: \(x=-19\)
b) \(27-\left(x+5\right)=-15+39=14\)
\(\Rightarrow x+5=27-14=13\)
\(\Rightarrow x=13-5=8\)
Vậy: \(x=8\)
c) \(28⋮x;42⋮x;70⋮x\Rightarrow x\inƯC_{\left(28;42;70\right)}\)
\(\Rightarrow x\in\left\{1;2;7;14\right\}\)
Mà: \(1< x< 10\) nên \(x\in\left\{1;2;7\right\}\)
Vậy: \(x=1;2;7\)
Ta có : 91-(27+x)=38
<=>91-27-x=38
<=>91-27-38=x
<=>x=26
39+(x-17)=72
<=>39+x-17=72
<=>x=72+17-39
<=>x=50
Tk mình nha!
\(91-\left(27+x\right)=28\)
\(27+x=91-28\)
\(27+x=63\)
\(x=63-27\)
\(x=36\)
\(39+\left(x-17\right)=72\)
\(x-17=72-39\)
\(x-17=33\)
\(x=33+17\)
\(x=50\)
hok tốt
3990- (463 x 72 - 39 x 39 ) : 15 =
= 3990 - ( 463 x 72 - 39 x 39 ) : 15
= 3990 - ( 33336 - 1521 ) : 15
= 3390 - 31815 : 15
= 3390 - 2121
= 1269