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\(b=a+b+c+d-\left(a+c+d\right)=1-2=-1\\ c=a+b+c+d-\left(a+b+d\right)=1-3=-2\\ d=a+b+c+d-\left(a+b+c\right)=1-4=-3\\ a=a+b+c+d-b-c-d=1+1+2+3=7\)
a, 10 ⋮ 3a+1 => 3a+1 ∈ Ư(10) => 3a+1 ∈ {1;2;5;10} => a ∈ { 0 ; 1 3 ; 4 3 ; 3 }. Vì a ∈ N, a ∈ {0;3}
b, a+6 ⋮ a+1 => a+1+5 ⋮ a+1 => 5 ⋮ a+1 => a+1 ∈ Ư(5) => a+1 ∈ {1;5} => a ∈ {0;4}
c, 3a+7 ⋮ 2a+3 => 2.(3a+7) - 3(2a+3) ⋮ 2a+3 => 5 ⋮ 2a+3 => 2a+3 ∈ Ư(5)
=> 2a+3 ∈ {1;5} => a = 1
d, 6a+11 ⋮ 2a+3 => 3.(2a+3)+2 ⋮ 2a+3 => 2 ⋮ 2a+3 => 2a+3 ∈ Ư(2)
=> 2a+3 ∈ {1;2} => a ∈ ∅
a/ \(a+3\inƯ\left(7\right)\)
\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow a\in\left\{-10;-4;-2;4\right\}\)
b/ \(2a\inƯ\left(-10\right)\)
\(Ư\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
\(\Rightarrow a\in\left\{-5;-1;1;5\right\}\)do \(a\inℤ\)
c/ \(a+1\inƯ\left(3a+7\right)\Rightarrow3a+7⋮a+1\)
\(\Rightarrow3a+7-3\left(a+1\right)⋮a+1\)
\(\Leftrightarrow4⋮a+1\)
\(Ư\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Rightarrow a\in\left\{-5;-3;-2;0;1;3\right\}\)
d/ \(2a+1\inƯ\left(3a+5\right)\Rightarrow3a+5⋮2a+1\)
\(\Rightarrow3a+5-\left(2a+1\right)⋮2a+1\)
\(\Leftrightarrow a+4⋮2a+1\)
\(\Rightarrow2\left(a+4\right)⋮2a+1\Leftrightarrow2a+8⋮2a+1\)
\(\Rightarrow2a+8-\left(2a+1\right)⋮2a+1\Leftrightarrow7⋮2a+1\)
\(Ư\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow a\in\left\{-4;-1;0;3\right\}\)
a: \(\Leftrightarrow6a-2+1⋮3a-1\)
\(\Leftrightarrow3a-1\in\left\{1;-1\right\}\)
hay \(a=0\)
b: \(\Leftrightarrow4a-5⋮a\)
\(\Leftrightarrow a\in\left\{1;-1;5;-5\right\}\)
c: \(\Leftrightarrow a-1\in\left\{1;-1;11;-11\right\}\)
hay \(a\in\left\{2;0;12;-10\right\}\)
1: C=4a+2a+10b-b
=6a+9b
=3(2a+3b)
=3*12=36
D=21a+9b-6a-4b
=15a+5b
=5(3a+b)
=5*18=90
B=5a+7a-4b-8b
=12a-12b
=12(a-b)
=12*8=96
4:
Gọi hai số cần tìm là a,b
Theo đề, ta có hệ phương trình:
a+b=38570 và a=3b+922
=>a=29158 và b=9412
1: C=4a+2a+10b-b
=6a+9b
=3(2a+3b)
=3*12=36
D=21a+9b-6a-4b
=15a+5b
=5(3a+b)
=5*18=90
B=5a+7a-4b-8b
=12a-12b
=12(a-b)
=12*8=96
a + b + c + d = -1
=> a + b + c = -1 - d
Mặt khác a + b + c = 4
=> -1 - d = 4
<=> d = -1 - 4
<=> d = -5
Tương tự ta có : c = 2; b = 1; a = -1 - 2 - 1 + 5 = 1
Vậy a = 1; b = 1; c = 2; d = -5
Ta có : \(\hept{\begin{cases}a+b+c=-3\\a+b+c+d=-1\end{cases}}\Rightarrow d=-1-\left(-3\right)=2\)
Vì \(\hept{\begin{cases}a+b+d=-3\\d=2\end{cases}}\Rightarrow a+b=2-\left(-3\right)=5\)
Ta có : \(\hept{\begin{cases}a+b+c=4\\a+b=5\end{cases}}\Rightarrow c=4-5=-1\)
Vì \(\hept{\begin{cases}a+c+d=-2\\d=2\\c=-1\end{cases}}\Rightarrow a=-2-2-\left(-1\right)=-3\)
Ta có : \(\hept{\begin{cases}a+b+c+d=-1\\a=-3;c=-1;d=2\end{cases}}\Rightarrow b=-1-\left(-3\right)-\left(-1\right)-2=1\)
\(\Rightarrow a=-3;b=1;c=-1;d=2\)