Câu 2/

Điều kiện xác định b tự làm nhé:

\(\frac{6}{x^2-9}+\frac{4}{x^2-11}-\frac{7}{x^2-8}-\frac{3}{x^2-12}=0\)

\(\Leftrightarrow x^4-25x^2+150=0\)

\(\Leftrightarrow\left(x^2-10\right)\left(x^2-15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=10\\x^2=15\end{cases}}\)

Tới đây b làm tiếp nhé.

a. ĐK: \(\frac{2x-1}{y+2}\ge0\)

Áp dụng bđt Cô-si ta có: \(\sqrt{\frac{y+2}{2x-1}}+\sqrt{\frac{2x-1}{y+2}}\ge2\)

\(\)Dấu bằng xảy ra khi  \(\frac{y+2}{2x-1}=1\Rightarrow y+2=2x-1\Rightarrow y=2x-3\) 

Kết hợp với pt (1) ta tìm được x = -1, y = -5 (tmđk)

b. \(pt\Leftrightarrow\left(\frac{6}{x^2-9}-1\right)+\left(\frac{4}{x^2-11}-1\right)-\left(\frac{7}{x^2-8}-1\right)-\left(\frac{3}{x^2-12}-1\right)=0\)

\(\Leftrightarrow\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}+\frac{1}{x^2-8}+\frac{1}{x^2-12}\right)=0\)

\(\Leftrightarrow x^2-15=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{15}\\x=-\sqrt{15}\end{cases}}\)

Bài 1:
$x^2y+4y=x+6$

$\Leftrightarrow y(x^2+4)=x+6$

$\Leftrightarrow y=\frac{x+6}{x^2+4}$

Để $y$ nguyên thì $\frac{x+6}{x^2+4}$ nguyên

$\Rightarrow x+6\vdots x^2+4(1)$

$\Rightarrow x^2+6x\vdots x^2+4$

$\Rightarrow (x^2+4)+(6x-4)\vdots x^2+4$

$\RIghtarrow 6x-4\vdots x^2+4(2)$

Từ $(1); (2)\Rightarrow 6(x+6)-(6x-4)\vdots x^2+4$

$\Rightarrow 40\vdots x^2+4$

$\Rightarrow x^2+4\in\left\{4; 5; 8; 10; 20;40\right\}$ (do $x^2+4$ là số nguyên $\geq 4$)

$\Rightarrow x\in\left\{0; \pm 1; \pm 2; \pm 4; \pm 6\right\}$

Đến đây thay vào tìm $y$ thôi.

Bài 2:
 

Lấy PT(1) trừ PT (2) theo vế thu được:

$3x=5y-2$
$\Leftrightarrow x=\frac{5y-2}{3}$

Thay vào PT(1) thì:

$(2.\frac{5y-2}{3}+1)(y+2)=9$

$\Leftrightarrow 10y^2+19y-29=0$

$\Leftrightarrow (y-1)(10y+29)=0$

$\Rightarrow y=1$ hoặc $y=\frac{-29}{10}$

Với $y=1\Rightarrow x=\frac{5y-2}{3}=1$

Với $y=\frac{-29}{10}\Rightarrow x=\frac{5y-2}{3}=\frac{-11}{2}$

a: Đặt \(a=\frac{1}{x-y+1};b=\frac{1}{x+y-2}\)

Theo đề, ta có: \(\begin{cases}-3a+b=12\\ 2a-3b=-1\end{cases}\Rightarrow\begin{cases}-9a+3b=36\\ 2a-3b=-1\end{cases}\)

=>\(\begin{cases}-9a+3b+2a-3b=36-1\\ -3a+b=12\end{cases}\Rightarrow\begin{cases}-7a=35\\ b=3a+12\end{cases}\)

=>\(\begin{cases}a=-5\\ b=3\cdot\left(-5\right)+12=-15+12=-3\end{cases}\Rightarrow\begin{cases}x-y+1=-\frac15\\ x+y-2=-\frac13\end{cases}\)

=>\(\begin{cases}x-y=-\frac15-1=-\frac65\\ x+y=-\frac13+2=\frac53\end{cases}\Rightarrow\begin{cases}x=\left(-\frac65+\frac53\right):2=\left(-\frac{18}{15}+\frac{25}{15}\right):2=\frac{7}{15}:2=\frac{7}{30}\\ y=\frac53-\frac{7}{30}=\frac{50}{30}-\frac{7}{30}=\frac{43}{30}\end{cases}\)

b: \(\begin{cases}x^2+2\left(y^2+2y\right)=10\\ 3x^2-\left(y^2+2y\right)=9\end{cases}\Rightarrow\begin{cases}3x^2+6\left(y^2+2y\right)=30\\ 3x^2-\left(y^2+2y\right)=9\end{cases}\)

=>\(\begin{cases}3x^2+6\left(y^2+2y\right)-3x^2+\left(y^2+2y\right)=30-9\\ 3x^2-\left(y^2+2y\right)=9\end{cases}\)

=>\(\begin{cases}7\left(y^2+2y\right)=21\\ 3x^2=\left(y^2+2y\right)+9\end{cases}\Rightarrow\begin{cases}y^2+2y=3\\ 3x^2=3+9=12\end{cases}\Rightarrow\begin{cases}y^2+2y-3=0\\ x^2=4\end{cases}\)

=>\(\begin{cases}\left(y+3\right)\left(y-1\right)=0\\ x\in\left\lbrace2;-2\right\rbrace\end{cases}\Rightarrow\begin{cases}y\in\left\lbrace-3;1\right\rbrace\\ x\in\left\lbrace2;-2\right\rbrace\end{cases}\)

c: ĐKXĐ; x>1; y>-2

\(\begin{cases}\frac{7}{\sqrt{x-1}}-\frac{5}{\sqrt{y+2}}=\frac92\\ \frac{3}{\sqrt{x-1}}+\frac{2}{\sqrt{y+2}}=4\end{cases}\Rightarrow\begin{cases}\frac{21}{\sqrt{x-1}}-\frac{15}{\sqrt{y+2}}=\frac92\cdot3=\frac{27}{2}\\ \frac{27}{\sqrt{x-1}}+\frac{18}{\sqrt{y+2}}=36\end{cases}\)

=>\(\begin{cases}\frac{21}{\sqrt{x-1}}-\frac{15}{\sqrt{y+2}}-\frac{21}{\sqrt{x-1}}-\frac{18}{\sqrt{y+2}}=\frac{27}{2}-36\\ \frac{7}{\sqrt{x-1}}-\frac{5}{\sqrt{y+2}}=\frac92\end{cases}\Rightarrow\begin{cases}-\frac{33}{\sqrt{y+2}}=\frac{27}{2}-\frac{72}{2}=\frac{-45}{2}\\ \frac{7}{\sqrt{x-1}}=\frac{5}{\sqrt{y+2}}+\frac92\end{cases}\)

=>\(\begin{cases}\sqrt{y+2}=33\cdot\frac{2}{45}=\frac{66}{45}=\frac{22}{15}\\ \frac{7}{\sqrt{x-1}}=5:\frac{22}{15}+\frac92=5\cdot\frac{15}{22}+\frac92=\frac{75}{22}+\frac{99}{22}=\frac{174}{22}=\frac{87}{11}\end{cases}\)

=>\(\begin{cases}y+2=\frac{484}{225}\\ x-1=\frac{5929}{7569}\end{cases}\Rightarrow\begin{cases}y=\frac{484}{225}-2=\frac{34}{225}\\ x=\frac{13498}{7569}\end{cases}\) (nhận)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+y^2+y=8\\\left(x^2+x\right)\left(y^2+y\right)=12\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+x=u\\y^2+y=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=8\\uv=12\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;2\right);\left(2;6\right)\)

TH1: \(\left\{{}\begin{matrix}x^2+x=6\\y^2+y=2\end{matrix}\right.\) \(\Rightarrow...\)

TH2: ... tương tự

cảm ơn thầy ạ 3>

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

TH1: \(\left\{{}\begin{matrix}x+y=3\\x-y=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+y=-\frac{3}{2}\\x-y=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{7}{4}\\y=-\frac{13}{4}\end{matrix}\right.\)

Câu 2:

\(\left\{{}\begin{matrix}5\left(x-y\right)^2+3\left(x-y\right)-8=0\\2x+3y=12\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x-y=1\\x-y=-\frac{8}{5}\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x-y=1\\2x+3y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Th2: \(\left\{{}\begin{matrix}x-y=-\frac{8}{5}\\2x+3y=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\frac{36}{25}\\y=\frac{76}{25}\end{matrix}\right.\)

đặt thui

cứ làm đi bạn

câu 2 có lẽ dễ nhất luôn :

tách x^2+(1+y)^2=1 thành x^2+1+2y+y^2=1   (1)

tách y^2+(1+x)^2=1 thành y^2+1+2x+x^2=1    (2)

lấy(1) trừ( 2)

==>>>> x=y 

tự làm tiếp nhé