theo t/c số nguyên tố: nếu tăng hoặc giảm 1 đơn vị 1 số NT >3 thì 1 trong 2 kết quả chia hết cho 6 => 1 trong 2 kết quả chia hết cho 3

=> p+1=6k => p=6k-1 hoặc p-1=6k => p=6k+1

+ Nếu p=6k-1 chia hết cho 6

(6k-1)²+3n+2=36k²-12k+1+3n+2=36k²-12k+3n+3 chia hết cho 3 nên p²+3n+2 là hợp số

+ Nếu p=6k+1 chia hết cho 6

(6k+1)²+3n+2=36k²+12k+1+3n+2=36k²+12k+3n+3 chia hết cho 3

nên p²+3n+2 là hợp số

\(A=\dfrac{1}{x-1}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{2x-4}{x-5}\left(x\ne1;2;5\right)\)

\(A=\dfrac{1}{x-1}+\dfrac{x+1}{x-5}-\dfrac{2x-4}{x-5}=\dfrac{1}{x-1}-\dfrac{x-5}{x-5}=\)

\(=\dfrac{1}{x-1}-1\)

Để A nguyên x-1 phải là ước của 1

\(\Rightarrow x-1=\left\{-1;1\right\}\Rightarrow x=\left\{0;2\right\}\) đối chiếu đk \(x\ne2\) => x=0

\(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{c}=0\Rightarrow2bc+ca+2ab=0\left(a;b;c\ne0\right)\)

\(\Rightarrow2bc=-a\left(2b+c\right);ca=-2b\left(a+c\right);2ab=-c\left(a+2b\right)\)

\(\Rightarrow A=-\dfrac{2b+c}{a}-\dfrac{a+c}{2b}-\dfrac{a+2b}{c}=\)

\(=\dfrac{-2bc\left(2b+c\right)-ac\left(a+c\right)-2ab\left(a+2b\right)}{2abc}=\)

\(=\dfrac{4b^2c+2bc^2+ac\left(a+c\right)+2a^2b+4ab^2}{-2abc}=\)

\(=\dfrac{2b\left(2bc+a^2+c^2+2ab\right)+ac\left(a+c\right)}{-2abc}=\)

\(=\dfrac{2b\left[\left(a^2+c^2\right)+2bc+2ab\right]+ac\left(a+c\right)}{-2abc}=\)

\(=\dfrac{2b\left[\left(a+c\right)^2-2ac+2bc+2ab\right]+ac\left(a+c\right)}{-2abc}=\)

\(=\dfrac{2b\left[\left(a+c\right)^2-3ac+2bc+ac+2ab\right]+ac\left(a+c\right)}{-2abc}=\)

\(=\dfrac{2b\left[\left(a+c\right)^2-3ac\right]+ac\left(a+c\right)}{-2abc}=\)

\(=\dfrac{2b\left(a+c\right)^2-6abc+ac\left(a+c\right)}{-2abc}=\)

\(=\dfrac{\left(a+c\right)\left[2b\left(a+c\right)+ac\right]-6abc}{-2abc}=\)

\(=\dfrac{\left(a+c\right)\left(2ab+2bc+ac\right)-6abc}{-2abc}=\dfrac{-6abc}{-2abc}=3\)

Dạo này khi viết công thức cứ mất các dấu ngoặc là sao nhỉ

=8.3-9x2+27x8-1

=24-18+216-1

=6+216-1

=222-1

=221

giúp cách nào khi bạn không biết bạn tìm CÁI gì?

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3abc-3ab\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\left(đpcm\right)\)

\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\dfrac{1}{2}\left(a+b+c\right).\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\left(đpcm\right)\)

\(\dfrac{2x-1}{2}+\dfrac{5-x}{6}=2-\dfrac{3\left(x+1\right)}{4}\)

\(\Leftrightarrow\dfrac{6\left(2x-1\right)}{12}+\dfrac{2\left(5-x\right)}{12}=\dfrac{24}{12}-\dfrac{9\left(x+1\right)}{12}\)

\(\Leftrightarrow6\left(2x-1\right)+2\left(5-x\right)=24-9\left(x+1\right)\)

\(\Leftrightarrow12x-6+10-2x=24-9x-9\)

\(\Leftrightarrow19x=11\)

`<=>x=11/19`

Vậy \(S=\left\{\dfrac{11}{19}\right\}\)

\(\dfrac{2x-1}{2}\) + \(\dfrac{5-x}{6}\) = 2 - \(\dfrac{3\left(x+1\right)}{4}\)

6(2x-1) + 2(5-x) = 24 - 9 (x+1)  (quy đồng mẫu số)

12x - 6 + 10 -2x = 24 - 9x - 9   (nhân phá ngoặc)

10x + 9x = 24 - 9  -10 + 6 ( chuyển vế đổi dấu)

19x = 11

x = 11/19