Câu 2: 

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

hay \(x\in\left\{1;7\right\}\)

Thi à :)?

Đề bài yêu cầu gì?

Bài 1: 

c: x=4

b: x=2

a) (x - 140) : 7 = 3³ - 2³ . 3

(x - 140) : 7 = 27 - 8 . 3 = 27 - 24 = 3

x - 140 = 3 x 7 = 21

x = 21 + 140 = 161

b) x³ . x² = 2⁸ : 2³

x⁵ = 2⁵

=> x = 2

c) (x + 2) . ( x - 4) = 0

x = -2 hoặc 4

d) 3^x-3 - 3² = 2 . 3² =

3^x-3 - 9 = 2 . 9 = 18

3^x-3 = 18 + 9 = 27

3^x-3 = 3³

=> x - 3 = 3

x = 3 + 3 = 6

Ta có: \(\frac{1}{x^2-4}=\frac{a}{x-2}+\frac{b}{x+2}\)

=>\(\frac{a\left(x+2\right)+b\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{1}{\left(x-2\right)\left(x+2\right)}\)

=>a(x+2)+b(x-2)=1

=>x(a+b)+2a-2b=1

=>a+b=0 và 2a-2b=1

=>a+b=0 và a-b=1/2

=>\(a=\left(0+\frac12\right):2=\frac12:2=\frac14;b=a=-\frac14\)

Ta có: \(\overline{a,5}\times\overline{3,bc}=7,85\)

=>\(\left(a+0,5\right)\times\left(3+0,1b+0,01c\right)=7,85\)

=>3a+0,1ab+0,01ac+1,5+0,05b+0,005c=7,85

=>a=2; b=1; c=4

TC: a/b=b/c

=>

\(1,\\ a,X=\left\{3;4\right\};\left\{2;3;4\right\};\left\{1;2;3;4\right\}\\ b,X=\left\{2;4\right\}\\ X=\left\{2\right\}\\ X=\left\{4\right\}\\ X=\varnothing\)

\(2,\\ a,A=\left\{-3;-2;0;1;2;3;4\right\}\\ B=\left\{0;1;2;3;4;6;9;10\right\}\\ b,A=\left\{1;2;3;4;5\right\}\\ B=\left\{1;2;3;6;9\right\}\)

\(a+b=132\)\(\left(1\right)\)

\(a-b=4\) \(\left(2\right)\)

lấy \(\left(1\right)-\left(2\right)\)ta có 

\(a+b-a+b=132-4\)

<=> \(2b=128\)

<=> \(b=64\)

=> \(a=4+b=4+64=68\)

\(a+b=132\)

\(a-b=4\)

\(\Rightarrow a=\left(132+4\right)\text{ : }2\)

\(\Rightarrow a=136\text{ : }2\)

\(\Rightarrow a=68\)

\(\Rightarrow b=68-4\)

\(\Rightarrow b=64\)

\(\text{Vậy : }a=68\)

            \(b=64\)

\(a+2⋮a-1\)

\(=>\left(a-1\right)+3⋮a-1\)

\(\)Vì \(a-1⋮a-1\) mà \(\left(a-1\right)+3⋮a-1\)

\(=>3⋮a-1\)

\(=>a\in\text{Ư}\left(3\right)=\left\{-3;-1;1;3\right\}\)

co a+2=a-1+3

de a+2 chia het cho a-1 thi 3 chia het cho a-1

=> a-1 thuoc uoc cua 3

ma U(3)∈{-1;1;-3;3}

ta co bang sau

vay...