This is olm ,not is Scratch ,do you know it ?And no ,i don' t study Scratch !!!

\(\hept{\begin{cases}xy-x+xy+y=6\\xy+x-y-1=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2xy-x+y=6\\2xy+2x-2y=4\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(2xy+2x-2y\right)-\left(2xy-x+y\right)=4-6\\\left(x-1\right)\left(y+1\right)=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}3x-3y=-2\\\left(x-1\right)\left(y+1\right)=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-\frac{2}{3}+y\\\left(-\frac{2}{3}+y-1\right)\left(y+1\right)=1\end{cases}}\)
Đến đây tự giải nhé bạn

Ta có:

\(\hept{\begin{cases}x\left(y-1\right)+y\left(x+1\right)=6\\\left(x-1\right)\left(y+1\right)=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2xy-x+y=6\\xy+x-y-1=1\end{cases}}\)

DAt \(\hept{\begin{cases}x-y=a\\xy=b\end{cases}}\). Khi do HPT tro thanh:

\(\hept{\begin{cases}2b-a=6\\b+a=2\end{cases}\Leftrightarrow}\hept{\begin{cases}3b=8\\a+b=2\end{cases}\Leftrightarrow}\hept{\begin{cases}b=\frac{8}{3}\\a=-\frac{2}{3}\end{cases}}\)

Theo cách dat: \(\hept{\begin{cases}x+y=-\frac{2}{3}\\xy=\frac{8}{3}\end{cases}\left(PTVN\right)}\)

Ta có: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x,y,z\) 

=> \(2\left(x^2+y^2+z^2\right)-2\left(xy+yz+zx\right)\ge0\) 

=> \(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\) 

=> \(3(x^2+y^2+z^2)\ge\left(x+y+z\right)^2\) 

=> \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\) 

Mà x+y+z=2007 nên \(x^2+y^2+z^2\ge\frac{2007^2}{3}=1342683\) 

Dấu = xảy ra khi \(x=y=z=669\)

DKXD: x\(\ge0\)

a) TA có:

\(\frac{1}{\sqrt{x+3}+\sqrt{x+2}}+\frac{1}{\sqrt{x+2}+\sqrt{x+1}}+\frac{1}{\sqrt{x+1}+\sqrt{x}}=1\)

\(\Leftrightarrow\sqrt{x+3}-\sqrt{x+2}+\sqrt{x+2}-\sqrt{x+1}+\sqrt{x+1}-\sqrt{x}\)=1

\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)

\(\Leftrightarrow\sqrt{x+3}=\sqrt{x}+1\Leftrightarrow x+3=x+2\sqrt{x}+1\)

\(\Leftrightarrow x=1\left(TMDKXD\right)\)

b) Ta thay:

\(\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=3\)

Do do:

\(\sqrt{x^2+3}-x=y+\sqrt{y^2+3}\)

\(\Leftrightarrow\left(y+x\right)+\left(\sqrt{y^2+3}-\sqrt{x^2+3}\right)=0\)

\(\Leftrightarrow\left(x+y\right)\left(1+\frac{y-x}{\sqrt{y^2+3}+\sqrt{x^2+3}}\right)\)=0

Bạn tu giải tiep nhé

bạn thay đoạn đẩu mình ko hiểu lắm nhờ bạn giúp đỡ

Ta có:

\(x^3-8=2x^2-6x+4\)

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

\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-2\left(x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-6x +6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x^2-6x+6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\\orbr{\begin{cases}x=3-\sqrt{3}\\x=3+\sqrt{3}\end{cases}}\end{cases}}\)

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