Câu 12:

a:n tia chung gốc sẽ có \(\dfrac{n\left(n-1\right)}{2}\)(góc)

Theo đề, ta có: \(\dfrac{n\left(n-1\right)}{2}=28\)

=>\(n^2-n=56\)

=>\(n^2-n-56=0\)

=>(n-8)(n+7)=0

=>\(\left[{}\begin{matrix}n=8\left(nhận\right)\\n=-7\left(loại\right)\end{matrix}\right.\)

b: Theo đề, ta có: \(\dfrac{n\left(n-1\right)}{2}=190\)

=>n(n-1)=380

=>\(n^2-n-380=0\)

=>(n-20)(n+19)=0

=>\(\left[{}\begin{matrix}n=20\left(nhận\right)\\n=-19\left(loại\right)\end{matrix}\right.\)

Câu 10:

a: \(\widehat{ABC}< \widehat{BAC}< \widehat{ACB}\)

b: \(\widehat{ABC}+\widehat{BAC}+\widehat{ACB}=180^0\)

Câu 11:

a: Chúng tạo ra 13 góc

Có 3 góc bẹt

b:n đường cắt nhau sẽ có 2n tia Số góc tạo thành là \(\dfrac{2n\left(2n-1\right)}{2}=n\left(2n-1\right)\left(góc\right)\)

Câu 9:

Trên hình vẽ có 6 góc: \(\widehat{AOB};\widehat{AOC};\widehat{AOD};\widehat{BOC};\widehat{BOD};\widehat{COD}\)

Câu 9:

Trên hình vẽ có 6 góc: \(\widehat{AOB};\widehat{AOC};\widehat{AOD};\widehat{BOC};\widehat{BOD};\widehat{COD}\)

Câu 11:

xy+2y-8=4x+35

=>xy+2y-4x=43

=>y(x+2)-4x-8=35

=>(x+2)(y-4)=35

=>\(\left(x+2\right)\left(y-4\right)=1\cdot35=35\cdot1=\left(-1\right)\cdot\left(-35\right)=\left(-35\right)\cdot\left(-1\right)=5\cdot7=7\cdot5=\left(-5\right)\cdot\left(-7\right)=\left(-7\right)\cdot\left(-5\right)\)

=>\(\left(x+2;y-4\right)\in\left\{\left(1;35\right);\left(35;1\right);\left(-1;-35\right);\left(-35;-1\right);\left(5;7\right);\left(7;5\right);\left(-5;-7\right);\left(-7;-5\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-1;39\right);\left(33;5\right);\left(-3;-31\right);\left(-37;3\right);\left(3;11\right);\left(5;9\right);\left(-7;-3\right);\left(-9;-1\right)\right\}\)

\(10A=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)

\(10B=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)

Ta có: \(10^{2022}+1< 10^{2023}+1\)

=>\(\dfrac{9}{10^{2022}+1}>\dfrac{9}{10^{2023}+1}\)

=>\(1+\dfrac{9}{10^{2022}+1}>1+\dfrac{9}{10^{2023}+1}\)

=>10A>10B

=>A>B

ĐKXĐ: n<>-1/2

Để B là số nguyên thì \(6n-8⋮2n+1\)

=>\(6n+3-11⋮2n+1\)

=>\(-11⋮2n+1\)

=>\(2n+1\in\left\{1;-1;11;-11\right\}\)

=>\(n\in\left\{0;-1;5;-6\right\}\)

ĐKXĐ: n<>7

Để A là số nguyên thì \(4n+1⋮n-7\)

=>\(4n-28+29⋮n-7\)

=>\(29⋮n-7\)

=>\(n-7\in\left\{1;-1;29;-29\right\}\)

=>\(n\in\left\{8;6;36;-22\right\}\)

?????????

Chọn B