Tập A là tập các số chia 3 dư 1

Tập B có dạng tổng quát 6m + 4 = 6m + 3 +1 => tập các số chia 3 dư 1

=> \(B\subset A\)

P/s

a) (x - 25) - 130 = 150

=> x - 25            = 150 + 130

=> x - 25            = 280

=> x                   = 280 + 25

=> x                   = 305

c) 435 + 5.x = 135

=>          5.x = 135 - 435

=>          5.x = - 300

=>             x = -300 : 5

=>             x = - 60

\(\left(x-25\right)-130=150\)

\(\left(x-25\right)=150+130\)

\(\left(x-25\right)=280\)

\(x=280+25\)

\(x=305\)

Ta có: \(a^3+b^3+c^3+d^3-a-b-c-d\)

\(=\left(a-1\right)a\left(a+1\right)+\left(b-1\right)b\left(b+1\right)+\left(c-1\right)c\left(c+1\right)+\left(d-1\right)d\left(d+1\right)\)chia hết cho 6 ( tích 3 số nguyên liên tiếp thì chia hết cho 6)

Mà \(a^3+b^3+c^3+d^3=2004⋮6\Rightarrow a+b+c+d⋮6\left(đpcm\right)\)

\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)

\(\Rightarrow3x-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)=0\)

\(\Rightarrow3x-\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)=0\)

\(\Rightarrow3x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)

\(\Rightarrow3x-\left(1-\frac{1}{99}\right)=0\)

\(\Rightarrow3x-\frac{98}{99}=0\)

\(\Rightarrow3x=0+\frac{98}{99}\)

\(\Rightarrow3x=\frac{98}{99}\)

\(\Rightarrow x=\frac{98}{99}:3\)

\(\Rightarrow x=\frac{98}{297}\)

\(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}=0\)

\(2\left(3x-\frac{1}{3}-\frac{1}{15}-\frac{1}{35}-\frac{1}{63}-\frac{1}{99}\right)=2.0\)

\(6x-\frac{2}{3}-\frac{2}{15}-\frac{2}{35}-\frac{2}{63}-\frac{2}{99}=0\)

\(6x-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)=0\)

\(6x-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)=0\)

\(6x-\left(1-\frac{1}{11}\right)=0\)

\(6x-\frac{10}{11}=0\)

\(6x=\frac{10}{11}\)

\(x=\frac{5}{33}\)

\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}\)\(-\sqrt{60}\)

\(=2\sqrt{3}.\sqrt{3}\)\(+\sqrt{5}.\sqrt{3}\)\(-\sqrt{4.15}\)

\(=2.3+\sqrt{15}-2\sqrt{15}\)

\(=6+\sqrt{15}.\left(1-2\right)\)

\(=6-\sqrt{15}\)

\(\left(2\sqrt{3}+\sqrt{5}\right).\sqrt{3}-\sqrt{60}\)

\(=2\sqrt{3}.\sqrt{3}+\sqrt{5}.\sqrt{3}-\sqrt{60}\)

\(=2.3+\sqrt{5.3}-\sqrt{60}\)

\(=6+\sqrt{15}-\sqrt{60}\)

\(=6-\sqrt{15}\)

Ta có:\(3x=2z-x\Rightarrow4x=2z\Rightarrow2x=z\)

\(x+y+z=60\Rightarrow z=60-x-y\Rightarrow2x=60-x-y\Rightarrow3x=60-y\)

                                                                                                                            \(\Rightarrow4y=60-y\Rightarrow5y=60\Rightarrow y=12\)

\(\Rightarrow4y=3x=12.4=48\Rightarrow x=\frac{48}{3}=16\)

Mà \(2x=z\Rightarrow z=16.2=32\)

Vậy\(x=16;y=12;x=32\)

Ta có :

3x = 4y = 2z => \(\frac{x}{4}=\frac{y}{3}\)và \(\frac{y}{2}=\frac{z}{4}\)=> \(\frac{x}{8}=\frac{y}{6}\)và \(\frac{y}{6}=\frac{z}{12}\)

=> \(\frac{x}{8}=\frac{y}{6}=\frac{z}{12}\). Áp dụng tính chất dãy tỉ số bằng nhau, ta có :

\(\frac{x}{8}=\frac{y}{6}=\frac{z}{12}=\frac{x+y+z}{8+6+12}=\frac{60}{24}\)

Suy ra : \(\frac{x}{8}=\frac{60}{24}\Rightarrow x=\frac{60}{24}.8=20\)

               \(\frac{y}{6}=\frac{60}{24}\Rightarrow y=\frac{60}{24}.6=15\)

               \(\frac{z}{12}=\frac{60}{24}\Rightarrow x=\frac{60}{24}.12=30\)

Vậy : x = 20 ; y = 15 ; z = 20