Đề thiếu phải ko?

bài mô ri nhật

cái kia phải là 1/(y+2) chứ nhỉ :v

\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}+\dfrac{2}{y+2}=4\\\dfrac{2x}{x-1}+\dfrac{1}{x+2}=5\end{matrix}\right.\)(ĐKXĐ: \(x\ne1;y\ne-2\))\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x\left(y+2\right)}{\left(x-1\right)\left(y+2\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(y+2\right)}=\dfrac{4\left(x-1\right)\left(y+2\right)}{\left(x-1\right)\left(y+2\right)}\\\dfrac{2x\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\dfrac{\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=\dfrac{5\left(x-1\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3xy+6x+2x-1=4xy+8x-4y-8\\2x^2+4x+x-1=5x^2+10x-5x-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4y-xy+7=0\\-3x^2+9=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}4y-xy+7=0\\x=\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4y-\sqrt{3}y+7=0\Leftrightarrow y\left(4-\sqrt{3}\right)y=-7\\x=\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{4-\sqrt{3}}\\x=\sqrt{3}\end{matrix}\right.\)

b) \(x^2+2\sqrt{3}x-6=0\)

\(\Leftrightarrow\) \(x^2+2\sqrt{3}x+3-9=0\)

\(\Leftrightarrow\) \(\left(x+\sqrt{3}\right)^2-9=0\)

\(\Leftrightarrow\) \(\left(x+\sqrt{3}-3\right).\left(x+\sqrt{3}+3\right)=0\)

\(\Leftrightarrow\) \(\left[\begin{array}{} x+\sqrt{3}-3=0 \\ x+\sqrt{3}+3=0 \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} x= 3-\sqrt{3} \\ x= -3-\sqrt{3} \end{array} \right.\)

Vậy phương trình có tập nghiệm là S={\(3-\sqrt{3};-3-\sqrt{3}\)}

\(\left\{{}\begin{matrix}\dfrac{8}{x-1}+\dfrac{15}{y+2}=1\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{8}{x-1}+\dfrac{15}{y+2}=1\\\dfrac{8}{x-1}+\dfrac{8}{y+2}=\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y+2}=\dfrac{1}{3}\\\dfrac{1}{x-1}+\dfrac{1}{y+2}=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=21\\\dfrac{1}{x-1}=\dfrac{1}{12}-\dfrac{1}{21}=\dfrac{1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=19\\x=29\end{matrix}\right.\)

x,y khác 0 đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}81a+105b=8\\54a+42b=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}9.9.2a+105.2.b=8.2\\9.6.3a+42.3b=4.3\end{matrix}\right.\)

\(\Leftrightarrow\left(105.2-42.3\right)b=8.2-4.3=4\left(4-3\right)=4\)

\(\Leftrightarrow\left(105-21.3\right)b=2\)

\(\Leftrightarrow3\left(35-21\right)b=2\Rightarrow b=\dfrac{2}{3.14}=\dfrac{1}{3.7}=\dfrac{1}{21}\)

\(54a+42.\dfrac{1}{21}=4\Leftrightarrow54a+2=4\)

\(\left\{{}\begin{matrix}b=\dfrac{1}{21}\\a=\dfrac{1}{27}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=27\\y=21\end{matrix}\right.\)