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a: =>-2x=90/91
hay x=-45/91
b: =>2x=-7
hay x=-7/2
c: ->-3x=-12
hay x=4
(x-7)/16=9/24=>(3x-21)/48=18/48
a) x/4=(x+1)/8=>2x/8=(x+1)/8
=>2x=x+1=>x=1
b) (2x-1)/15=3/5=>(2x-1)/15=9/15
=)2x-1=9=)2x=10=>x=5
Quy đồng mẫu số \(\frac{x-7}{16}=\frac{9}{24}\)
\(\frac{9}{24}=\frac{3}{8}\)
=> \(\frac{\left(x-7\right):2}{16:2}=\frac{3}{8}\)
=> \(\left(x-7\right):2=3\)
\(x-7=3\cdot2=6\)
\(x=6+7=13\)
=> \(\frac{13-7}{16}=\frac{9}{24}=\frac{3}{8}\)( Vừa bằng nhau vừa có mẫu = 8 đấy nhé )
Tìm số nguyên x thỏa mãn
a) \(\frac{x}{4}=\frac{x+1}{8}\)
\(\Rightarrow8x=4\left(x+1\right)\)
\(\Rightarrow8x=4x+4\)
\(\Rightarrow8x-4x=4\)
\(\Rightarrow4x=4\Rightarrow x=1\)
b) \(\frac{2x-1}{15}=\frac{3}{5}\)
\(\Rightarrow\frac{\left(2x-1\right):3}{15:3}=\frac{3}{5}\)
\(\Rightarrow\left(2x-1\right):3=3\)
\(\Rightarrow2x-1=9\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
d) Ta có: \(n^2+5n+9⋮n+3\)
\(\Leftrightarrow n^2+3n+2n+6+3⋮n+3\)
\(\Leftrightarrow n\left(n+3\right)+2\left(n+3\right)+3⋮n+3\)
mà \(n\left(n+3\right)+2\left(n+3\right)⋮n+3\)
nên \(3⋮n+3\)
\(\Leftrightarrow n+3\inƯ\left(3\right)\)
\(\Leftrightarrow n+3\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{-2;-4;0;-6\right\}\)
Vậy: \(n\in\left\{-2;-4;0;-6\right\}\)
d) Ta có: n2+5n+9⋮n+3n2+5n+9⋮n+3
⇔n2+3n+2n+6+3⋮n+3⇔n2+3n+2n+6+3⋮n+3
⇔n(n+3)+2(n+3)+3⋮n+3⇔n(n+3)+2(n+3)+3⋮n+3
mà n(n+3)+2(n+3)⋮n+3n(n+3)+2(n+3)⋮n+3
nên 3⋮n+33⋮n+3
⇔n+3∈Ư(3)⇔n+3∈Ư(3)
⇔n+3∈{1;−1;3;−3}
a) x = 1.
b) x = 6.
c)x = - ll.