2003 + a x 15 = 2003 + 10 x 3
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A) 95.(a-75)=95.(100-75)
=>a-75=100-75
vậy a =100
B) 2003+a.15=2003.10.3.5
a.15=10.3.5
a 95 x { a - 75 } = 95 x { 100 - 75 }
=>a-75=100-75
vậy a=100
b } 2003 + a x 15 = 2003 x 10 x 3 x5
a x 15 = 10 x 3 x5
a x 15 = 10x(3x5)
a x 15=10x15
Vậy a=10
a: \(\left(15-x\right)+\left(x-12\right)=7-\left(x-5\right)\)
=>7-x+5=15-x+x-12
=>12-x=3
hay x=9
b: \(\Leftrightarrow x-\left\{57-\left[42-23-x\right]\right\}=13-\left\{47+25-32+x\right\}\)
\(\Leftrightarrow x-\left\{57-19+x\right\}=13-\left\{40+x\right\}\)
=>x-38-x=13-40-x
=>-27-x=-38
=>x+27=38
hay x=11
e: \(x^2+3x+9⋮x+3\)
\(\Leftrightarrow x\left(x+3\right)+9⋮x+3\)
\(\Leftrightarrow x+3\in\left\{1;-1;9;-9;3;-3\right\}\)
hay \(x\in\left\{-2;-4;6;-12;0;-6\right\}\)
a =2004.10+1992+2002+2004
= 2004(10+1)+3994
= 2004.11+3994=26038
b =2003(1+493+1520)=2003.2024=4054072
\(=2003\times\left(58+52-10\right)+335\times100\)
\(=100\times\left(2003+335\right)=233800\)
(2003 x 58 + 52 x 2003 - 10 x 2003) + (670-335) x 100
= 2003 x ( 58 + 52 - 10 ) + 335 x 100
= 2003 x 100 + 335 x 100
= ( 2003 + 335 ) x 100
= 2338 x 100
233 800
( 2003 x 58 + 52 x 2003 - 10 x 2003 ) + ( 670 - 335 ) x 100
= 2003 x ( 58 + 52 -10 ) + 315 x 100
= 2003 x 100 + 315 x 100
= 200300 + 31500
= 231800
( 2003 x 58 + 52 x 2003 - 10 x 2003 ) + ( 670 - 355 ) x 100
= 200300 + 315 x100
= 200300 + 31500
= 231800
2003 + a x 15 = 2003 + 10 x 3
2003 + a x 15 = 2003 + 10 x 3
a x 15 = 10 x 3
= 30
a = 30 : 15
a = 2
\(\Rightarrow2003+15a=2003+30\)
\(\Rightarrow2003-2003+15a=30\)
\(\Rightarrow15a=30\)
\(vây\)\(a=2\)