2 giờ 10 phút = \(\frac{13}{6}\) giờ = 2,166666666... giờ

2 giờ 10 phút = \(2\frac{1}{6}\) giờ 

học tốt!!!

mấy cái dấu kẻ dọc ngắn là giá trị tuyệt đối

\(\left(\frac{-2}{5}+\frac{3}{10}\right)\div\left(\frac{2}{-5}+\left|-\frac{4}{3}-2\right|\right)\)

\(=\frac{-1}{10}\div\left(\frac{-2}{5}+\left|\frac{-10}{3}\right|\right)\)

\(=\frac{-1}{10}\div\left(\frac{-2}{5}+\frac{10}{3}\right)\)

\(=\frac{-1}{10}\div\frac{44}{15}\)

\(=\frac{-3}{88}\)

\(3x^2+y^2+4xy-8x-2y=0\)

\(\Leftrightarrow4x^2+4xy+y^2-4x-2y+1-x^2-4x-4=-3\)

\(\Leftrightarrow\left(2x+y-1\right)^2-\left(x+2\right)^2=-3\)

\(\Leftrightarrow\left(2x+y-1-x-2\right)\left(2x+y-1+x+2\right)=-3\)

\(\Leftrightarrow\left(x+y-3\right)\left(3x+y+1\right)=-3\)

Do \(x,y\in Z\Rightarrow x+y-3;3x+y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Bạn lập bảng xét ước rồi tìm ra x,y thỏa mãn

Vậy \(\left(x,y\right)=\left(0,2\right);\left(-4,8\right);\left(-4;10\right);\left(0,0\right)\)

1/ \(=3^n.3^2+3^n=3^n\left(3^2+1\right)=10.3^n⋮10\)

2/ \(100.x+\left(1+2+3+...+100\right)=7450\)

Đến đây bạn tự làm nốt nhé

1. Ta có: \(3^{n+2}+3^n=3^n.\left(3^2+1\right)=3^n.\left(9+1\right)=3^n.10⋮10\)( đpcm )

2. \(\left(x+1\right)+\left(x+2\right)+.......+\left(x+100\right)=7450\)

\(\Leftrightarrow x+1+x+2+........+x+100=7450\)

\(\Leftrightarrow100x+\frac{100.101}{2}=7450\)

\(\Leftrightarrow100x+5050=7450\)

\(\Leftrightarrow100x=2400\)\(\Leftrightarrow x=24\)

Vậy \(x=24\)

1000+3,64×x+6,36×x=2015

3,64×x+6,36×x=2015-1000

3,64×x+6,36×x=1015

x×(3,64+6,36)=1015

x×10=1015

x=1015:10

x=101,5

vậy x=101,5

tớ cũng cần giúp nhưng bai đấy dễ ẹt

\(10^{35}:\left(5^{30}.4^{15}\right)=10^{35}:\left[5^{30}.\left(2^2\right)^{15}\right]=10^{35}:\left(5^{30}.2^{30}\right)\)

\(=10^{35}:10^{30}=10^5\)

a, \(\Leftrightarrow x\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(x=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

b, \(\Leftrightarrow x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

a) Ta có : \(8⋮x\)

\(\Rightarrow x\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Rightarrow x=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

b) Ta có : \(6⋮x-1\)

\(\Rightarrow x-1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

\(\Rightarrow x=\left\{0;2;-1;3;-2;4;-5;7\right\}\)