Chọn B

B

\(n^2+n-7=\left(n^2-2n\right)+\left(3n-6\right)-1\\ =n\left(n-2\right)+3\left(n-2\right)-1\\ =\left(n-2\right)\left(n+3\right)-1\)

Để: \(\left(n^2+n-7\right)⋮\left(n-2\right)\Rightarrow\left[\left(n-2\right)\left(n+3\right)-1\right]⋮\left(n-2\right)\\ \Rightarrow1⋮\left(n-2\right)\) (Vì: \(\left(n-2\right)\left(n+3\right)⋮\left(n-2\right)\forall n\inℤ\) )

\(\Rightarrow n-2\in\left\{1;-1\right\}\Rightarrow n\in\left\{3;1\right\}\)

Ta có:

\(n^2+n-7\\ =\left(n^2-2n\right)+\left(3n-6\right)-1\\ =n\left(n-2\right)+3\left(n-2\right)-1\\ =\left(n-2\right)\left(n+3\right)-1\)

Để `n^2+n-7` chia hết cho n - 2 thì: 

1 ⋮ n - 2

=> n - 2 ∈ Ư(1) = {1; -1}

=> n ∈ {3; 1} 

\(a,14364:9+20020\)

\(=1596+20020\)

\(=21616\)

\(b,\left(10356\times5-780\right):6\)

\(=\left(51780-780\right):6\)

\(=51000:6\)

\(=8500\)

`c`

\(=51000:6\)

Vậy `9276 : 39 = 237`(dư`33`)

\(3^{x+2}-5.3^x\\ =3^x\left(3^2-5\right)\\ =3^x.\left(9-5\right)\\ =4.3^x\)

bạn ơi mình thiếu =36

\(x^2+xy-2x-2y\\ =\left(x^2+xy\right)-\left(2x+2y\right)\\ =x\left(x+y\right)-2\left(x+y\right)\\ =\left(x-2\right)\left(x+y\right)\)

\(P\left(x\right)=5x^2+x+2=5\left(x^2+\dfrac{1}{5}x\right)+2\\ =5\left(x^2+2.x.\dfrac{1}{10}+\left(\dfrac{1}{10}\right)^2\right)-5.\left(\dfrac{1}{10}\right)^2+2\\ =5\left(x+\dfrac{1}{10}\right)^2+\dfrac{39}{20}\)

Nhận xét: \(\left(x+\dfrac{1}{10}\right)^2\ge0\forall x\inℝ\\ \Rightarrow5\left(x+\dfrac{1}{10}\right)^2\ge0\\ \Rightarrow P\left(x\right)=5\left(x+\dfrac{1}{10}\right)^2+\dfrac{39}{20}\ge\dfrac{39}{20}\)

\(Min_{P\left(x\right)}=\dfrac{39}{20}\) tại \(\left(x+\dfrac{1}{10}\right)^2=0\Leftrightarrow x+\dfrac{1}{10}=0\Leftrightarrow x=-\dfrac{1}{10}\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k=>\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\left(1\right)\\ \dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\dfrac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\dfrac{b^2}{d^2}\left(2\right)\)

Từ (1) và (2) => \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

\(1-2.\left(3-2\right)\)

\(=1-2.1\)

\(=1-2\)

\(=1+\left(-2\right)\)

\(=-1\)

1 - 2.(3 - 2)

= 1 - 2.1

= 1 - 2 

= -1