Trả lời:

5 - (10 - x) = 7

10 - x        = -2

x               = 10 - (-2)

x               = 12 (thỏa mãn \(x\inℤ\))

Vậy x = 12

Hok tốt!

Vuong Dong Yet

a) I-9 - xI = 12 + 5

I-9 - xI = 17

\(\Rightarrow\hept{\begin{cases}-9-x=17\\x=-9-17\\x=-26\end{cases}hoac\hept{\begin{cases}-9-x=-17\\x=-9-\left(-17\right)\\x=8\end{cases}}}\)

Vậy\(x\in\left\{-26;8\right\}\)

Chúc bạn học tốt !

b) 11 - (x + 84) = 97

11 - x - 84 = 97

11 - 84 - 97 = x

-170 = x

Vậy x = -170

c) - (x + 84) + 213 = -16

-x - 84 + 213 = -16

-84 + 213 + 16 = x

145 = x

Vậy x = 145

e) -2x - (-17) = 15

-2x + 17 = 15

17 - 15 = 2x

2 = 2x

x = 2 : 2

x = 1

Vậy x = 1

Chúc bạn học tốt !

\(A=\frac{3}{2}\times\left(\frac{1}{13\times11}+\frac{1}{13\times15}+\frac{1}{15\times17}+.....+\frac{1}{97\times99}\right)\)

\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+......+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=\frac{3}{2}\times\left(\frac{1}{11}-\frac{1}{99}\right)\)

\(A=\frac{3}{2}\times\frac{8}{99}\)

\(A=\frac{4}{33}\)

b] \(\frac{A}{5}=\frac{4}{31.35}+\frac{6}{35.41}+\frac{9}{41.50}+\frac{7}{50.57}\)

\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)

\(\frac{A}{5}=\frac{1}{31}-\frac{1}{57}\)

\(\Rightarrow A=5\left(\frac{1}{31}-\frac{1}{57}\right)=\frac{130}{1767}\)

c] Ta đặt \(\left(8n+5,6n+4\right)=d\)

\(\Rightarrow\frac{8n+5\div d}{6n+4\div d}\Rightarrow4\times\left(6n+4\right)-3\times\left(8n+5\right)=\left(24n+16\right)-\left(24n+15\right):d\)\(\Rightarrow d=1\)

Vậy \(\frac{8n+5}{6n+4}\)là phân số tối giản

Answer:

a) 

Có:

\(15=1.15=\left(-1\right).\left(-15\right)=3.5=\left(-3\right).\left(-5\right)\)

b) 

Có:

\(M=1+3+3^2+3^3+3^4+...+3^{98}+3^{99}+3^{100}\)

\(=\left(1+3\right)+\left(3^2+3^3+3^4\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=4+3^2.\left(1+3+3^2\right)+...+3^{98}.\left(1+3+3^2\right)\)

\(=4+3^2.13+3^{98}.13\)

\(=4+13.\left(3^2+...+3^{98}\right)\)

Vậy M chia 13 dư 4

Có:

\(M=1+3+3^2+3^3+3^4+...+3^{99}+3^{100}\)

\(=1+\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)

\(=1+3.40+3^5.40+...+3^{97}.40\)

\(=1+40.\left(3+3^5+...+3^{97}\right)⋮40\)

Mà \(40\left(3+3^5+...+3^{97}\right)⋮40\)

Vậy M chia 40 dư 1

\(3^x.3^{x+1}.3^{x+2}\ge729\)

\(\Leftrightarrow3^x.3^x.3.3^x.3^2\ge729\)

\(\Leftrightarrow\left(3^x\right)^3\ge729\)

\(\Leftrightarrow3^x\ge27=3^3\)

\(\Leftrightarrow x\ge1\).

Gọi số sách cần tìm là a (a thuộc N*)

Nếu số sách xếp thành từng bó 12,.15, 18 thì vừ đủ và 300<400

=> a là BC (12,15,18)

12=3.2² 15=3.5 18=3².2

=> BCNN(12,15,18)= 3².2².5=360

=> BC(12,15,18)=B(360)={0,360,720.....}

Vậy a=360 (TMĐK)