Ta có:

\(3a+2b⋮17\\ \Leftrightarrow30a+20b⋮17\\ 30a+20b-17b⋮17\\ \Leftrightarrow30a+3b⋮17\\ \Leftrightarrow3\left(10a+b\right)⋮17\)

Vì \(3⋮̸17\Rightarrow10a+b⋮17\left(dpcm\right)\)

1500 : 5=300

Bài 1:

\(a,8.6+288.\left(x+3\right)^2=50\\ \Leftrightarrow48+288\left(x+3\right)^2=50\\ \Leftrightarrow\left(x+3\right)^2=\dfrac{1}{144}\\ \Leftrightarrow x+3\in\left\{-\dfrac{1}{12};\dfrac{1}{12}\right\}\\ \Leftrightarrow x\in\left\{-\dfrac{37}{12};-\dfrac{35}{12}\right\}\\ Vậy.....\)

\(b,\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5750\)

=>Số lượng số hạng của tổng trên là (x+100-x-1):1+1=100(số hạng)

\(\Rightarrow\dfrac{\left(2x+101\right).100}{2}=5750\\ \Rightarrow2x+101=\dfrac{5750.2}{100}\\ \Rightarrow2x+101=115\\ \Rightarrow2x=14\\ \Rightarrow x=7\\ Vậy........\)

\(\left(3n-4\right)\left(6n+21\right)\\ =18n^2+63n-24n-84\\ =18n^2+39n-84⋮3\\ \Rightarrow\left(3n-4\right)\left(6n+21\right)⋮3\left(dpcm\right)\)

Gọi dạng của số a là 12k+8\(\left(k\in N^{sao}\right)\)

Ta có:a=12k+8

Vì\(12k⋮4;8⋮4\Rightarrow12k+8⋮4\Rightarrow a⋮4\)

\(1,x^3-x=x\left(x^2-1\right)=x\left(x-1\right)\left(x+1\right)\\ 2,4ax^2-ax=ax\left(4-x\right)\\ 3,x^3-2x^2+5x=x\left(x^2-2x+5\right)\\ 4,y-4xy+4x^2y\\ =y\left(4x^2-4x+1\right)=y\left(2x-1\right)^2\)

a) (-2x5 + 3x2 – 4x3) : 2x2 = (-2/2)x5 – 2 + 3/2x2 – 2 + (-4/2)x3 – 2 = – x3 + 3/2 – 2x.

b) (x3 – 2x2y + 3xy2) : (-1/2x) = (x3 : – 1/2x) + (-2x2y : – 1/2x) + (3xy2 : – 1/2x) = -2x2 + 4xy – 6y2 = -2x(x + 2y + 3y2)

c)(3x2y2 + 6x2y3 – 12xy) : 3xy = (3x2y2 : 3xy) + (6x2y2 : 3xy) + (-12xy : 3xy) = xy + 2xy2 – 4.

\(a,4b^2c^2-\left(b^2+c^2-a^2\right)^2\\ =\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\\ =\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\\ =-\left(\left(b-c\right)^2-a^2\right).\left(\left(b+c\right)^2-a^2\right)\\ =\left(-b+c+a\right)\left(b-c+a\right)\left(b+c-a\right)\left(b+c+a\right)\)

Các câu sau hầu như bn dùng HĐT số 2 nhóm vào

\(a,A=\left\{1;2;3;4;6;9;12;18;36\right\}\\ B=\left\{0;3;6;9\right\}\\ B\subset A\\ b,E=\left\{1;2;4;12;18;36\right\}\\ c,C=\left\{0;3\right\}\)

\(a,\dfrac{3n+7}{n+2}=\dfrac{3n+6+1}{n+2}=3+\dfrac{1}{n+2}\)

Để 3n+7 chia hết cho n+2 thì

\(1⋮n+2\)

\(\Rightarrow n+2=1\\ \Rightarrow n=-1\left(loai\right)\\ Vayn\in\varnothing\)

\(b,\dfrac{5n+9}{n+1}=5+\dfrac{4}{n+1}\)

Để

\(5n+9⋮n+1\Rightarrow4⋮n+1\\ \Rightarrow n+1\in\left\{1;2;4\right\}\\ \Rightarrow n\in\left\{0;1;3\right\}\)

Vậy.......