a: \(\dfrac{3}{4}xy^2z^3\cdot\dfrac{-2}{7}x^4y^2z=-\dfrac{3}{14}x^5y^4z^4\)

Bậc là 13

b: \(-\dfrac{4}{5}x^6z^3y\cdot\dfrac{-10}{17}y^4z^6=\dfrac{8}{17}x^6y^5z^7\)

Bậc là 18

\(\left(2x-1\right)\left(y^2-5\right)=12\)

\(\left(2x-1\right)\left(y^2-5\right)=12=1.12=12.1=2.6=6.2=3.4=4.3\)

\(\Rightarrow\orbr{\begin{cases}2x-1=1\\y^2-5=12\end{cases}\Rightarrow}\orbr{\begin{cases}2x=2\\y^2=17\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\y=\sqrt{17}\end{cases}}}\)( loại )

\(\Rightarrow\orbr{\begin{cases}2x-1=12\\y^2-5=1\end{cases}\Rightarrow\orbr{\begin{cases}2x=13\\y^2=6\end{cases}\Rightarrow}}\orbr{\begin{cases}x=\frac{13}{2}\\y=\sqrt{6}\end{cases}}\)( loại )

Đến đây bạn tự làm típ nha :))))

Để M nguyên thì \(x-3\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{4;2;5;1\right\}\)

Để M nguyên thì $x-3\in \left\{1;-1;2;-2\right\}$

hay $x\in \left\{4;2;5;1\right\}$

\(\Leftrightarrow\left(x-1\right)\left(y-3\right)=7\)

\(\Leftrightarrow\left(x-1;y-3\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(4;10\right);\left(8;4\right);\left(0;-4\right);\left(-6;2\right)\right\}\)

\(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-1,3-y\in Z\\x-1,3-y\inƯ\left(-7\right)=\left\{\pm1;\pm7\right\}\end{matrix}\right.\)

Ta có bảng:

Vậy \(\left(x,y\right)\in\left\{\left(0;-4\right);\left(2;10\right);\left(-6;2\right);\left(8;4\right)\right\}\)

\(M=\dfrac{3x\left(x-3\right)+2}{x-3}\)

\(\Rightarrow M=\dfrac{3x\left(x-3\right)}{x-3}+\dfrac{2}{x-3}\)

\(\Rightarrow M=3x+\dfrac{2}{x-3}\)

Để M nguyên thì \(\dfrac{2}{x-3}\in Z\Rightarrow x-3\inƯ\left(2\right)=\left\{\pm1;\pm3\right\}\)

Ta có bảng:

Vậy \(x\in\left\{2;0;4;6\right\}\)

Để M nguyên thì \(x-3\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{4;2;5;1\right\}\)

= 11/22 + 3/22 = 14/22

=\(\frac{7}{11}\)

k mình nha

chúc các bạn học giỏi

=7/11

A=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)

A=\(\frac{1.2.3.4...2015}{2.3.4...2016}=\frac{1}{2016}\)

Hok tốt

A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2015}\right).\left(1-\frac{1}{2016}\right)\)

= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)

= \(\frac{1}{2016}\)

Vậy ...

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2015}\right)\left(1-\frac{1}{2016}\right)\)

\(\Rightarrow A=\frac{1.2.3.4....2015}{2.3.4....2016}\)

\(\Rightarrow A=\frac{1}{2016}\)

#hok tốt