a) 10 ⋮ (x - 1)

⇒ x - 1 ∈ Ư(10) = {-10; -5; -2; -1; 1; 2; 5; 10}

⇒ x ∈ {-9; -4; -1; 0; 2; 3; 6; 11}

b) x + 5 = x - 2 + 7

Để (x - 5) ⋮ (x - 2) thì 7 ⋮ (x - 2)

⇒ x - 2 ∈ Ư(7) = {-7; -1; 1; 7}

⇒ x ∈ {-5; 1; 3; 9}

c) 3x + 8 = 3x - 3 + 11

= 3(x - 1) + 11

Để (3x + 8) ⋮ (x - 3) thì 11 ⋮ (x - 3)

⇒ x - 3 ∈ Ư(11) = {-11; -1; 1; 11}

⇒ x ∈ {-8; 2; 4; 14}

a) 25 chia hết cho n + 2

=> n + 2 ∈ Ư(25) 

=> n + 2 ∈ {1; -1; 5; -5; 25; -25}

=> n ∈ {-1; -3; 3; -7; 23; -27}  

b) 2n + 4 chia hết cho n - 1

=> (2n - 2) + 6 chia hết chi n - 1

=> 2(n - 1) + 6 chia hết cho n - 1

=> 6 chia hết cho n - 1

=> n - 1 ∈ Ư(6) = {1; -1; 2; -2; 3; -3; 6; -6}

=> n ∈ {2; 0; 3; -1; 4; -2; 7; -5} 

a) 25 ⋮ (n + 2)

⇒ n + 2 ∈ Ư(25) = {-25; -5; -1; 1; 5; 25}

⇒ n ∈ {-27; -7; -3; -1; 3; 23}

b) 2n + 4 = 2n - 2 + 6

= 2(n - 1) + 6

Để (2n + 4) ⋮ (n - 1) thì 6 ⋮ (n - 1)

⇒ n - 1 ∈ Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}

⇒ n ∈ {-5; -2; -1; 0; 2; 3; 4; 7}

Bài 11.2 

\(a,A=3+3^2+3^3+....+3^{99}\\ =\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+....+\left(3^{97}+3^{98}+3^{99}\right)\\ =3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+...+3^{97}\cdot\left(1+3+9\right)\\ =13\cdot\left(3+3^4+....+3^{97}\right)⋮13\\ b,B=5+5^2+...+5^{50}\\ =\left(5+5^2\right)+\left(5^3+5^4\right)+..+\left(5^{49}+5^{50}\right)\\ =5\cdot\left(1+5\right)+5^3\cdot\left(1+5\right)+....+5^{49}\cdot\left(1+5\right)\\ =6\cdot\left(5+5^3+...+5^{49}\right)⋮6\)

Bài 11.4:

a: \(10⋮x-1\)

=>\(x-1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

=>\(x\in\left\{2;0;3;-1;6;-4;11;-9\right\}\)

b: 

\(x+5⋮x-2\)

=>\(x-2+7⋮x-2\)

=>\(7⋮x-2\)

=>\(x-2\in\left\{1;-1;7;-7\right\}\)

=>\(x\in\left\{3;1;9;-5\right\}\)

c: \(3x+8⋮x-1\)

=>\(3x-3+11⋮x-1\)

=>\(11⋮x-1\)

=>\(x-1\in\left\{1;-1;11;-11\right\}\)

=>\(x\in\left\{2;0;12;-10\right\}\)

Bài 11.5:

a: (x+4)(y-1)=13

=>\(\left(x+4;y-1\right)\in\left\{\left(1;13\right);\left(13;1\right);\left(-1;-13\right);\left(-13;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-3;14\right);\left(9;2\right);\left(-5;-12\right);\left(-17;0\right)\right\}\)

b: xy-3x+y=20

=>x(y-3)+y-3=17

=>(x+1)(y-3)=17

=>\(\left(x+1;y-3\right)\in\left\{\left(1;17\right);\left(17;1\right);\left(-1;-17\right);\left(-17;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(0;20\right);\left(16;4\right);\left(-2;-14\right);\left(-18;2\right)\right\}\)

`-12(x - 15) + 7(3 - x) = 15`

`=> -12x + 180 + 21 - 7x - 15 = 0`

`=> -19x + 186 = 0`

`=> -19x = -186`

`=> x = -186 : (-19) `

`=> x = 186/19`

Vậy ...

-------------------------

`(x - 1) . (x + 2) . (-x - 3) = 0`

`=> x - 1 = 0` hoặc `x + 2 = 0` hoặc `-x - 3 = 0`

`=> x =1` hoặc `x = -2` hoặc `x = -3`

Vậy ...

\(=-125.75-125.\left(-43\right)+75.125+75.43\)

\(=\left(-125.75+75.125\right)+125.43+75.43\)

\(=0+43.\left(125+75\right)\)

\(=43.200\)

\(=8600\)

`124 . (-49) + 62 . (-102)`

`= 124 . (-49) + 62 . 2.(-51) `

`=124. (-49) +124. (-102)`

`= 124 . (-49 + (-102))`

`= 124 . (-100)`

`= -12400`

\(=124.\left(-49\right)+62.2.\left(-51\right)\)

\(=124.\left(-49\right)+124.\left(-51\right)\)

\(=124.\left(-49+\left(-51\right)\right)\)

\(=124.\left(-100\right)\)

\(=-12400\)

`2/5 + 3/4 :x= -1/2 `

`=> 3/4 :x = -1/2 - 2/5`

`=> 3/4 : x = -9/10`

`=> x = 3/4 : (-9/10)`

`=> x = 3/4 . (-10/9) `

`=> x = -5/6`

Vậy `x = -5/6`

------------------

`5/7 - 2/3 x = 4/5`

`=> 2/3 x = 5/7 - 4/5`

`=> 2/3 x = -3/35`

`=> x = -3/35 : 2/3`

`=> x = -3/35 . 3/2`

`=> x = -9/70`

Vậy `x = -9/70`

------------------

`1/2 x + 3/5 x = -2/3`

`=> (1/2 + 3/5) x = -2/3`

`=> 11/10 x = -2/3`

`=> x = -2/3 : 11/10`

`=> x = -2/3 . 10/11`

`=> x = -20/33`

Vậy ` x = -20/33`

------------------

`4/7 x - x = -9/14`

`=> (4/7 - 1) x = -9/14`

`=> -3/7 x = -9/14`

`=> 3/7 x = 9/14`

`=> x = 9/14 : 3/7`

`=> x = 9/14 . 7/3`

`=> x = 3/2`

Vậy `x = 3/2`

\(\left(2x-\dfrac{1}{3}\right)^2=-\dfrac{8}{15}\times\dfrac{15}{27}:\left(2x-\dfrac{1}{3}\right)\)

\(\left(2x-\dfrac{1}{3}\right)^2\times\left(2x-\dfrac{1}{3}\right)=-\dfrac{8}{27}\)

\(\left(2x-\dfrac{1}{3}\right)^3=\left(-\dfrac{2}{3}\right)^3\)

\(2x-\dfrac{1}{3}=-\dfrac{2}{3}\)

\(2x=\dfrac{1}{3}-\dfrac{2}{3}\)

\(2x=-\dfrac{1}{3}\)

\(x=-\dfrac{1}{6}\)

Áp dụng t/c dãy tỉ số bằng nhau: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

Từ \(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+b^3}{c^3+d^3}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có: \(VT=\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{bk^3+b^3}{dk^3+d^3}=\dfrac{b.\left(k+1\right)^3}{d.\left(k+1\right)^3}=\dfrac{b}{d}\)

\(VP=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}=\dfrac{\left(bk+b\right)^3}{\left(dk+d\right)^3}=\dfrac{b.\left(k+1\right)^3}{d.\left(k+1\right)^3}=\dfrac{b}{d}\)

Vậy \(VT=VP\left(đpcm\right)\)

____________

VT = vế trái

VP = vế phải

\(#NqHahh\)

x+2x+1=16

=>3x=16-1=15

=>\(x=\dfrac{15}{3}=5\)

180-(x-45):2=120