Yêu cầu đề bài là gì vậy bạn?

thiếu yêu cầu đề bài à bạn

a) \(\dfrac{5x-2}{15}+\dfrac{2x+2}{15}\)

\(=\dfrac{5x-2+2x+2}{15}\)

\(=\dfrac{7x}{15}\)

b) \(\dfrac{2-2x}{6x^3y}+\dfrac{3+2y}{6x^3y}+\dfrac{2x-5}{6x^3y}\)

\(=\dfrac{2-2x+3+2y+2x-5}{6x^3y}\)

\(=\dfrac{2y}{6x^3y}\)

\(=\dfrac{1}{3x^3}\)

c) \(\dfrac{x^2}{x+y}+\dfrac{y^2}{y+z}+\dfrac{-y^2}{x+y}+\dfrac{-z^2}{y+z}\)

\(=\dfrac{x^2-y^2}{x+y}+\dfrac{y^2-z^2}{y+z}\)

\(=\dfrac{\left(x+y\right)\left(x-y\right)}{x+y}+\dfrac{\left(y+z\right)\left(y-z\right)}{y+z}\)

\(=x-y+y-z\)

\(=x-z\)

\(\left|2x-1\right|=16\)

\(\Rightarrow\) Ta có: \(\left\{{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\)

               \(\left\{{}\begin{matrix}2x=16+1\\2x=-16+1\end{matrix}\right.\)

               \(\left\{{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\)

               \(\left\{{}\begin{matrix}x=17:2\\x=-15:2\end{matrix}\right.\)

          \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{17}{2}\\x=\dfrac{-15}{2}\end{matrix}\right.\)

   \(\Rightarrow x\in\left\{{}\begin{matrix}\dfrac{17}{2}\\\dfrac{-15}{2}\end{matrix}\right.\)

|2\(x\) - 1| = 16

\(\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{17}{2}\\x=-\dfrac{15}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-\(\dfrac{15}{2}\); \(\dfrac{17}{2}\)}

a) \(M=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\dfrac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\)

\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}:\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}}.\dfrac{\sqrt{x}+1}{2.\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b) Ta có:

\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)

Để M là số nguyên thì \(2⋮\left(\sqrt{x}-1\right)\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)

Mà \(\sqrt{x}-1\ge-1\)

\(\Rightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{0;2;3\right\}\)

\(\Rightarrow x\in\left\{0;4;9\right\}\)

Vậy \(x\in\left\{0;4;9\right\}\) thì M là số nguyên

phần  2 nha

Lời giải:

Giả sử vòi 1 và vòi 2 chảy 1 mình sau $a,b$ giờ sẽ đầy bể.

Trong 1 giờ, vòi 1 chảy được $\frac{1}{a}$ bể, vòi 2 chảy được $\frac{1}{b}$ bể.

Theo bài ra ta có:

$\frac{1}{a}+\frac{1}{b}=\frac{3}{4}$

$\frac{1}{4a}+\frac{1}{3b}=\frac{5}{4}$
Đến đây bạn chỉ cần giải hệ là được.