ẢO THẬT ĐẤY!  <:0

\(a,67\cdot56+12\cdot67+68\cdot33\\ =67\cdot\left(56+12\right)+68\cdot33\\ =67\cdot68+68\cdot33\\ =68\cdot\left(67+33\right)\\ =68\cdot100\\ =6800\)

Đặt A=25+29+...+101

Số số hạng là \(\dfrac{101-25}{4}+1=20\left(số\right)\)

Tổng của dãy số là \(\left(101+25\right)\cdot\dfrac{20}{2}=1260\)

\(\left(25-x\right)+\left(29-x\right)+...+\left(101-x\right)=128\)

=>1260-20x=128

=>20x=1260-128=1132

=>x=1132:20=56,6

\(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)\cdot462-x=19\)

=>\(\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\cdot462-x=19\)

=>\(\left(\dfrac{1}{11}-\dfrac{1}{21}\right)\cdot462-x=19\)

=>\(462\cdot\dfrac{10}{231}-x=19\)

=>20-x=19

=>x=20-19=1

a: M nằm trên đoạn AB

=>MA+MB=AB

=>MB=10-4=6(cm)

b: A là trung điểm của MN

=>\(MN=2\cdot MA=2\cdot4=8\left(cm\right)\)

MN=8cm

MB=6cm

mà 8>6

nên MN>MB

\(\left(3^{85}+2^{41}\right)\cdot\left(9^{17}-5^{10}\right)\cdot\left(18-6^2:2^1\right)\\ =\left(3^{85}+2^{41}\right)\cdot\left(9^{17}-5^{10}\right)\cdot\left(18-36:2\right)\\ =\left(3^{85}+2^{41}\right)\cdot\left(9^{17}-5^{10}\right)\cdot\left(18-18\right)\\ =\left(3^{85}+2^{41}\right)\cdot\left(9^{17}-5^{10}\right)\cdot0\\ =0\)

Để `(x+3)\vdots(x+1),` ta có:

`(x+3)\vdots(x+1)`

`=> (x+1)+2\vdots(x+1)`

Vì: `(x+1)\vdots(x+1)` \(\rightarrow\) `(x+1)` thuộc `Ư(2) = {+-1;+-2}`

`=> x = {0;-2;1;-3}`

Vậy: `x={0;-2;1;-3}` thì `(x+3)\vdots(x+1)`

(x+3)⋮(x+1)

x+1+2⋮x+1

2⋮x+1 (Vì x+1⋮x+1)

=> x+1 thuộc Ư(2) = {-1; 1; 2; -2}

=> x thuộc {-2; 0; 1; -3}

Vậy x thuộc {-2, 0; 1; -3}

\(\left(3\cdot4\cdot2^{16}\right)^2:\left(11\cdot2^{13}\cdot4^{11}-16^9\right)\\ =\left(3\cdot2^2\cdot2^{16}\right)^2:\left(11\cdot2^{13}\cdot2^{22}-2^{36}\right)\\ =3^2\cdot2^4\cdot2^{32}:\left(11\cdot2^{35}-2^{36}\right)\\ =3^2\cdot2^{36}:\left[2^{35}\cdot\left(11-2\right)\right]\\ =9\cdot2^{36}:\left(2^{35}\cdot9\right)\\ =9\cdot2^{36}:2^{35}:9\\ =2\)

D={0;4;8;12;16;20}

D={x\(⋮\)4; x<21}

Cách 1:

\(D=\left\{0;4;8;12;16;20\right\}\)

Cách 2:

\(D=\left\{x\in N|x⋮4,x< 21\right\}\)

Để `5n+22 \vdots n+3,` ta có:

`5n +22 \vdots n+3`

`=> 5n + 15 + 7 \vdots n + 3`

`=> 5 (n + 3)  + 7 \vdots n + 3`

Vì:: `5 ( n + 3)\vdots n + 3 -> n  + 3 in Ư(7)={+-1;+-7}`

`=> n = {-2;-4;4;-10}`

Vậy: `n = {-2;-4;4;-10}` thì `5n + 22 \vdots n+3`

\(5n+22⋮n+3\\ \Leftrightarrow5n+15+7⋮n+3\\ \Leftrightarrow7⋮n+3\text{ }\left(\text{Vì 5n + 14 ⋮ n + 3}\right)\\ \Leftrightarrow n+3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Leftrightarrow\left[{}\begin{matrix}n+3=1\\n+3=-1\\n+3=7\\n+3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=-2\\n=-4\\n=4\\n=-10\end{matrix}\right.\)

Vậy \(n\in\left\{-2;-4;4;-10\right\}\)