Ta có x - y = 1 => x = y + 1 

\(\dfrac{x+2}{9}=\dfrac{1}{y+2}\Rightarrow\left(x+2\right)\left(y+2\right)=9\)

\(\Leftrightarrow\left(3+y\right)\left(y+2\right)=9\Leftrightarrow y^2+5y-3=0\Leftrightarrow y=\dfrac{-5\pm\sqrt{37}}{2}\)

thay vào tìm x 

ps nhưng số xấu quá bạn ạ, kiểm tra lại đề nhé 

ĐKXĐ:\(y\ne-2\)

\(\left\{{}\begin{matrix}\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y+2}\\x-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y+1-1}{9}+\dfrac{3}{9}=\dfrac{1}{y+2}\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y+3}{9}=\dfrac{1}{y+2}\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(y+3\right)\left(y+2\right)=9\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y^2+5y+6-9=0\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y^2+5y-3=0\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=\dfrac{-5+\sqrt{37}}{2}\\y=\dfrac{-5-\sqrt{37}}{2}\end{matrix}\right.\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{-3+\sqrt{37}}{2}\\y=\dfrac{-5+\sqrt{37}}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{-3-\sqrt{37}}{2}\\y=\dfrac{-5-\sqrt{37}}{2}\end{matrix}\right.\end{matrix}\right.\)

Ta có:

\(x-y=1\Rightarrow x=1+y\)

Thay vào 

\(\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y}+2\) \(\left(đk:y\ne0\right)\)

\(\dfrac{x+2}{9}=\dfrac{2y+1}{y}\)

\(\Leftrightarrow\dfrac{y+3}{9}=\dfrac{2y+1}{y}\)

\(\Leftrightarrow y^2+3y=18y+9\)

\(\Leftrightarrow y^2-15y-9=0\)

\(\Leftrightarrow\)\(\left(y-\dfrac{15}{2}\right)^2=\dfrac{261}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}y-\dfrac{15}{2}=\dfrac{\sqrt{261}}{2}\\y-\dfrac{15}{2}=-\dfrac{\sqrt{261}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{\sqrt{261}+15}{2}\\y=\dfrac{15-\sqrt{261}}{2}\end{matrix}\right.\)

\(x\times5=100\)

\(x=20\)

x.(2+3)=100

x.5=100

x=20

x X (2+3)=100

x X 5=100

x=20

\(\Rightarrow x^3+3x^2+3x+1=0\\ \Rightarrow\left(x+1\right)^3=0\Rightarrow x+1=0\Rightarrow x=-1\)

phân tích các số sau ra thừa số nguyên tố 48;56;84;105;360

Bài 1:

a) \(A=-\left(2x-5\right)^2+6\left|2x-5\right|+4=-\left[\left(2x-5\right)^2-6\left|2x-5\right|+9\right]+13=-\left(\left|2x-5\right|-3\right)^2+13\le13\)

\(maxA=13\Leftrightarrow\) \(\left[{}\begin{matrix}2x-5=3\\2x-5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)

b) \(B=-x^2-y^2+2x-6y+9=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+19=-\left(x-1\right)^2-\left(y+3\right)^2+19\le19\)

\(maxC=19\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

Bài 2:

\(A=2\left(x^3-y^3\right)-3\left(x+y\right)^2=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=4\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

bài 2
\(A=2\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=2.2\left(x^2+xy+y^2\right)-3\left(x^2+2xy+y^2\right)\)
\(A=\left(4x^2+4xy+4y^2\right)+\left(-3x^2-6xy-3y^2\right)\)
\(A=x^2-2xy+y^2=\left(x-y\right)^2=2^2=4\)

+) \(P=\sqrt{7x+9}+\sqrt{7y+9}+\sqrt{7z+9}\)

\(P^2\le3\left(7x+7y+7z+27\right)=102\)
\(P\le\sqrt{102}\)

\(MaxP=102\Leftrightarrow x=y=z=\dfrac{1}{3}\)

+) \(x,y,z\in[0;1]\)\(\Rightarrow\left\{{}\begin{matrix}x\ge x^2\\y\ge y^2\\z\ge z^2\end{matrix}\right.\)

\(P\ge\sqrt{x^2+6x+9}+\sqrt{y^2+6y+9}+\sqrt{z^2+6z+9}\)

\(=x+y+z+9=10\)

\(MinP=10\Leftrightarrow\left(x;y;z\right)=\left(0;0;1\right)\text{và các hoán vị}\)

\(\left(x-2\right)\left(x^2+2x+4\right)+3x-4=\left(x+2\right)\left(x^2-2x+4\right)-x+1\)

\(\Rightarrow\left(x^3-8\right)+3x-4=\left(x^3+8\right)-x+1\)

\(\Rightarrow x^3-8+3x-4=x^3+8-x+1\)

\(\Rightarrow x^3-x^3+3x+x=8+8+4+1\)

\(\Rightarrow4x=21\)

\(\Rightarrow x=\dfrac{21}{5}\)

\(\frac{x}{8}=\frac{-2}{5}\cdot\frac{3}{16}\)

\(\frac{x}{8}=\frac{-3}{40}\)

\(\Rightarrow x=\frac{8.\left(-3\right)}{40}=\frac{-3}{5}\)

\(\frac{x+4}{5}=\frac{-1}{7}\cdot\frac{14}{15}\)

\(\frac{x+4}{5}=\frac{-2}{15}\)

\(\Rightarrow x+4=\frac{5.\left(-2\right)}{15}=\frac{-2}{3}\)

\(\Rightarrow x=\frac{-14}{3}\)

x=2/3

\(3x-2x=x+1\)

\(-2x+3=x+1\)

\(-2x+3-3=x+1-3\)

\(-2x+3-3=x+1-3\)

\(-2x=x+1-3\)

\(-2x=x-2\)

\(-2x-x=x-2-x\)

\(-3x=x-2-x\)

\(-3x=-2\)

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