Tìm n để thoả mãn điều gì thế em???

P = (a - b) - (b - c + a) - c

= a - b - b + c - a - c

= ( a - a ) + ( -b - b) + ( c - c )

= -2b

=> Đáp án B

Phần kl anh viết ngược chỗ (0;20) rồi ạ.

\(\dfrac{5}{x}-\dfrac{y}{3}=\dfrac{1}{6}\\ \Rightarrow\dfrac{15-xy}{3x}=\dfrac{1}{6}\\ \Rightarrow6\left(15-xy\right)=3x\\ \Rightarrow60-6xy=3x\\ \Rightarrow3x+6xy=60\\ \Rightarrow3x\left(1+2y\right)=60\)

x,y nguyên => \(1+2y\) luôn lẻ  

Mà: 1 + 2y \(\inƯ\left(60\right)\) \(\Rightarrow1+2y\in\left\{\pm1;\pm3;\pm5;\pm15\right\}\) 

Ta có bảng sau:

Vì x,y là các số nguyên nên: (x;y)={(20;0);(-20;-1);(2;2);(-2;-3)}

a) 733 - (33 + 320)

= 733 - 33 - 320

= 700 - 320

= 380

b) 945 + (372 - 245)

= 945 + 372 - 245

= (945 - 245) + 372

= 700 + 372

= 1072

c) 717 - (617 - 225)

= 717 - 617 + 225

= 100 + 225 

= 325

d) 1732 - (732 - 257)

= 1732 - 732 + 257

= 1000 + 257

= 1257

Câu 8:

a) -13,1

456,6

-8865,1

66,7

b) -13

457

-8865

67

c) -13,15

456,58

-8865,12

66,66

Câu 9:

\(\dfrac{3}{10}=\dfrac{30}{100}=30\%\\ \dfrac{4}{5}=\dfrac{80}{100}=80\%\\ \dfrac{30}{8}=\dfrac{15}{4}=\dfrac{375}{100}=375\%\)

B8: 

a) 

\(-13,145\approx-13,1\\ 456,578\approx456,6\\ -8865,123\approx-8865,1\\ 66,66\approx66,7\)

b) 

\(-13,145\approx-13\\ 456,578\approx457\\ -8865,123\approx-8865\\ 66,66\approx67\)

c) 

\(-13,145\approx-13,15\\ 456,578\approx456,58\\ -8865,123\approx-8865,12\\ 66,66\approx66,66\) 

B9: \(\dfrac{3}{10}=30\%;\dfrac{4}{5}=80\%;\dfrac{30}{8}=375\%\)

\(25< 5^{2x-1}< 5^5\)

\(5^2< 5^{2x-1}< 5^5\)

\(2< 2x-1< 5\)

\(2+1< 2x< 5+1\)

\(3< 2x< 6\)

\(\dfrac{3}{2}< x< 3\)

x=2

\(A=\dfrac{6n-1}{2n+3}=\dfrac{3\left(2n+3\right)-10}{2n+3}\\ =3-\dfrac{10}{2n+3}\)

Để A nguyên thì: \(\dfrac{10}{2n+3}\) nguyên

\(\Rightarrow10⋮\left(2n+3\right)\)

\(\Rightarrow2n+3\inƯ\left(10\right)=\left\{1;-1;2;-2;5;-5;10;-10\right\}\\ \Rightarrow2n\in\left\{-2;-4;-1;-5;2;-8;7;-13\right\}\\ \Rightarrow n\in\left\{-1;-2;-\dfrac{1}{2};-\dfrac{5}{2};1;-4;\dfrac{7}{2};-\dfrac{13}{2}\right\}\)

\(\left(x-3\right)+\left(x-5\right)+\left(x-7\right)+...+\left(x-19\right)=3618\\ x-3+x-5+x-7+...+x-19=3618\\ 9x-\left(3+5+7+...+19\right)=3618\\ 9x-99=3618\\ 9x=3618+99\\ 9x=3717\\ x=3717:9\\ x=413\)

Vậy...

a: x+|-2|=0

=>x+2=0

=>x=-2

b: \(4x-20=2^5:2^3\)

=>\(4x-20=2^2=4\)

=>\(4x=20+4=24\)

=>\(x=\dfrac{24}{4}=6\)

a) x+|-2| = 0

⇒|x| = 0-(-2)

⇒|x| = 2⇒x= 2 hoặc x= -2

Vậy x = 2 hoặc x = -2 ϵ z

b) 4x - 20 = 2⁵ : 2³

⇒ 4x - 20 = 32 : 8

⇒ 4x - 20 = 4

⇒ 4x = 4+20

⇒4x = 24

⇒ x = 6

Vậy x = 6 ϵ z