Câu 2: 

a: \(\Leftrightarrow x+2\in\left\{3;9\right\}\)

hay \(x\in\left\{1;7\right\}\)

Thi à :)?

a+b+c=(1+3+2):2=3.vậy a=3-3=0;b=1-0=1;c=3-1=2

b + c - (a + c) = 3 - 2 = 1

b + c - a - c = 1

b -a = 1

a = (1+1)/2 = 1

b = 1 - 1 = 0

c = 3-  0 = 3

Vậy (a,b,c)=  (1,  0 ,3)

a=0

b=1

c=2

\(C=2x^2-y+xy-x^2y^2\)

a)\(C=2x^2-y+xy+x^2y^2\)

b)\(C=3y-x^2y^2-xy-2\)

\(B=2x^2-y+xy-x^2y^2+x2 - 2y + xy + 1 \)

\(=3x^2-3y+2xy-x^2y^2+1\)

a) (x - 1)² = 1.

<=> x - 1 = 1 hoặc x - 1 = -1.

<=> x = 2 hoặc x = 0.      

b) 7^{2x - 6} = 49.

<=> 7^{2x - 6} = 7².   

<=> 2x - 6 = 2.

<=> x = 4.  

c) (2x - 16)⁷ = 128.

<=> (2x - 16)⁷ = 2⁷.

<=> 2x - 16 = 2.

<=> x = 9.

ngay mai e nop rui

TC: a/b=b/c

=>

a) Ta có:\(n-6⋮n-1\)

\(\Leftrightarrow n-1-5⋮n-1\)

mà \(n-1⋮n-1\)

nên \(-5⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(-5\right)\)

\(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{2;0;6;-4\right\}\)

Vậy: \(n\in\left\{2;0;6;-4\right\}\)

b) Ta có: \(3n+2⋮n-1\)

\(\Leftrightarrow3n-3+5⋮n-1\)

mà \(3n-3⋮n-1\)

nên \(5⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(5\right)\)

\(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{2;0;6;-4\right\}\)

Vậy: \(n\in\left\{2;0;6;-4\right\}\)

c) Ta có: \(n^2+5⋮n+1\)

\(\Leftrightarrow n^2+2n+1-2n+4⋮n+1\)

\(\Leftrightarrow\left(n+1\right)^2-2n-2+6⋮n+1\)

mà \(\left(n+1\right)^2⋮n+1\)

và \(-2n-2⋮n+1\)

nên \(6⋮n+1\)

\(\Leftrightarrow n+1\inƯ\left(6\right)\)

\(\Leftrightarrow n+1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(n\in\left\{0;-2;1;-3;2;-4;5;-7\right\}\)

Vậy: \(n\in\left\{0;-2;1;-3;2;-4;5;-7\right\}\)

Sao cho gì bạn

ở câu A (n-6):(n-1) hay là (n-6)⋮(n-1) vậy bn???