\(\left(2x+3\right)^3=\left(-25\right).\left(-5\right)\)

\(\Leftrightarrow\left(2x+3\right)^3=125\)

\(\Leftrightarrow\left(2x+3\right)^3=5^3\)

\(\Leftrightarrow2x+3=5\)

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

\(\left(2x+3\right)^3=\left(-25\right).\left(-5\right)\)

\(\Leftrightarrow\left(2x+3\right)^3=5^3\)

\(\Leftrightarrow2x+3=5\)

\(\Leftrightarrow2x=2\)

\(\Leftrightarrow x=1\)

(2x+3)³ = 125

(2x+3)³ = 5³

=> 2x + 3 = 5

2x= 5+3

2x=8

x=4

\(\Leftrightarrow x\left(y+1\right)+y+1=31\\ \Leftrightarrow\left(x+1\right)\left(y+1\right)=31=31\cdot1\)

\(TH_1:\left\{{}\begin{matrix}x+1=31\\y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=0\end{matrix}\right.\rightarrow\left(30;0\right)\\ TH_2:\left\{{}\begin{matrix}x+1=1\\y+1=31\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=30\end{matrix}\right.\rightarrow\left(0;30\right)\)

Vậy \(\left(x;y\right)\in\left\{\left(30;0\right);\left(0;30\right)\right\}\)

\(\Leftrightarrow xy+x+y+1=31\Leftrightarrow\left(x+1\right)\left(y+1\right)=31\)

Do x, y nguyên dương nên xảy ra các TH sau:

TH1: \(\left\{{}\begin{matrix}x+1=1\\y+1=31\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=30\end{matrix}\right.\)(LOẠI)

TH2:\(\left\{{}\begin{matrix}x+1=31\\y+1=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=0\end{matrix}\right.\)(LOẠI)

Vậy không có x, y nguyên dương thoả mãn đề bài.

b: T=(-1)+(-1)+...+(-1)

=-1011

a)x=-2

b) y=10

c)x=-5

d) x=-12

\(a,\dfrac{x}{3}=\dfrac{6}{-9}\\ \Rightarrow x=-\dfrac{2}{3}.3\\ \Rightarrow x=-2\\ b,\dfrac{4}{y}=\dfrac{-2}{-5}\\ \Rightarrow y=4:\dfrac{2}{5}\\ \Rightarrow y=10\\ c,\dfrac{-2}{3}=\dfrac{x-1}{6}\\ \Rightarrow3x-3=-12\\ \Rightarrow3x=-9\\ \Rightarrow x=-3\\ d,\dfrac{3}{x}=\dfrac{6}{-24}\\ \Rightarrow x=3:-\dfrac{1}{4}\\ \Rightarrow x=-12\)

a)x=-2

b) y=10

c)x=-5

d) x=-12 .

\(a,\dfrac{3}{4}=\dfrac{15}{20}\\ \dfrac{7}{10}=\dfrac{14}{20}\\ b,\dfrac{-5}{14}=\dfrac{-55}{154}\\ \dfrac{9}{22}=\dfrac{63}{154}\\ \dfrac{-4}{7}=\dfrac{-36}{63}\\ \dfrac{8}{9}=\dfrac{56}{63}\\ \dfrac{-10}{21}=\dfrac{-30}{63}\)