\(\left(3x+4\right)⋮\left(x-1\right)\)

vì \(\left(x-1\right)⋮\left(x-1\right)\)

=> \(\left(3x-3\right)⋮\left(x-1\right)\)

=> \(\left(3x+4\right)-\left(3x-3\right)⋮\left(x-1\right)\)

=> \(\left(3x+4-3x+3\right)⋮\left(x-1\right)\)

=> \(7⋮\left(x-1\right)\)

=> \(\left(x-1\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

ta có bảng sau

vậy x\(\in\left\{-6;0;2;8\right\}\)

\(\dfrac{a}{b}=\dfrac{c}{d}\) => \(\dfrac{a}{c}=\dfrac{b}{d}\)

áp dụng tính chất dãy tỉ số = nhau ta có

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+3b}{2c+3d}=\dfrac{2a-3b}{2c-3d}\)

= \(\dfrac{2a+3b}{2a-3b}=\dfrac{2c+3d}{2c-3d}\) (đpcm)

dấu:là dấu\(⋮\) đúng ko vậy

7¹⁰¹=...5

dư 5 hay sao ấy, mik ko chắc

chtt

b)3n+40$⋮$n+4

=> vì \(\left(n+4\right)⋮\left(n+4\right)\)

=> \(\left(3n+12\right)⋮\left(n+4\right)\)

=> \(\left(3n+40-4x-12\right)⋮\left(n+4\right)\)

=> \(28⋮\left(n+4\right)\)

=> \(n+4\inƯ\left(28\right)=\left\{1;2;4;7;14;28\right\}\)

ta có bảng sau

vậy \(n\in\left\{0;3;10;24\right\}\)

147,6

ai tích mk lên 900 mk tích lại cho 

A=\(\dfrac{7}{2-4\left(x-2\right)^2}\)

do \(\left(x-2\right)^2\ge0\forall x\)

=> \(2-4\left(x-2\right)^2\ge2-4.0\)

=> \(2-4\left(x-2\right)^2\ge2\)

=> \(\dfrac{7}{2-4\left(x-2\right)^2}\le\dfrac{7}{2}\)

=> A\(\le\dfrac{7}{2}\)

GTLN A =\(\dfrac{7}{2}\) khi x-2=0

=> x=2

vậy ...

\(\left(n-5\right)⋮\left(n-2\right)\)

=> \(\left(n-5\right)-\left(n-2\right)⋮\left(n-2\right)\)

=> \(\left(n-5-n+2\right)⋮\left(n-2\right)\)

=> \(-3⋮\left(n-2\right)\)

=> n-2\(\inƯ\left(-3\right)\) ={\(\pm1,\pm3\) }

ta có bảng sau

vậy n\(\in\left\{1;3;5\right\}\)