Bài 4: Giải hệ phương trình bằng phương pháp cộng đại số

\(=6\sqrt{13}+\dfrac{1}{5}\cdot5\sqrt{13}-4\sqrt{13}=3\sqrt{13}\)

\(3\sqrt{52}+\dfrac{1}{5}\sqrt{325}-\sqrt{208}\)

\(\Leftrightarrow3.2\sqrt{13}+\dfrac{1}{5}.5\sqrt{13}-4\sqrt{13}\)

\(\Leftrightarrow\) \(6\sqrt{13}+\sqrt{13}-4\sqrt{13}\)

\(\Leftrightarrow3\sqrt{13}\)

Trl: = 6 √ 13 + 15 ⋅ 5 √ 13 − 4 √ 13 = 3 √ 13

Hiện nay mạng điện trong nhà có lắp bảng điện, ổ cắm điện thấp hơn 1,3-1,5m. Vì sao?

Đặt 1/y-1=b

\(\Leftrightarrow\left\{{}\begin{matrix}2x+3b=5\\4x-3b=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\b=\dfrac{7}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{16}{7}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15}{x+1}-10y=-5\\\dfrac{15}{x+1}+9y=33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8x-2\left|y+2\right|=6\\x+2\left|y+2\right|=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y\in\left\{1;-5\right\}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+\sqrt{x+1}=4\\2\left(x+y\right)-6\sqrt{x+1}=-10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x+y=1\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(3;-2\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+2\sqrt{y-1}=5\\8\sqrt{x}-2\sqrt{y-1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x-3}{x-1}+\dfrac{3}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{2x-2}{x-1}+\dfrac{2}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x-1}-\dfrac{2}{y+2}=1\\\dfrac{4}{x-1}+\dfrac{2}{y+2}=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=1\\y+2=1\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(2;-1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x+y}+\dfrac{2}{y-1}=10\\\dfrac{1}{x+y}-\dfrac{2}{y-1}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\y-1=1\end{matrix}\right.\Leftrightarrow\left(x,y\right)=\left(-1;2\right)\)

\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{1}{y}=2.\\\dfrac{6}{x}-\dfrac{2}{y}=1.\end{matrix}\right.\left(ĐK:x\ne0;y\ne0\right).\)

Đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b\left(a\ne0;b\ne0\right).\)

\(\Rightarrow\left\{{}\begin{matrix}2a+b=2.\\6a-2b=1.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4a+2b=4.\\6a-2b=1.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2a+b=2.\\10a=5.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=1.\\a=\dfrac{1}{2}.\end{matrix}\right.\left(TM\right).\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=1.\\\dfrac{1}{x}=\dfrac{1}{2}.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=1.\\x=2.\end{matrix}\right.\) \(\left(TM\right).\)

Vậy hệ phương trình có nghiệm duy nhất \(\left(x;y\right)=\left(2;1\right).\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}3x+3+2x+4y=4\\4x+4-x-2y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+4y=1\\3x-2y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+4y=1\\6x-4y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)