(2x+1)(3y-2)=-55

xảy ra các t/hợp

2x(3y−2)+(3y−2)=−55

(2x + 1)(3y - 2) = -55

tới đây xét bảng rồi ra thôi, dễ mà

giúp mk nha mấy bạn (làm ơn nhanh lên) ! xin cảm ơn^-^

Ta có

\(\left|2x-0,\left(24\right)\right|+\left|3y+0,1\left(5\right)\right|=0\)

\(\Rightarrow\left|2x-\frac{24}{99}\right|+\left|3y+0,\left(5\right)-0,4\right|=0\)

\(\Rightarrow\left|2x-\frac{8}{33}\right|+\left|3y+\frac{5}{9}-\frac{4}{5}\right|=0\)

Ta có

\(\begin{cases}\left|2x-\frac{8}{33}\right|\ge0\\\left|3y+\frac{5}{9}-\frac{2}{5}\right|\ge0\end{cases}\)

\(\Rightarrow\begin{cases}2x-\frac{8}{33}=0\\3y+\frac{5}{9}-\frac{2}{5}=0\end{cases}\)

\(\Rightarrow\begin{cases}2x=\frac{8}{33}\\3y=\frac{7}{45}\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{4}{33}\\y=\frac{7}{135}\end{cases}\)

Vậy \(\left(x;y\right)=\left(\frac{4}{45};\frac{7}{135}\right)\)

0,1(55) cũng giống 0,1(5) bạn nhé

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)

\(\Rightarrow b\left(b^2+1\right)-3a^2=\left(a^2+1\right)a-3b^2\)

\(\Rightarrow a^3-b^3+3a^2-3b^2+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)\left(3a+3b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3a+3b+1\right)=0\)

\(\Leftrightarrow a=b\Rightarrow\sqrt{2x+3}=\sqrt{y}\)

\(\Rightarrow y=2x+3\)

\(\Rightarrow M=x\left(2x+3\right)+3\left(2x+3\right)-4x^2-3\) tới đây chắc chỉ cần bấm máy

a) \(x\left(x^2-2x\right)+\left(x-2x\right)=x^2\left(x-2\right)+x\left(x-2\right)=\left(x-2\right)\left(x^2+x\right)⋮x-2\forall x,y\in Z\)

b) \(x^3y^2-3yx^2+xy=xy\left(x^2y-3x+1\right)⋮xy\forall x,y\in Z\)

c) \(x^3y^2-3x^2y^3+xy^2=xy^2\left(x^2-3xy+1\right)⋮\left(x^2-3xy+1\right)\forall x,y\in Z\)