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Chúc Nguyễn Thúy Diệp  năm mới vui vẻ, vạn sự như ý

Ta có:

\(9xy+3x+3y=51 \)

\(\Leftrightarrow9xy+3x+3y+1=52 \)

\(\Leftrightarrow3x(3y+1)+(3y+1)=52 \)

\(\Leftrightarrow\)\((3y+1)(3x+1)=52\)

Do \(x,y\in N^{\text{*}}\) nên \(3x+1\) , \(3y+1\) là các stn lớn hơn \(1\) và chia cho \(3\)cũng dư \(1\).

Mặt khác: \(52=4.13\)

- TH1:

\(\left\{{}\begin{matrix}3x+1=4\\3y+1=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=4\end{matrix}\right.\)

- TH2:

\(\left\{{}\begin{matrix}3x+1=13\\3y+1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)

tick dùm mình nhá

Đăng ít thôi.

\(\dfrac{x^3+y^3+z^3-3xyz}{\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2}\)

\(=\dfrac{\left(x+y\right)^3+z^3-\left(3x^2y+3xy^2+3xyz\right)}{x^2-2xy+y^2+z^2-2xz+z^2+y^2-2yz+z^2}\)

\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz-z^2\right)-3xy\left(x+y+z\right)}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)

\(=\dfrac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)}{2\left(x^2+y^2+z^2-xy-xz-yz\right)}\)

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

Giải

Độ dài đoạn CD là:

9-(1+4)= 4 (cm)

Đ/s: 4cm

1.

a) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)

\(=\dfrac{3x+6-x}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)

b) \(x^2+1-\dfrac{x^4-3x^2+2}{x^2-1}\)

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

\(=\dfrac{x^4-1-x^4+3x^2-2}{x^2-1}\)

\(=\dfrac{-3+3x^2}{x^2-1}\)