Vd \(1\cdot2=2\) là số nguyên tố

=>Đề sai rồi bạn

Gọi 2 số nguyên tô đó lần lượt là `a;b`

Ta có: Tích `2` số nguyên tố là `ab`

Do `a vdots a; b vdots b => ab vdots a` và `b`

Mà `ab vdots 1` và `ab`

`=> ab` có nhiều hơn `2` ước (đpcm)

\(2022^0+\left[100-\left(3^2+1\right)^2\right]\)

\(=1+100-10^2\)

=1

làm sao để ra 10² vậy bạn🤡

\(\left|x-y+1\right|>=0\forall x,y\)

=>\(-2\left|x-y+1\right|< =0\forall x,y\)

\(\left|y-2\right|>=0\forall y\)

=>\(-3\left|y-2\right|< =0\forall y\)

Do đó: \(-2\left|x-y+1\right|-3\left|y-2\right|< =0\forall x,y\)

=>\(C=-2\left|x-y+1\right|-3\left|y-2\right|-4< =-4\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y+1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y-1=2-1=1\end{matrix}\right.\)

Sao mình không nhìn thấy đề bài bạn nhỉ?

Tổng các số là:

\(\dfrac{6}{3}.111.\left(6+5+3\right)=3108\)

D={T;R;U;O;N;G;H;C;H;A;I;E;S}

=>D có 13 phần tử

\(6^x+6^{x+1}=2^{x+1}+2\cdot2^{x+2}+4\cdot2^x\)

=>\(6^x+6^x\cdot6=2^x\cdot2+4\cdot2^x+4\cdot2^x\)

=>\(6^x\cdot7=2^x\cdot10\)

=>\(3^x=\dfrac{10}{7}\)

=>\(x=log_3\left(\dfrac{10}{7}\right)\)

6^\(x\) + 6^\(x+1\) = 2^\(x+1\) + 2.2^\(x+2\) + 4.2\(^x\) (\(x\in\) N)

6\(^x\)(1 + 6) = 2\(^x\).(2 + 2.2² + 4)

6\(^x\).7 = 2\(^x\).(2+ 8 + 4)

6^\(x\).7 = 2\(^x\).(10 + 4)

6\(^x\).7 = 2\(^x\).14

6\(^x\)      = 2\(^x\).14 : 7

6\(^x\)      = 2^\(x\).2

6\(^x\) : 2\(^x\) = 2

   3\(^x\) = 2 ⇒ 3\(^x\) ⋮ 2 (vô lý) Vậy pt vô nghiệm hay 

\(x\in\) \(\varnothing\) 

\(7-\left(x-1\right)=15+3\left(x+1\right)\\ 7-x+1=15+3x+3\\ 8-x=18+3x\\ 3x+x=8-18\\ 4x=-10\\ x=-\dfrac{10}{4}\\ x=\dfrac{-5}{2}\)

Vậy: ... 

\(A=\left\{n^2\text{ }|\text{ }n\in N,\text{ }1\le n\le7\text{ }\right\}\)

Hoặc:

\(A=\left\{x\text{ }|\text{ }\text{x là số chính phương},\text{ }0< x< 50\right\}\) 

\(1)-\dfrac{3}{7}+\dfrac{5}{13}-\dfrac{4}{7}+\dfrac{8}{13}\\ =\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)+\left(\dfrac{5}{13}+\dfrac{8}{13}\right)\\ =\dfrac{-7}{7}+\dfrac{13}{13}\\ =-1+1\\ =0\\ 2)-\dfrac{5}{14}-\dfrac{2}{14}+\dfrac{1}{8}+\dfrac{1}{8}\\ =\left(\dfrac{-5}{14}-\dfrac{2}{14}\right)+\left(\dfrac{1}{8}+\dfrac{1}{8}\right)\\ =\dfrac{-7}{14}+\dfrac{2}{8}\\ =\dfrac{-1}{2}+\dfrac{1}{4}\\ =\dfrac{-1}{4}\\ 3)\dfrac{-5}{22}-1+\dfrac{3}{2}-\dfrac{6}{22}\\ =\left(\dfrac{-5}{22}-\dfrac{6}{22}\right)+\left(\dfrac{3}{2}-1\right)\\ =\dfrac{-11}{22}+\dfrac{1}{2}\\ =\dfrac{-1}{2}+\dfrac{1}{2}\\ =0\)

\(4,\dfrac{7}{16}+\dfrac{5}{9}+\dfrac{-3}{16}+\dfrac{-2}{9}\\ =\left(\dfrac{7}{16}-\dfrac{3}{16}\right)+\left(\dfrac{5}{9}-\dfrac{2}{9}\right)\\ =\dfrac{4}{16}+\dfrac{3}{9}\\ =\dfrac{1}{4}+\dfrac{1}{3}\\ =\dfrac{7}{12}\\ 5,\dfrac{2}{5}-\dfrac{3}{11}-\dfrac{7}{35}+\dfrac{14}{11}-\dfrac{1}{5}\\ =\left(-\dfrac{3}{11}+\dfrac{14}{11}\right)+\left(\dfrac{2}{5}-\dfrac{1}{5}\right)-\dfrac{7}{35}\\ =\dfrac{11}{11}+\dfrac{1}{5}-\dfrac{7}{35}\\ =1+\dfrac{1}{5}-\dfrac{1}{5}\\ =1\\ 6,\dfrac{3}{4}-\dfrac{3}{17}+\dfrac{-5}{6}+\dfrac{20}{17}-\dfrac{1}{4}\\ =\left(\dfrac{3}{4}-\dfrac{1}{4}\right)+\left(\dfrac{-3}{17}+\dfrac{20}{17}\right)+\dfrac{-5}{6}\\ =\dfrac{1}{2}+1+\dfrac{-5}{6}\\ =\dfrac{3}{2}-\dfrac{5}{6}\\ =\dfrac{9}{6}-\dfrac{5}{6}\\ =\dfrac{4}{6}=\dfrac{2}{3}\)

\(A=\left(x+1\right)+\left(x+\dfrac{5}{45}\right)+\left(x+\dfrac{5}{117}\right)+\left(x+221\right)=10\\ \Rightarrow x+1+x+\dfrac{1}{9}+x+\dfrac{5}{117}+x+221=10\\ \Rightarrow4x+\left(1+\dfrac{1}{9}+\dfrac{5}{117}+221\right)=10\\ \Rightarrow4x+\dfrac{2888}{13}=10\\ \Rightarrow4x=10-\dfrac{2888}{13}\\ \Rightarrow4x=-\dfrac{2758}{13}\\ \Rightarrow x=-\dfrac{1379}{26}\)

Cho hỏi 1/9 ở đâu v ak