x = 6;7

6;7 k nhé bạn

x = 6,7 nha bn

..........

k mk nhé

pải làm đủ nka

ko đủ mk tét đấy

\(x^2+4x+y^2-2xy+x^2+4=0\)

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

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

vì \(\left(x-y\right)^2\ge0;\left(x+2\right)^2\ge0\)nên

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-y=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\y=-2\end{cases}\Rightarrow}x=y=-2}\)

x=25 y=9 k cho mình nha bạn 

5/x = 1/5 = y/45

x=25;y=9

Ta có:

\(\frac{5}{x}=\frac{1}{5}\)\(\Leftrightarrow x=5\times5:1=25\)

\(\frac{y}{45}=\frac{1}{5}\)\(\Leftrightarrow y=45\times1:5=9\)

\(\frac{3x}{2\cdot5}+\frac{3x}{5\cdot8}+\frac{3x}{8\cdot11}+\frac{3x}{11\cdot14}=\frac{1}{21}\)

\(=>\frac{3x}{3}\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)

\(=>x\left[\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}\right]=\frac{1}{21}\)

\(=>x\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{11}-\frac{1}{14}\right]=\frac{1}{21}\)

\(=>x\left[\frac{1}{2}-\frac{1}{14}\right]=\frac{1}{21}\)

\(=>x\cdot\frac{3}{7}=\frac{1}{21}\Leftrightarrow x=\frac{1}{9}\)

\(PTK_{hc}=1,9375\times PTK_S=1,9375\times32=62\text{đ}vC\)

\(2\times NTK_X+1\times NTK_O=62\text{đ}vC\)

\(2\times NTK_X+16=62\)

\(2\times NTK_X=62-16\)

\(2\times NTK_X=46\)

\(NTK_X=\frac{46}{2}\)

\(NTK_X=23\text{đ}vC\)

=> Na

PTK_{hợp chất}=1,9375 * 32 =   62  (đvC)

PTKoxi = 16 (đvC) => PTK_2x = 62 - 16 = 46 (đvC)

=> PTK_x = 46 : 2 = 23 (đvC) => NTK = 23 (đvC)

=> X là nguyên tố NATRI

\(\left(x+\frac{2012}{2013}\right)^6=0\)

=> \(\left(x+\frac{2012}{2013}\right)^6=0^6\)

=> \(x+\frac{2012}{2013}=0\)

=> \(x=\frac{-2012}{2013}\)

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