Gọi `x` là số học sinh ở lớp `6A (x > 10) `

Do phần thưởng nhận được chia đều cho mỗi em nên:

\(\left\{{}\begin{matrix}129⋮x\\215⋮x\end{matrix}\right.\)

`=> x` thuộc `ƯC(129;215) `

Mà 

`129 = 3 . 43`

`215 = 43 . 5`

`=> ƯC(129;215) = 43`

Hay `x = 43` (Thỏa mãn)

Vậy lớp `6A` có `43` học sinh

a: \(-\dfrac{2}{3}\cdot x=\dfrac{4}{15}\)

=>\(x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-12}{30}=-\dfrac{2}{5}\)

b: \(-\dfrac{7}{19}\cdot x=\dfrac{-13}{24}\)

=>\(x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{13}{24}\cdot\dfrac{19}{7}=\dfrac{247}{168}\)

\(-\dfrac{2}{3}x=\dfrac{4}{15}\)

<=> \(x=\dfrac{4}{15}:\left(-\dfrac{2}{3}\right)\)

<=> \(x=\dfrac{4}{15}.\left(-\dfrac{3}{2}\right)\)

<=> \(x=-\dfrac{2}{5}\)

\(-\dfrac{7}{19}.x=-\dfrac{13}{24}\)

=> \(x=\left(-\dfrac{13}{24}\right):\left(-\dfrac{7}{19}\right)\)

=> \(x=\dfrac{13}{24}.\dfrac{19}{7}\)

=> \(x=\dfrac{247}{168}\)

bạn ghi thế không hiểu đâu,bạn có thể dùng chữ ích xì ( x )

ok thế bạn giải hộ mình đi

\(\dfrac{8^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{2^{33}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{2^{33}\cdot3^{17}}{3^{60}}=\dfrac{2^{33}}{3^{43}}\)

\(\dfrac{8^{11}.3^{17}}{27^{10}.9^{15}}\\ =\dfrac{\left(2^3\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}\\ =\dfrac{2^{33}.3^{17}}{3^{30}.3^{30}}\\ =\dfrac{2^{33}.3^{17}}{3^{60}}\\ =\dfrac{2^{33}}{3^{43}}\)

\(\left[195-\left(3.x-27\right)\right].39=4212\\ \Rightarrow195-\left(3.x-27\right)=4212:39\\ \Rightarrow195-\left(3.x-27\right)=108\\ \Rightarrow3.x-27=195-108\\ \Rightarrow3.x-27=87\\ \Rightarrow3.x=87+27\\ \Rightarrow3.x=114\\ \Rightarrow x=114:3\\ \Rightarrow x=38\)
Vậy x=38

A và B cùng dấu nên AB>0

=>\(2x^3\cdot\left(-3\right)x^4>0\)

=>\(x^7< 0\)

=>x<0

có nhân 2 với -3 k ạ

\(10⋮x-4\)

=>\(x-4\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

=>\(x\in\left\{5;3;6;2;9;-1;14;-6\right\}\)

Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(2\cdot\left(\widehat{IBC}+\widehat{ICB}\right)=180^0-\widehat{BAC}\)

=>\(\widehat{IBC}+\widehat{ICB}=90^0-\dfrac{1}{2}\cdot\widehat{BAC}\)

Xét ΔBIC có \(\widehat{BIC}+\widehat{IBC}+\widehat{ICB}=180^0\)

=>\(\widehat{BIC}+90^0-\dfrac{1}{2}\widehat{BAC}=180^0\)

=>\(\widehat{BIC}=180^0-90^0+\dfrac{1}{2}\cdot\widehat{BAC}=90^0+\dfrac{1}{2}\cdot\widehat{BAC}\)