a/ Bạn tự giải

b/ ĐKXĐ:...

Cộng vế với vế: \(\frac{x-y}{y+12}=3\Rightarrow x-y=3y+36\Rightarrow x=4y+36\)

Thay vào pt đầu: \(\frac{4y+36}{y}-\frac{y}{y+12}=1\)
Đặt \(\frac{y+12}{y}=a\Rightarrow4a-\frac{1}{a}=1\Rightarrow4a^2-a-1=0\)

\(\Rightarrow a=\frac{1\pm\sqrt{17}}{8}\) \(\Rightarrow\frac{y+12}{y}=\frac{1\pm\sqrt{17}}{8}\)

\(\Rightarrow\left[{}\begin{matrix}y+12=y\left(\frac{1+\sqrt{17}}{8}\right)\\y+12=y\left(\frac{1-\sqrt{17}}{8}\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left(\frac{-7+\sqrt{17}}{8}\right)y=12\\\left(\frac{-7-\sqrt{17}}{8}\right)y=12\end{matrix}\right.\) \(\Rightarrow y=...\)

Chắc bạn ghi sai đề, nghiệm quá xấu

3/ \(\Leftrightarrow\left\{{}\begin{matrix}3x^2+y^2=5\\3x^2-9y=3\end{matrix}\right.\) \(\Rightarrow y^2+9y=2\Rightarrow y^2+9y-2=0\Rightarrow y=...\)

4/ ĐKXĐ:...

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

\(\Rightarrow5\sqrt{3x-1}=15\Rightarrow\sqrt{3x-1}=3\Rightarrow x=\frac{10}{3}\)

\(\sqrt{2y+1}=\sqrt{3x-1}-1=3-1=2\Rightarrow2y+1=4\Rightarrow y=\frac{3}{2}\)

Bài 2:

ĐK: ..........
Đặt $\sqrt{x+\frac{1}{y}}=a; \sqrt{x+y-3}=b$ $(a,b\geq 0$)

HPT \(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2+3=8\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2=5\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ (a+b)^2-2ab=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ ab=2\end{matrix}\right.\)

Áp dụng định lý Vi-et đảo thì $a,b$ là nghiệm của pt $X^2-3X+2=0$

$\Rightarrow (a,b)=(2,1); (1,2)$

Nếu $(a,b)=(2,1)$

\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y-3=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y=4\end{matrix}\right.\Rightarrow y=\frac{1}{y}\Rightarrow y=\pm 1\)

$y=1\rightarrow x=3$

$y=-1\rightarrow y=5$

Nếu $(a,b)=(1,2)$

\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y-3=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y=7\end{matrix}\right.\Rightarrow y-\frac{1}{y}=6\)

\(\Rightarrow y^2-6y-1=0\Rightarrow y=3\pm \sqrt{10}\)

Nếu $y=3+\sqrt{10}\rightarrow x=4-\sqrt{10}$

Nếu $y=3-\sqrt{10}\rightarrow x=4+\sqrt{10}$

Vậy...........

Bài 1:

Đặt $\sqrt[4]{y^3-1}=a; \sqrt{x}=b$ $(a,b\geq 0$)

Khi đó hệ PT trở thành:

\(\left\{\begin{matrix} a+b=3\\ b^4+a^4+1=82\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^4+b^4=81\end{matrix}\right.\)

Có: \(a^4+b^4=81\)

\(\Leftrightarrow (a^2+b^2)^2-2a^2b^2=81\)

\(\Leftrightarrow [(a+b)^2-2ab]^2-2a^2b^2=81\)

\(\Leftrightarrow (9-2ab)^2-2a^2b^2=81\)

\(\Leftrightarrow 2a^2b^2-36ab=0\)

\(\Leftrightarrow ab(ab-18)=0\Rightarrow \left[\begin{matrix} ab=0\\ ab=18\end{matrix}\right.\)

Nếu $ab=0$. Kết hợp với $a+b=3$ suy ra $(a,b)=(3,0); (0,3)$

$\Rightarrow (x,y)=(0, \sqrt[4]{82}); (9, 1)$

Nếu $ab=18$. Kết hợp với $a+b=3$ và định lý Vi-et đảo suy ra $a,b$ là nghiệm của pt: $X^2-3X+18=0$

Dễ thấy pt này vô nghiệm nên loại

Vậy......

Bài 2:

ĐK: ..........
Đặt $\sqrt{x+\frac{1}{y}}=a; \sqrt{x+y-3}=b$ $(a,b\geq 0$)

HPT \(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2+3=8\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ a^2+b^2=5\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} a+b=3\\ (a+b)^2-2ab=5\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a+b=3\\ ab=2\end{matrix}\right.\)

Áp dụng định lý Vi-et đảo thì $a,b$ là nghiệm của pt $X^2-3X+2=0$

$\Rightarrow (a,b)=(2,1); (1,2)$

Nếu $(a,b)=(2,1)$

\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y-3=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=4\\ x+y=4\end{matrix}\right.\Rightarrow y=\frac{1}{y}\Rightarrow y=\pm 1\)

$y=1\rightarrow x=3$

$y=-1\rightarrow y=5$

Nếu $(a,b)=(1,2)$

\(\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y-3=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x+\frac{1}{y}=1\\ x+y=7\end{matrix}\right.\Rightarrow y-\frac{1}{y}=6\)

\(\Rightarrow y^2-6y-1=0\Rightarrow y=3\pm \sqrt{10}\)

Nếu $y=3+\sqrt{10}\rightarrow x=4-\sqrt{10}$

Nếu $y=3-\sqrt{10}\rightarrow x=4+\sqrt{10}$

Vậy...........

\(2,\left\{{}\begin{matrix}x^3-2x^2y-15x=6y\left(2x-5-4y\right)\left(1\right)\\\frac{x^2}{8y}+\frac{2x}{3}=\sqrt{\frac{x^3}{3y}+\frac{x^2}{4}}-\frac{y}{2}\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(2y-x\right)\left(x^2-12y-15\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2y=x\\y=\frac{x^2-15}{12}\end{matrix}\right.\)

Ta xét các trường hợp sau:

Trường hợp 1:

\(y=\frac{x^2-15}{12}\) thay vào phương trình \(\left(2\right)\) ta được:

\(\frac{3x^2}{2\left(x^2-15\right)}+\frac{2x}{3}=\sqrt{\frac{4x^3}{x^2-15}+\frac{x^2}{4}}-\frac{x^2-15}{24}\)

\(\Leftrightarrow\frac{36x^2}{x^2-15}-12\sqrt{\frac{x^2}{x^2-15}\left(x^2+16x-15\right)}+\left(x^2+16x-15\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\6\sqrt{\frac{x^2}{x^2-15}}=\sqrt{\left(x^2+16x-15\right)}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36\frac{x^2}{x^2-15}=x^2+16x-15\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+16x-15\ge0\\36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\left(3\right)\end{matrix}\right.\)

Ta xét phương trình \(\left(3\right):36x^2=\left(x^2-15\right)\left(x^2+16x-15\right)\)

Vì: \(x=0\) Không phải là nghiệm. Ta chia cả hai vế p.trình cho \(x^2\) ta được:

\(36=\left(x-\frac{15}{x}\right)\left(x+16-\frac{15}{x}\right)\)

Đặt: \(x-\frac{15}{x}=t\Rightarrow t^2+16t-36=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-18\end{matrix}\right.\)

+ Nếu như:

\(t=2\Leftrightarrow x-\frac{15}{x}=2\Leftrightarrow x^2-2x-15=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)\(\Leftrightarrow x=5\)

+ Nếu như:

\(t=-18\Leftrightarrow x-\frac{15}{x}=-18\Leftrightarrow x^2+18x-15=0\Leftrightarrow\left[{}\begin{matrix}x=-9-4\sqrt{6}\\x=-9+4\sqrt{6}\end{matrix}\right.\Leftrightarrow x=-9-4\sqrt{6}\)

Trường hợp 2:

\(x=2y\) thay vào p.trình \(\left(2\right)\) ta được:

\(\Leftrightarrow\frac{x^2}{4x}+\frac{2x}{3}=\sqrt{\frac{2x^3}{3x}+\frac{x^2}{4}}-\frac{x}{4}\Leftrightarrow\frac{7}{6}x=\sqrt{\frac{11x^2}{12}}\Leftrightarrow x=0\left(ktmđk\right)\)

Vậy nghiệm của hệ đã cho là: \(\left(x,y\right)=\left(5;\frac{5}{6}\right),\left(-9-4\sqrt{6};\frac{27+12\sqrt{6}}{2}\right)\)

Năm mới chắc bị lag @@ tớ sửa luôn đề câu 3 nhé :v

3, \(\left\{{}\begin{matrix}8\left(x^2+y^2\right)+4xy+\frac{5}{\left(x+y\right)^2}=13\left(1\right)\\2xy+\frac{1}{x+y}=1\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow8\left[\left(x+y\right)^2-2xy\right]+4xy+\frac{5}{\left(x+y\right)^2}=13\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow8\left(a^2-2b\right)+4b+\frac{5}{a^2}=13\)

\(\Leftrightarrow8a^2-12b+\frac{5}{a^2}=13\)

Ta cũng có \(\left(2\right)\Leftrightarrow2b+\frac{1}{a}=1\)

\(\Leftrightarrow2b=1-\frac{1}{a}\)

Thay vào (1) ta được :

\(8a^2+\frac{5}{a^2}-6\cdot\left(1-\frac{1}{a}\right)=13\)

\(\Leftrightarrow8a^2+\frac{5}{a^2}-6+\frac{6}{a}=13\)

\(\Leftrightarrow8a^2+\frac{5}{a^2}+\frac{6}{a}=19\)

Giải pt được \(a=1\)

Khi đó \(b=\frac{1-\frac{1}{1}}{2}=0\)

Ta có hệ :

\(\left\{{}\begin{matrix}x+y=1\\xy=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\end{matrix}\right.\)

Vậy...

a/ ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\frac{1}{x-1}=u\\\frac{1}{2y+1}=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2u+v=\frac{6}{5}\\3u-2v=\frac{11}{10}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=\frac{1}{2}\\v=\frac{1}{5}\end{matrix}\right.\)

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

b/ ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}x+y=u\\\sqrt{x+1}=v\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2u+v=4\\u-3v=-5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=1\\v=2\end{matrix}\right.\)

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

\(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)

ĐK: \(x>4,y\ne-2\)

Đặt a=\(\frac{1}{\sqrt{x-4}}\left(a>0\right)\),\(b=\frac{1}{y+2}\)

Vậy \(\left\{{}\begin{matrix}\frac{3}{\sqrt{x-4}}+\frac{4}{y+2}=7\\\frac{5}{\sqrt{x-4}}-\frac{1}{y-2}=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3a+4b=7\\5a-b=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}3a+4b=7\\20a-4b=16\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}23a=23\\5a-b=4\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)(tm)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\frac{1}{\sqrt{x-4}}=1\\\frac{1}{y+2}=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x-4}=1\\y+2=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=5\\y=-1\end{matrix}\right.\)(tm)

Vậy (x;y)=(5;-1)